CBSE Class 12 Physics Dual Nature And Radiation Boards Questions Worksheet

Question. Cathode ray consists of
a. Photons
b. Electrons
c. Protons
d. α-particles
Answer. B

Question. In which of the following, emission of electrons does not take place?
a. Thermionic emission
b. X-rays emission
c. Photoelectric emission
d. Secondary emission
Answer. B

Question. Which of the following when falls on a metal will emit photo electrons?
a. UV radiations
b. Infrared radiation
c. Radio waves
d. Microwaves
Answer. A

Question. The slope of stopping potential vs frequency of the incident light graph is
a. e/h
b. h/e
c. h/c
d. c/h
Answer. B

Question. Photoelectric effect shows
a. wave like behaviour of light
b. particle like behaviour of light
c. both wavelike and particle like behavior
d. neither wave like nor particle like behaviour of light.
Answer. B

Question. An electron and a proton have the same de Broglie wave length. Which of them have greater velocity.
a. Electron
b. proton.
c. both a and b
d. none of the above
Answer. A

1 Marks Questions

Question. If the wavelength of an electromagnetic radiation is doubled what will happen to the energy of photons?
Answer. Energy of photon reduces to one half.

Question. Two metals A and B have work functions 4 eV and 10 eV, respectively. Which metal has higher threshold wavelength?
Answer. Wavelength is inversely proportional to work function, so metal A with lower work function has higher threshold wavelength.

Question. What is the momentum of a photon of energy 1 MeV?
Answer. Energy E = 1 MeV = 1.6 x 10 ^-13J, p = E/c= 5.33x 10^-22Kg m/s

Question. Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (i) red light, (ii) blue light?
Answer. Since electron ejection is difficult from copper than sodium, so copper has greater work function than sodium. As threshold wavelength is inversely related with work function, so sodium has higher threshold wavelength than copper.

Question. A proton, a neutron, an electron and an α particle have same energy. Then their de- Broglie wavelengths compare as?
Answer. λ α 1/√m m_α > m_p = m_n > m_e, λ_e < λ_p = λ_n > λ_α

Question. Why do we not observe the phenomenon of photoelectric effect with non-metals?
Answer. Non-metals have high work function.

2 Marks Questions

Question. An electron is accelerated through a potential difference of 100 volt. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?
Answer. 
λ = 1.227/√V nm = 1.227/√100 =1.227A⁰
X rays

Question. The graph shows variation of stopping potential Vo verses frequency of incident radiation ϑ for two photosensitive metals A and B
Answer. 
(i) From Einstein photoelectric equation,
eV₀ = h𝜈- ϕ0 or
V₀ =h𝜈/e − ϕ0/e
It is an equation of straight line as shown by line of A and B in figure
∴ slope of the line AB = ΔV₀/Δ𝜈 = h/e also threshold frequency is value from origin to point where line meets /cuts frequency axis.
Hence from the graph, the threshold frequency of Metal A is greater than the Metal B, therefore the work function of Metal A is more than Metal B
(ii) Intercept on potential axis = − ϕ₀/e from the equation Where, Work function = ϕ₀, e = charge of electron

Question. What is the effect of wavelength of incident photons on velocity of photoelectrons?
A beam of monochromatic radiation is incident on a photosensitive surface. Do the emitted photoelectrons have the same kinetic energy ? Explain
Answer. No, the different electrons belong to different energy level in conduction band. They need different energies to come out of the metal surface. For the same incident radiation, electrons knocked off from different energy levels come out with different energies.

Question. In an experiment on photoelectric emission, following observations were made 1) Wavelength of the incident light = 2 × 10^–7 m ²) Stopping potential = 3 V Find (i) kinetic energy of photoelectrons with maximum speed (ii) work function
Answer. 
Vs = 3 V and Kmax = eVo, so Kmax = 3 eV
(ii) λ = 2000 Å = 2 × 10^–7m.
Energy of incident photon = hc/ λ = 6.6 x 10^-34x 3 x 10⁸ / 2 × 10^–7 = 6.20 eV
W = E – Kmax = 3.2 eV

3 Marks Questions

Question. The work function of Caesium metal is 2.14eV. When light of frequency 6 x 10¹⁴Hz is incident on the metal surface photoemission of electrons occurs.
a. What is the maximum kinetic energy of the emitted photoelectrons
b. stopping potential
c. maximum speed of the emitted photoelectrons
Answer. Kmax = hv – ϕ₀
i) 6.63x10^-34x 6 x10¹⁴/1.6x10^-19 – 2.14eV = 0.314eV
ii) eV₀ = Kmax = 0.314eV V₀ = 0.314V
iii) 345.8 X 10³ m/s

Question. Define the terms (i) cut-off voltage and (ii) threshold frequency in relation to the Phenomenon of photoelectric effect. Using Einstein’s photoelectric equation show how the cut -off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot.
Answer. Definitions.
From Einstein’s photoelectric equation, eV₀ = hν – φ₀ for ν > ν₀.
Cut off voltage V₀ = (h/e) ν – (φ0/e )
h/e is the slope of the graph.
ν₀ = φ₀/h

Question. The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer. E = hc/λ =6.6×10⁻³⁴×3×10⁸ / 330 x10^-9 = 3.767eV ϕ_o = 4.2eV since E < W₀ no photoelectric emission

Question. X-rays of wavelength fall on a photo sensitive surface emitting electrons. Assume that the work function of the metal can be neglected, prove that the de-Broglie wave length of the emitted electron will be √hλ/2mc.
Answer. As x ray photon of wavelength λ is incident on the metal surface, so KE of electrons KEmax= hc/λ (since, work function is negligible)
De Broglie wavelength of emitted electron, λ’= h / √ (2mKE_max) But, KE_max = hc/λ
So, λ’=h/ √(2m(hc/λ))=√(hλ/2mc)

Question. An electron and a photon each have a wavelength 10^-9 m. Find (i) Their momenta (ii) The energy of the photon and (iii) The kinetic energy of electron.
Answer. p= h/λ = 6.63x10^-25 m (ii) 𝐸= hc/λ= 1243 eV (iii) 𝐸= 𝑝2/2m= 1.52 eV.

Question. Light of wavelength 2000 A0 falls on an aluminum surface. In aluminum 4.2 eV are required to remove an electron. What is the kinetic energy of (a) fastest (b) the slowest photoelectron?
Answer. Given wavelength is λ = 2000A^o = 2×10⁻⁷ mϕ_o= 4.2eV.
(a) The kinetic energy is K.E_max = 1/2mv²max = hv−½ mV²max= hc/λ−ϕo = (6.6×10⁻³⁴×3×10⁸ /2×10⁻⁷)−4.2
½ mV²_max = 2eV This is the K.E of the fastest electron is 2eV
(b) The velocity of the slowest electron would be zero, hence the kinetic energy it possesses is also zero.

Question. A beam of monochromatic radiation is incident on a photosensitive Surface. Answer the following questions giving reasons.
a. Do the emitted photoelectrons have the same kinetic energy?
b. Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation?
c. On what factors does the number of emitted photoelectrons depend?
Answer. 
a) No kinetic energy or photoelectrons depends on the energy level from which it comes out. Electrons from different energy levels beer different kinetic energies.
b) No kinetic energy depends on the energy of each photon only and not on the number of photons (i.e. intensity of light)
c) The number of photoelectrons depends on the intensity of incident light.

CASE BASED QUESTIONS:

1. According to wave theory of light, the light of any frequency can emit electrons from metallic surface provided the intensity of light be sufficient to provide necessary energy for emission of electrons, but according to experimental observations, the light of frequency less than threshold frequency cannot emit electrons; whatever be the intensity of incident light. Einstein also proposed that electromagnetic radiation is quantized.
If photoelectrons are ejected from a surface when light of wavelength λ1 = 550 nm is incident on it. The stopping potential for such electrons is Vs =0.19. If photoelectrons are ejected from a surface when light of wavelength λ1 = 550 nm is incident on it. The stopping potential for such electrons is Vs =0.19. Suppose the radiation of wavelength λ2 = 190 nm is incident on the surface.

Question. Photoelectric effect supports quantum nature of light because
a. there is a minimum frequency of light below which no photoelectrons are emitted.
b. the maximum K.E. of photoelectric depends only on the frequency of light and not on its intensity
c. even when the metal surface is faintly illuminated, the photo electrons leave the surface immediately.
d. electric charge of the photoelectrons is quantized.
Answer. A

Question. In photoelectric effect, electrons are ejected from metals, if the incident light has a certain minimum
a. Wavelength
b. Frequency
c. Amplitude
d. angle of incidence
Answer. B

Question. Calculate the stopping potential Vs2 of surface.
a. 4.47
b. 3.16
c. 2.76
d. 5.28
Answer. A

Question. Calculate the work function of the surface
a. 3.75
b. 2.07
c. 4.20
d. 3.60
Answer. B

Question. Calculate the threshold frequency for the surface
a. 500 x 10¹² Hz
b. 480 x 10¹³ Hz
c. 520 x 10¹¹ Hz
d. 460 x 10¹³ Hz
Answer. A

2. Lenard observed that when ultraviolet radiations were allowed to fall on the emitter plate of an evacuated glass tube, enclosing two electrodes (metal plates), current started flowing in the circuit connecting the plates. As soon as the ultraviolet radiations were stopped, the current flow also stopped. These observations proved that it was ultraviolet radiations, falling on the emitter plate, that ejected some charged particles from the emitter and the positive plate attracted them.

Question. Alkali metals like Li, Na, K and Cs show photo electric effect with visible light but metals like Zn, Cd and Mg respond to ultraviolet light. Why?
a. Frequency of visible light is more than that for ultraviolet light
b. Frequency of visible light is less than that for ultraviolet light
c. Frequency of visible light is same for ultraviolet light
d. Stopping potential for visible light is more than that for ultraviolet light
Answer. B

Question. Why do we not observe the phenomenon of photoelectric effect with non-metals?
a. For non-metals the work function is high
b. Work function is low
c. Work function can’t be calculated
d. For non-metals, threshold frequency is low
Answer. A

Question. What is the effect of increase in intensity on photoelectric current?
a. Photoelectric current increases
b. Decreases
c. No change
d. Varies with the square of intensity
Answer. A

Question. Name one factor on which the stopping potential depends
a. Work function
b. Frequency
c. Current
d. Energy of photon
Answer. B

Question. How does the maximum K.E of the electrons emitted vary with the work function of metal?
a. It doesn’t depend on work function
b. It decreases as the work function increases
c. It increases as the work function increases
d. It’s value is doubled with the work function
Answer. A

3. According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle or a wave is associated with moving material particle which controls the particle in every respect. The wave associated with moving material particle is called matter wave or de-Broglie wave whose wavelength called de-Broglie wavelength, is given by λ = h/mv

Question. The dual nature of light is exhibited by
a. diffraction and photo electric effect
b. photoelectric effect
c. refraction and interference
d. diffraction and reflection
Answer. A

Question. If the momentum of a particle is doubled, then its de-Broglie wavelength will
a. remain unchanged
b. become four times
c. become two times
d. become half
Answer. D

Question. If an electron and proton are propagating in the form of waves having the same λ , it implies that they have the same
a. Energy
b. Momentum
c. Velocity
d. angular momentum
Answer. B

Question. Velocity of a body of mass m, having de-Broglie wavelength λ , is given by relation
a. v = λ h/m
b. v = λm/h
c. v = λ/hm
d. v = h/ λm
Answer. D

Question. Moving with the same velocity, which of the following has the longest de Broglie wavelength?
a. ᵦ -particle
b. α -particle
c. proton
d. neutron
Answer. A

Question 701. What is Photoelectric effect ? 
Answer: The photoelectric effect is the phenomenon in which electrons are ejected from the surface of a metal when electromagnetic radiation (such as ultraviolet light or X-rays) of a sufficiently high frequency is incident upon it.
In simple words: The photoelectric effect is when light shining on a metal surface knocks electrons out of it.

Exam Tip: Mention that the incident light must be of "suitable/sufficiently high frequency" (above the threshold frequency) to get full marks.

Question 702. Define the term Work function of a photoelectric surface. 
Answer: The work function (\( W_0 \)) of a photosensitive metal surface is defined as the minimum energy required by an electron to escape from the metallic surface into vacuum: \[ W_0 = h \nu_0 = \frac{h c}{\lambda_0} \] where \( \nu_0 \) is the threshold frequency and \( \lambda_0 \) is the threshold wavelength.
In simple words: Work function is the minimum cover charge (energy) an electron must pay to break free from the metal's surface.

Exam Tip: Write down the mathematical formula \( W_0 = h \nu_0 \) to support your verbal definition.

Question 703. Define the term (i) cut off frequency & (ii) Threshold wavelength in photoelectric emission. 
Answer: The definitions are:
(i) **Cut-off Frequency (Threshold Frequency, \( \nu_0 \)):** It is the minimum frequency of incident radiation below which no photoelectric emission can occur, regardless of how intense the light is.
(ii) **Threshold Wavelength (\( \lambda_0 \)):** It is the maximum wavelength of incident radiation above which no photoelectric emission can occur.
In simple words: (i) Threshold frequency is the lowest frequency (pitch) of light that can knock electrons out. (ii) Threshold wavelength is the longest wavelength (stretch) of light that can still do the job.

Exam Tip: Note the opposite limits: threshold frequency is the *minimum* limit, whereas threshold wavelength is the *maximum* limit.

Question 704. Define the term ‘intensity of radiation’ in photon picture and write its S.I. unit. 
Answer: In the photon picture of radiation, the intensity of radiation is defined as the total number of photons striking a unit area of a surface perpendicularly per unit time.
* **SI Unit:** Watt per square meter (\( \text{W/m}^2 \)).
In simple words: Intensity is a measure of how crowded the stream of light particles (photons) is, representing the number of photons hitting a unit area every second.

Exam Tip: State both the photon definition ("number of photons per unit area per second") and the power-based SI unit to answer completely.

Question 705. Define the term “stopping potential” or “Cut-off Potential” in relation to photoelectric effect. 
Answer: The stopping potential (\( V_0 \)) is defined as the minimum negative (retarding) potential applied to the anode at which the photoelectric current drops to zero. At this potential, even the fastest emitted photoelectrons are stopped from reaching the collector plate: \[ e V_0 = K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 \]
In simple words: Stopping potential is the negative voltage needed on the collector plate to completely turn back even the fastest electrons, cutting the electrical current down to zero.

Exam Tip: State the mathematical relationship \( e V_0 = K_{\text{max}} \) to illustrate the physical significance.

Question 706. Name the phenomenon which shows the quantum nature of electromagnetic radiation.
Answer: The phenomenon that experimentally demonstrates the quantum (particle) nature of electromagnetic radiation is the **Photoelectric Effect**.
In simple words: The photoelectric effect is the classic experiment proving that light is made of discrete energy packets (particles) rather than continuous waves.

Exam Tip: Provide a single, direct answer: "Photoelectric effect" to ensure full marks.

Question 707. What is the stopping potential applied to a photocell if the maximum kinetic energy of a photoelectron is 5 eV ? 
Answer: The relationship between the stopping potential \( V_0 \) and the maximum kinetic energy \( K_{\text{max}} \) is: \[ e V_0 = K_{\text{max}} \] Given \( K_{\text{max}} = 5\text{ eV} \): \[ e V_0 = 5\text{ eV} \implies V_0 = -5\text{ V} \] Thus, the stopping (retarding) potential applied to the photocell is **\( -5\text{ V} \)** (with magnitude \( 5\text{ V} \)).
In simple words: Since stopping potential is measured in Volts and matches the maximum kinetic energy in electron-Volts, a peak energy of 5 eV requires a retarding potential of minus 5 Volts.

Exam Tip: Always specify the negative sign for the potential value as it acts as a "retarding" voltage.

Question 708. The stopping potential in an experiment is 1.5 V. What is the maximum K.E. of photoelectrons emitted ? 
Answer: The maximum kinetic energy \( K_{\text{max}} \) of the emitted photoelectrons is related to the stopping potential \( V_0 \) by: \[ K_{\text{max}} = e V_0 \] Given \( V_0 = 1.5\text{ V} \): \[ K_{\text{max}} = e (1.5\text{ V}) = 1.5\text{ eV} \]
In simple words: A stopping potential of 1.5 Volts means the fastest electrons have a kinetic energy of exactly 1.5 electron-Volts.

Exam Tip: You can express the answer in both \( \text{eV} \) and Joules (\( \text{J} \)) to show comprehensive understanding.

Question 709. Two metals A and B have work functions 4 eV and 10 eV respectively. Which metal has the highest threshold wavelength ? 
Answer: The work function \( W_0 \) is inversely proportional to the threshold wavelength \( \lambda_0 \): \[ W_0 = \frac{h c}{\lambda_0} \implies \lambda_0 \propto \frac{1}{W_0} \] Comparing the two metals:
* Work function of A: \( W_{0,A} = 4\text{ eV} \)
* Work function of B: \( W_{0,B} = 10\text{ eV} \)
Since metal A has a smaller work function, it will have a larger threshold wavelength: \[ \lambda_{0,A} > \lambda_{0,B} \] Thus, **Metal A** has the highest threshold wavelength.
In simple words: Since threshold wavelength is inversely proportional to the work function, the metal with the smaller work function (Metal A, 4 eV) will have a longer (higher) threshold wavelength.

Exam Tip: State the inverse proportionality relation \( \lambda_0 = \frac{hc}{W_0} \) to justify your conclusion clearly.

Question 710. Two metals X and Y, when illuminated with appropriate radiations emit photoelectrons. The work function of X is higher than that of Y. Which metal will have higher value of cut off frequency & why ?
Answer: **Metal X** will have the higher cut-off frequency.
**Reason:**
The work function \( W_0 \) is directly proportional to the cut-off frequency (threshold frequency) \( \nu_0 \): \[ W_0 = h \nu_0 \implies \nu_0 \propto W_0 \] Since the work function of X is higher than that of Y (\( W_{0,X} > W_{0,Y} \ )), the cut-off frequency of X must also be higher: \[ \nu_{0,X} > \nu_{0,Y} \]
In simple words: Since the work function is directly proportional to the threshold frequency, the metal with the higher work function (Metal X) requires a higher-frequency light to eject electrons.

Exam Tip: Write down the formula \( W_0 = h \nu_0 \) to support your logical reasoning.

Question 711. A photosensitive surface emits photoelectrons when red light falls on it. Will the surface emit photoelectrons when blue light is incident on it ? Give reason. 
Answer: Yes, the surface will definitely emit photoelectrons.
**Reason:**
The frequency of blue light is higher than that of red light (\( \nu_{\text{blue}} > \nu_{\text{red}} \)). Since red light has a frequency high enough to cause photoemission (which means the frequency of red light exceeds the threshold frequency, \( \nu_{\text{red}} > \nu_0 \)), the higher frequency of blue light will easily exceed the threshold frequency as well: \[ \nu_{\text{blue}} > \nu_{\text{red}} > \nu_0 \] This ensures that photoemission occurs.
In simple words: Yes. Blue light has a higher frequency (and more energy per photon) than red light. If red light already has enough energy to knock out electrons, the more energetic blue light will easily do so as well.

Exam Tip: Relate frequencies using the inequality \( \nu_{\text{blue}} > \nu_{\text{red}} > \nu_0 \) to present a mathematically sound reason.

Question 712. For a photosensitive surface, threshold wavelength is \lambda_0, Does photoemission occure, if the wavelength (\lambda) of the incident radiation is (i) more than \lambda_0 (ii) less than \lambda_0. Justify your answer. 
Answer: The condition for photoelectric emission is that the energy of the incident photon must exceed the metal's work function, which requires: \[ \lambda \le \lambda_0 \]
(i) **When \( \lambda > \lambda_0 \):** **No**, photoemission will not occur.
**Justification:** The wavelength is larger than the threshold wavelength, meaning the photon energy is less than the work function (\( E < W_0 \)).
(ii) **When \( \lambda < \lambda_0 \):** **Yes**, photoemission will occur.
**Justification:** The wavelength is shorter than the threshold, meaning the photon energy exceeds the work function (\( E > W_0 \)).
In simple words: (i) No, because longer wavelengths carry less energy than the minimum needed to free an electron. (ii) Yes, because shorter wavelengths carry more than enough energy.

Exam Tip: State the energy-wavelength inverse relationship \( E = \frac{hc}{\lambda} \) to mathematically back up both parts.

Question 713. Electrons are emitted from a photosensitive surface when it is illuminated by green light but does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (i) red light, and (ii) blue light ? 
Answer: The frequencies of the colors in the visible spectrum follow the order: \[ \nu_{\text{red}} < \nu_{\text{yellow}} < \nu_{\text{green}} < \nu_{\text{blue}} \] Since green light causes photoemission but yellow light does not, the threshold frequency \( \nu_0 \) lies between green and yellow: \[ \nu_{\text{yellow}} < \nu_0 < \nu_{\text{green}} \]
(i) **For Red Light:** **No**, because its frequency is even lower than yellow light, so \( \nu_{\text{red}} < \nu_0 \).
(ii) **For Blue Light:** **Yes**, because its frequency is higher than green light, so \( \nu_{\text{blue}} > \nu_{\text{green}} > \nu_0 \).
In simple words: (i) No, red light has even less energy than yellow light, which already failed. (ii) Yes, blue light has a higher frequency and carries more energy than green light, which succeeded.

Exam Tip: Use the visible spectrum frequency hierarchy (VIBGYOR) to support your reasoning.

Question 714. Red light however bright is it, cannot produce the emission of electrons from a clean zinc surface but even a weak Ultraviolet radiation can do so. Why ? 
Answer: Photoelectric emission depends solely on the energy of individual photons (frequency), not on the total number of photons (brightness/intensity).
The photon energy of red light is less than the work function of the zinc surface (\( E_{\text{red}} < W_0 \)), so no emission can occur even if the beam is extremely bright. In contrast, the photon energy of ultraviolet radiation is greater than the work function (\( E_{\text{UV}} > W_0 \)), making emission possible even with a very weak UV beam.
In simple words: Photoelectric emission depends on the punch (energy) of each individual photon, not how many photons hit the surface. Red photons are too weak to free an electron, while UV photons have more than enough energy to do so instantly.

Exam Tip: State that "photoelectric effect is a single-photon-single-electron collision process" to show excellent conceptual understanding.

Question 715. Work function of sodium is 2.3 eV. Does sodium show photoelectric emission for light of wavelength 6800 A0 ? 
Answer: The energy \( E \) of an incident photon of wavelength \( \lambda = 6800\text{ \AA} = 6800 \times 10^{-10}\text{ m} \) is: \[ E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6800 \times 10^{-10}} \approx 2.925 \times 10^{-19}\text{ J} \] Converting this energy into electron-Volts (\( \text{eV} \)): \[ E = \frac{2.925 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.8\text{ eV} \] Since the incident photon energy (\( E = 1.8\text{ eV} \)) is less than the work function of sodium (\( W_0 = 2.3\text{ eV} \)): \[ E < W_0 \] photoelectric emission **will not take place**.
In simple words: The energy of the incoming light is only 1.8 eV, which is less than the 2.3 eV work function of sodium. Since the light is too weak to free the electrons, no emission occurs.

Exam Tip: Show the mathematical conversion from Joules to \( \text{eV} \) step-by-step to justify your comparison.

Question 716. If the intensity of the incident radiation on a photosensitive surface is doubled, how does the kinetic energy of emitted electrons get affected ? 
Answer: There is **no change** in the kinetic energy of the emitted photoelectrons.
**Reason:** The maximum kinetic energy of emitted photoelectrons depends solely on the frequency of the incident radiation and the work function of the metal. It is completely independent of the intensity of the incident radiation (which only determines the number of electrons emitted per second).
In simple words: The speed and energy of the ejected electrons do not change when you make the light brighter. Doubling the brightness only doubles the number of electrons kicked out, not their speed.

Exam Tip: State clearly that "kinetic energy depends on frequency, not intensity."

Question 717. Ultraviolet light is incident on two photosensitive materials having work functions W1 and W2 (W1 > W2). In which case will the kinetic energy of the emitted electrons be greater ? Why ? 
Answer: The kinetic energy of the emitted photoelectrons will be greater for the metal having the **smaller work function \( W_2 \)**.
**Reason:** According to Einstein's photoelectric equation: \[ K_{\text{max}} = h \nu - W \] Since both materials are exposed to the same ultraviolet light (same frequency \( \nu \) and photon energy \( h \nu \)), the material with the smaller work function (\( W_2 \)) will subtract less energy, yielding a larger maximum kinetic energy: \[ K_{\text{max},2} > K_{\text{max},1} \quad (\text{since } W_2 < W_1) \]
In simple words: The metal with the lower work function (\( W_2 \)) requires less energy to free its electrons. This leaves more leftover energy from the incoming light to be converted into the speed (kinetic energy) of the ejected electrons.

Exam Tip: Write down Einstein's photoelectric equation \( K_{\text{max}} = h\nu - W \) to support your reasoning.

Question 718. Ultraviolet radiations of different frequencies \nu1 and \nu2 are incident on two photosensitive materials having work functions W1 and W2 (W1 > W2) respectively. The kinetic energy of the emitted electrons is same in both the cases. Which one of the two radiations will be of higher frequency and why?
Answer: The radiation with frequency **\( \nu_1 \)** will have the higher frequency.
**Reason:** According to Einstein's photoelectric equation: \[ h \nu = K_{\text{max}} + W \] Since the kinetic energy \( K_{\text{max}} \) is identical in both cases: \[ h \nu \propto W \] Because the first material has a larger work function than the second (\( W_1 > W_2 \)), the corresponding incident photon energy (and thus frequency) must be higher to compensate and produce the same kinetic energy: \[ \nu_1 > \nu_2 \]
In simple words: Since the first metal has a tighter grip (higher work function \( W_1 \)) on its electrons, we must hit it with higher-frequency (higher-energy) light to knock them out with the same exit speed as the second metal.

Exam Tip: Express the relation \( h\nu = K_{\text{max}} + W \) mathematically to prove the positive correlation between \( \nu \) and \( W \).

Question 719. The threshold frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photo-electrons is v1. When the frequency of the incident radiation is increased to 5f, the maximum velocity of photo-electrons is v2. Find the ratio v1:v2. 
Answer: According to Einstein's photoelectric equation, the maximum kinetic energy is: \[ K_{\text{max}} = \frac{1}{2} m v^2 = h \nu - W_0 = h \nu - h \nu_0 \] Given the threshold frequency \( \nu_0 = f \).
**Case 1 (Incident frequency \( \nu_1 = 2f \)):** \[ \frac{1}{2} m v_1^2 = h(2f) - h f = h f \] ----- (1)
**Case 2 (Incident frequency \( \nu_2 = 5f \)):** \[ \frac{1}{2} m v_2^2 = h(5f) - h f = 4 h f \] ----- (2)
Taking the ratio of equation (1) to equation (2): \[ \frac{v_1^2}{v_2^2} = \frac{h f}{4 h f} = \frac{1}{4} \] Taking the square root on both sides: \[ \frac{v_1}{v_2} = \frac{1}{2} \] Thus, the ratio \( v_1 : v_2 \) is **\( 1 : 2 \)**.
In simple words: Subtracting the threshold frequency from the light's frequency shows that the kinetic energy quadruples (from 1hf to 4hf) when you increase the frequency from 2f to 5f. Since velocity is proportional to the square root of energy, the speed doubles, giving a ratio of 1 to 2.

Exam Tip: Show the substitution steps clearly for both Case 1 and Case 2 to make your derivation of the 1:2 ratio easy to follow.

Question 720. The graph below shows variation of photocurrent with collector plate potential for different frequencies of incident radiation.
(i) Which physical parameter is kept constant for the three curves ? 
(ii) Which frequency ( \nu1, \nu2 or \nu3 ) is the highest ?
Answer: The configurations are:
(i) **Intensity of incident radiation** is kept constant (since all three curves saturate at the same value of saturation photocurrent).
(ii) **\( \nu_1 \)** is the highest frequency.
**Reason:** The stopping potential for curve 1 is the most negative, indicating that the photoelectrons emitted have the highest maximum kinetic energy. Since kinetic energy is directly proportional to frequency, curve 1 corresponds to the highest frequency: \[ \nu_1 > \nu_2 > \nu_3 \]

In simple words: (i) The brightness (intensity) of the light is kept constant. (ii) Frequency \( \nu_1 \) is the highest because its curve requires the strongest negative stopping potential to turn off the current.

Exam Tip: Connect "same saturation current" to "constant intensity" and "highest stopping potential" to "highest frequency."

Question 721. The given graph shows the variation of photoelectric current (I) with applied voltage (V) for two different materials and for two different intensities of the incident radiations. Identify the pair of curves that corresponds to (i) different materials but same intensity of incident radiation
(ii) different intensities but same material.
Answer: The identifications are:
(i) **Different materials but same intensity:** The pairs are **(1, 2)** and **(3, 4)**.
**Reason:** Curves with the same intensity must saturate at the same value of photoelectric current. Curves representing different materials must have different stopping potentials (since different metals have different work functions).
(ii) **Different intensities but same material:** The pairs are **(1, 3)** and **(2, 4)**.
**Reason:** Curves representing the same material must have the exact same stopping potential. Since they have different intensities, they must reach different values of saturation photocurrent.
In simple words: (i) Curves 1 and 2 represent different metals because they have different stopping points, but they share the same brightness because they level off at the same current. (ii) Curves 1 and 3 represent the same metal because they start at the same stopping point, but they have different brightnesses.

Exam Tip: Note the rules: "same material \( \rightarrow \) same stopping potential" and "same intensity \( \rightarrow \) same saturation current."

Question 722. (i) Plot a graph showing the variation of photoelectric current with intensity of light.
(ii) Show the variation of photocurrent with collector plate potential for different intensity but same frequency of incident radiation
(iii) Show the variation of photocurrent with collector plate potential for different frequency but same intensity of incident radiation 
Answer: The requested graphs are:
(i) **Current vs Intensity:** A straight line passing through the origin (directly proportional).
(ii) **Current vs Potential (Variable Intensity):** Curves have different saturation currents but merge at the same stopping potential.
(iii) **Current vs Potential (Variable Frequency):** Curves have different stopping potentials but merge at the same saturation current.

In simple words: (i) Photocurrent grows in a straight line with light intensity. (ii) Brighter light yields higher current levels but stops at the same voltage. (iii) Different frequencies require different stopping voltages but reach the same maximum current.

Exam Tip: Be careful to label your axes correctly: "photoelectric current" on the vertical axis and "collector potential / intensity" on the horizontal axis.

Question 723. Two monochromatic beams, one red and other blue, have the same intensity. In which case- 
(i) the number of photons per unit area per second is larger,
(ii) the maximum kinetic energy of the photoelectrons is more ? Justify your answer.
Answer: Let the common intensity be \( I \).
**(i) Number of photons per unit area per second (\( n \)):**
Since intensity \( I = n \cdot h \nu \), the number of photons is: \[ n = \frac{I}{h \nu} = \frac{I \lambda}{h c} \implies n \propto \lambda \] Because red light has a longer wavelength than blue light (\( \lambda_{\text{red}} > \lambda_{\text{blue}} \)), the **red beam** contains a larger number of photons per unit area per second.
**(ii) Maximum Kinetic Energy:**
According to Einstein's equation, \( K_{\text{max}} = \frac{hc}{\lambda} - W_0 \). Since blue light has a shorter wavelength (\( \lambda_{\text{blue}} < \lambda_{\text{red}} \)), its photons carry more energy, meaning the **blue beam** will eject photoelectrons with greater maximum kinetic energy.
In simple words: (i) Red light has less energy per photon, so to match the same total brightness (intensity), the red beam must contain more photons. (ii) Blue photons have more energy, so they eject electrons with much higher kinetic energy.

Exam Tip: Show the relation \( n \propto \lambda \) to mathematically justify why the red beam contains more photons.

Question 724. How does the stopping potential in photoelectric emission depends upon- 
(i) intensity of the incident radiation
(ii) frequency of incident radiation
(iii) distance between light source and cathode in a photocell ?
Answer: The dependencies are described below:
(i) **Intensity:** Stopping potential is completely **independent** of the intensity of the incident radiation.
(ii) **Frequency:** Stopping potential is **directly proportional** to the frequency of the incident radiation (above the threshold frequency); higher frequency requires a higher stopping potential.
(iii) **Distance:** Stopping potential is completely **independent** of the distance between the light source and the cathode.
In simple words: Stopping potential only depends on the frequency (color) of the light. Making the light brighter (intensity) or moving the light source closer (distance) has absolutely no effect on it.

Exam Tip: Structure your answer into clear (i), (ii), and (iii) headings to match the sub-parts.

Question 725. A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following questions giving reasons :-
(i) Do the emitted photoelectrons have the same kinetic energy ? 
(ii) Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation ?
(iii) On what factors does the number of emitted photoelectrons depend ?
Answer: The details are:
(i) **No**, the emitted photoelectrons do not all have the same kinetic energy.
* **Reason:** Electrons are bound at different depths and energy levels inside the metal. While some require only the minimum work function energy to escape, others lose additional kinetic energy through internal atomic collisions on their way out, resulting in a range of kinetic energies from zero up to a maximum value (\( K_{\text{max}} \)).
(ii) **No**, the kinetic energy does not depend on the intensity.
* **Reason:** Intensity only determines the rate of photons arriving, not the energy of individual photons.
(iii) **Factors affecting the number of emitted photoelectrons:**
* It is directly proportional to the **intensity of the incident radiation**, provided the frequency of the light exceeds the threshold frequency (\( \nu > \nu_0 \)).
In simple words: (i) No, because some electrons are buried deeper and lose energy colliding with atoms on their way out, while surface electrons escape at full speed. (ii) No, brightness doesn't change the energy of individual light packets. (iii) The number of ejected electrons is directly proportional to the brightness (intensity) of the light.

Exam Tip: Explain that kinetic energy varies because of "internal collisions within the metallic lattice."

Question 726. Write two characteristic features observed in photoelectric effect which support the photon picture of electromagnetic radiation.
Answer: Two features supporting the particle (photon) model of light are:
1. **Instantaneous Emission:** Photoelectrons are emitted immediately (within \( 10^{-9}\text{ s} \)) upon illumination, without any measurable time lag, indicating a direct, localized particle-like collision between a photon and an electron.
2. **Existence of Threshold Frequency:** Emission only occurs if the light frequency exceeds a minimum threshold (\( \nu > \nu_0 \)), which cannot be explained by wave energy accumulation over time.
In simple words: 1. The immediate release of electrons with no delay proves that individual light particles hit them like billiard balls. 2. The existence of a threshold frequency shows that light energy is delivered in discrete, indivisible packets rather than gradual waves.

Exam Tip: "No time lag" and "threshold frequency" are the two best features to list here.

Question 727. State three important properties of photon which are used to write Einstein’s photoelectric equation. 
Answer: Three key properties of photons used to formulate the equation are:
1. A photon of frequency \( \nu \) carries a discrete packet of energy: \[ E = h \nu \]
2. During a collision, a photon transfers its entire energy to a single bound electron in an all-or-nothing interaction.
3. The intensity of a light beam is directly proportional to the number of photons crossing a unit area per second.
In simple words: 1. Each photon carries a set packet of energy equal to \( h\nu \). 2. A photon transfers its entire energy to exactly one electron in a single collision. 3. Brighter light simply means more photons, not more energetic ones.

Exam Tip: List these as distinct numbered points to match the "three important properties" requirement.

Question 728. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation. 
Answer: Three features that wave theory fails to explain but Einstein's equation solves are:
1. **Instantaneous Emission:** Wave theory predicts a long delay as the wave slowly builds up enough energy, but emission is actually instantaneous.
2. **Existence of Threshold Frequency:** Wave theory predicts that any light of sufficient intensity should eventually eject electrons, but emission stops completely below \( \nu_0 \).
3. **Independence of Kinetic Energy on Intensity:** Wave theory predicts that more intense (brighter) light should transfer more energy and eject faster electrons, but kinetic energy is actually unaffected by intensity.
In simple words: Wave theory wrongly predicts that: 1. there should be a delay as energy builds up, 2. any light can eject electrons if bright enough, and 3. brighter light ejects faster electrons. Einstein's model explains why these predictions are wrong.

Exam Tip: Explicitly contrast the wave theory prediction with the actual experimental observation for each point.

Question 729. Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies \nu_A > \nu_B. 
Answer: The stopping potential \( V_0 \) increases linearly with frequency \( \nu \) according to Einstein's relation: \[ e V_0 = h(\nu - \nu_0) \implies V_0 = \left(\frac{h}{e}\right)\nu - \left(\frac{h\nu_0}{e}\right) \] The slope of both lines is the constant ratio \( \frac{h}{e} \), so the two lines are parallel. Since \( \nu_A > \nu_B \), the line for material A starts further to the right on the frequency axis.

In simple words: The graph shows two parallel straight lines. Since material A has a higher threshold frequency, its line starts further to the right along the horizontal axis.

Exam Tip: Draw both lines as strictly parallel to show that their slopes (\( h/e \)) are identical and independent of the material.

Question 730. The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which of the two has higher value of work function ? Justify your answer. 
Answer: **Metal A** has the higher work function.
**Justification:**
From the graph, the threshold frequency of metal A (\( \nu_{0,A} \)) is greater than that of metal B (\( \nu_{0,B} \)). Since the work function is directly proportional to the threshold frequency: \[ W_0 = h \nu_0 \] A higher threshold frequency directly implies a larger work function: \[ W_{0,A} > W_{0,B} \]
In simple words: Metal A has its starting point further to the right on the frequency axis, meaning it has a higher threshold frequency. Since it needs higher-frequency light to start emitting, it has a larger work function.

Exam Tip: Note that the y-intercept of the graph also represents \( -W_0/e \), confirming that the line starting further to the right intercepts the y-axis lower down, indicating a larger work function.

Question 731. State de-Broglie hypothesis. 
Answer: The de-Broglie hypothesis states that a moving material particle (such as an electron, proton, or atom) behaves like a wave under appropriate conditions, and has an associated wavelength given by: \[ \lambda = \frac{h}{p} = \frac{h}{m v} \] where \( h \) is Planck's constant, \( m \) is the mass, and \( v \) is the velocity of the particle.
In simple words: The de-Broglie hypothesis says that any moving particle has a wave associated with it. The wavelength of this "matter wave" is Planck's constant divided by the particle's momentum.

Exam Tip: Write down the core formula \( \lambda = \frac{h}{mv} \) alongside the verbal definition to secure full marks.

Question 732. What reasoning led de-Broglie to put forward the concept of matter waves ? 
Answer: de-Broglie was led to this concept by two main symmetry-based arguments:
1. **Symmetry of Nature:** Nature is highly symmetrical, and since the universe consists of two major entities - matter and radiation - they must exhibit similar dual characteristics.
2. **Dual Nature of Radiation:** If radiation (which is wave-like) behaves as a particle under certain conditions, then matter (which is particle-like) must also display wave-like properties during its propagation.
In simple words: Nature loves symmetry. If electromagnetic waves can act like particles (photons), then physical particles must also be able to act like waves.

Exam Tip: Use the key phrase "symmetry of nature" as it is the central philosophical reason de-Broglie proposed his hypothesis.

Question 733. Name the two quantities which determine the wavelength and frequency of de-Broglie wave associated with moving electron. 
Answer: The two physical quantities are:
1. **Momentum (or Velocity / Mass):** Which determines the wavelength via \( \lambda = \frac{h}{p} \).
2. **Kinetic Energy:** Which determines the frequency of the matter wave via \( E = h \nu \).
In simple words: The wavelength is determined by the electron's momentum, and its frequency is determined by its kinetic energy.

Exam Tip: Match "momentum" to "wavelength" and "energy" to "frequency" to be precise.

Question 734. Draw a schematic diagram of a localized wave describing the wave nature of moving electron. 
Answer: A localized wave packet representing a moving electron is depicted as a wave of varying amplitude confined within a small region of space:

In simple words: An electron is represented as a "wave packet" - a short, localized group of waves whose amplitude peaks in the middle and dies out at the edges.

Exam Tip: Draw a wave packet envelope (a dotted oval shape) around your sine wave to represent localization accurately.

Question 734. Why are de-Broglie waves associated with a moving football not visible ?
Answer: According to the de-Broglie relation, the wavelength is: \[ \lambda = \frac{h}{m v} \] Because a football has an extremely large mass \( m \) compared to subatomic particles, the calculated de-Broglie wavelength \( \lambda \) is incredibly small (on the order of \( 10^{-34}\text{ m} \)). Such an tiny wavelength is completely outside our experimental detection limits and does not show any noticeable wave behavior.
In simple words: A football is massive compared to an electron. Its large mass makes its associated wavelength so incredibly tiny that it is completely invisible and has no observable effect on its motion.

Exam Tip: Emphasize the relation \( \lambda \propto \frac{1}{m} \) to show that large mass leads to an imperceptibly small wavelength.

Question 735. In what manner wave velocity of matter waves is different from that of light ?
Answer: The velocity of matter waves (\( v_w \)) is given by: \[ v_w = \frac{h}{m \lambda} \] This velocity depends on the wavelength \( \lambda \) of the wave even when propagating through a vacuum.
In contrast, light waves propagate through a vacuum at a constant, fixed speed \( c \) that is entirely independent of their wavelength.
In simple words: Light travels through empty space at a single constant speed regardless of its color (wavelength). Matter waves, however, change their speed depending on their wavelength.

Exam Tip: Mention that "matter waves are dispersive in a vacuum" while "light waves are non-dispersive."

Question 736. de-Broglie waves are also called matter waves. Why ? 
Answer: de-Broglie waves are called matter waves because they are associated with any moving material particle, regardless of whether the particle carries an electric charge or is completely neutral.
In simple words: They are called matter waves because they accompany any moving bit of matter, whether it has an electric charge (like an electron) or is neutral (like a neutron).

Exam Tip: Emphasize that "charge is not a requirement" for these waves to exist, unlike electromagnetic waves.

Question 737. de-Broglie waves cannot be electromagnetic waves. Why ? 
Answer: de-Broglie waves cannot be electromagnetic waves because:
1. Electromagnetic waves are only generated by accelerating charged particles, whereas de-Broglie waves are associated with any moving particle, charged or uncharged.
2. Electromagnetic waves can travel through a vacuum at the speed of light \( c \), whereas matter waves travel at slower, variable speeds.
In simple words: Electromagnetic waves can only be made by wiggling electric charges. Matter waves don't care about charge - they are created by the simple motion of any physical particle.

Exam Tip: Note the difference: "EM waves are associated with accelerated charges, whereas matter waves are associated with any moving matter."

Question 738. In what way wave nature of electrons helps us to increase the resolving limit of electron microscope ?
Answer: The resolving limit of a microscope is inversely proportional to the wavelength of the radiation used.
An electron accelerated through a potential difference of \( 50\text{ kV} \) will have a de-Broglie wavelength of about \( 0.0055\text{ nm} \), which is nearly \( 10^5 \) times smaller than the wavelength of visible light. This extremely short wavelength allows an electron microscope to resolve incredibly fine details that are completely blurred in optical microscopes.
In simple words: The resolution of a microscope depends on the wavelength of the beam. Because fast electrons have wavelengths \( 100,000 \) times shorter than visible light, electron microscopes can zoom in and resolve much smaller details.

Exam Tip: State the relation \( \text{Resolution} \propto \frac{1}{\lambda} \) to mathematically justify the superior resolving power.

Question 739. (i) Name an experiment which shows wave nature of electrons. 
(ii) Which phenomenon was observed in this experiment using electron beam ?
(iii) Also name the important hypothesis that was confirmed by this experiment.
Answer: The details are:
(i) **Davisson-Germer Experiment.**
(ii) **Electron Diffraction** (by a nickel crystal).
(iii) **de-Broglie's hypothesis** of matter waves.
In simple words: (i) The Davisson-Germer experiment proved electrons are waves. (ii) They saw electrons diffract (bend) when hitting a crystal. (iii) This confirmed de-Broglie's matter wave hypothesis.

Exam Tip: List all three answers clearly as (i), (ii), and (iii) to match the sub-parts.

Question 740. Write briefly the underlying principle used in Davison-Germer experiment to verify wave nature of electrons experimentally. 
Answer: The underlying principle is that a beam of moving electrons behaves as a wave and undergoes **diffraction** when scattered by the regularly spaced atoms of a crystal lattice. The resulting diffraction pattern satisfies Bragg's law of constructive interference: \[ 2d \sin\theta = n \lambda \] proving the wave nature of the electrons.
In simple words: The experiment relies on the fact that waves bend and interfere constructively (diffract) when hitting a grid. Seeing electrons form a classic diffraction pattern when scattered off a nickel crystal proves they act as waves.

Exam Tip: Mention "Bragg's diffraction law" and "nickel crystal" to write an excellent answer.

Question 741. Mention the significance of Davisson and Germer experiment. 
OR
With what purpose was famous Davisson- Germer experiment with electrons performed ? 
Answer: The key significance of this experiment is that it provided the very first direct experimental proof that moving electrons possess wave-like properties. This verified de-Broglie's hypothesis of matter waves, establishing wave-particle duality as a physical reality.
In simple words: This famous experiment was performed to test if moving electrons behave as waves. Its success proved that physical matter indeed has a wave nature, confirming de-Broglie's theories.

Exam Tip: Focus on the phrase "first experimental proof of the wave nature of electrons."

Question 742. Write the expression for the de-Broglie wavelength associated with a charged particle having charge q and mass m, when it is accelerated by potential V. 
Answer: The de-Broglie wavelength \( \lambda \) associated with an accelerated charged particle is given by: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m q V}} \] where \( h \) is Planck's constant, \( m \) is the mass, \( q \) is the charge of the particle, and \( V \) is the accelerating potential difference.
In simple words: The wavelength of an accelerated charged particle is Planck's constant divided by the square root of twice the product of its mass, charge, and the accelerating voltage.

Exam Tip: Define each variable (\( h \), \( m \), \( q \), \( V \)) used in the equation to show complete understanding.

Question 743. If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with the electron changed ? 
Answer: The de-Broglie wavelength of an electron is inversely proportional to the square root of the accelerating potential \( V \): \[ \lambda \propto \frac{1}{\sqrt{V}} \] If the potential difference is doubled (\( V' = 2V \)), the new wavelength \( \lambda' \) becomes: \[ \lambda' \propto \frac{1}{\sqrt{2V}} \implies \lambda' = \frac{\lambda}{\sqrt{2}} \] Thus, the de-Broglie wavelength decreases by a factor of **\( \frac{1}{\sqrt{2}} \)** (or becomes \( 0.707 \) times the original value).
In simple words: Since wavelength is inversely proportional to the square root of voltage, doubling the voltage shrinks the wavelength to \( 1/\sqrt{2} \) (about 70.7%) of its original size.

Exam Tip: Write the final factor \( 1/\sqrt{2} \) clearly to make it straightforward to grade.

Question 744. (i) Show on a graph the variation of the de-Broglie wavelength (\lambda) associated with an electron with the square root of accelerating potential V. 
(ii) Show graphically the variation of the de-Broglie wavelength (\lambda) with the potential (V) through which an electron is accelerated from rest.
Answer: The variations are plotted below:
(i) **Wavelength vs \( \frac{1}{\sqrt{V}} \):** Since \( \lambda \propto \frac{1}{\sqrt{V}} \), the graph of \( \lambda \) versus \( \frac{1}{\sqrt{V}} \) is a straight line passing through the origin.
(ii) **Wavelength vs Potential (\( V \)):** Plotting wavelength directly against voltage gives a curved line that drops off quickly because higher voltages create shorter wavelengths.

In simple words: (i) Plotting wavelength against one over the square root of voltage gives a straight line. (ii) Plotting wavelength directly against voltage gives a curved line that drops off quickly because higher voltages create shorter wavelengths.

Exam Tip: Pay attention to whether the axis is labeled \( V \), \( \sqrt{V} \), or \( 1/\sqrt{V} \) to draw the correct shape.

Question 745. (i) Plot a graph showing variation of de-Broglie wavelength \lambda versus \frac{1}{\sqrt{V}}, where V is accelerating potential for two particles A and B carrying same charge but of masses m1, m2 (m1 > m2).
(ii) Which one of the two graphs represents a particle of smaller mass and why ?
Answer: The details are:
The slope of the \( \lambda \text{ vs } \frac{1}{\sqrt{V}} \) graph is: \[ \text{slope} = \lambda \sqrt{V} = \frac{h}{\sqrt{2mq}} \] Since both particles carry the same charge \( q \), the slope is inversely proportional to the square root of the mass: \[ \text{slope} \propto \frac{1}{\sqrt{m}} \]
(i) Therefore, the line with the steeper slope represents the particle with the smaller mass.
(ii) **Line B** represents the particle with the smaller mass (\( m_2 \)) because its graph has a steeper slope, which mathematically corresponds to a smaller mass value.

In simple words: Since slope is inversely proportional to the square root of mass, the steeper line (Line B) represents the lighter particle (\( m_2 \)).

Exam Tip: Show the relation \( \text{slope} = \frac{h}{\sqrt{2mq}} \) to justify your conclusion mathematically.

Question 746. An electron is accelerated through a potential difference of 100 Volts. What is the de-Broglie wavelength associated with it ? To which part of the electromagnetic spectrum does this value of wavelength corresponds ? 
Answer: The de-Broglie wavelength \( \lambda \) of an electron accelerated through potential \( V = 100\text{ V} \) is: \[ \lambda = \frac{12.27}{\sqrt{V}}\text{ \AA} = \frac{12.27}{\sqrt{100}}\text{ \AA} = \frac{12.27}{10} = 1.227\text{ \AA} \] This wavelength of \( 1.227\text{ \AA} \) (or \( 0.123\text{ nm} \)) lies within the **X-ray region** of the electromagnetic spectrum.
In simple words: Using the electron shortcut formula, we find the wavelength is 1.227 Angstroms. This short wavelength sits in the X-ray band of the electromagnetic spectrum.

Exam Tip: Memorize the shortcut formula \( \lambda = \frac{12.27}{\sqrt{V}}\text{ \AA} \) for electrons as it is highly valued on exams.

Question 747. What is the de-Broglie wavelength of an electron with kinetic energy (K.E.) 120 eV ? 
Answer: A kinetic energy of \( 120\text{ eV} \) corresponds to an electron accelerated through a potential difference \( V = 120\text{ V} \). Using the shortcut formula: \[ \lambda = \frac{12.27}{\sqrt{V}}\text{ \AA} = \frac{12.27}{\sqrt{120}}\text{ \AA} \approx \frac{12.27}{10.95}\text{ \AA} \approx 1.12\text{ \AA} \] Thus, the wavelength is approximately \( 1.12\text{ \AA} \) (or \( 0.112\text{ nm} \)).
In simple words: Since the kinetic energy is 120 eV, the equivalent accelerating voltage is 120 Volts. Plucked into our formula, this gives a de-Broglie wavelength of 1.12 Angstroms.

Exam Tip: Show clearly that \( \text{Energy in eV} \implies \text{Potential in Volts} \) before carrying out the division.

Question 748. An \alpha-particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of Their de-Broglie wavelengths associated with them.
Answer: The de-Broglie wavelength formula for potential \( V \) is: \[ \lambda = \frac{h}{\sqrt{2mqV}} \] Since both particles are accelerated through the same potential \( V \): \[ \lambda \propto \frac{1}{\sqrt{mq}} \] Taking the ratio of the wavelength of the proton (\( p \)) to that of the alpha particle (\( \alpha \)): \[ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} \] Using the relations \( m_\alpha = 4 m_p \) and \( q_\alpha = 2 q_p \): \[ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4 m_p \times 2 q_p}{m_p \times q_p}} = \sqrt{8} = 2\sqrt{2} \] Thus, the ratio of their wavelengths is **\( 2\sqrt{2} : 1 \)**.
In simple words: Since both the mass and charge of the alpha particle are larger than the proton's, its wavelength is much smaller, yielding a proton-to-alpha wavelength ratio of \( 2\sqrt{2} \) to 1.

Exam Tip: Use the explicit substitutions \( m_\alpha = 4m_p \) and \( q_\alpha = 2q_p \) to make your mathematical steps rigorous.

Question 749. A proton and electron have same kinetic energy. Which one has greater de-Broglie wavelength and why ? 
Answer: The de-Broglie wavelength in terms of kinetic energy \( E_k \) is: \[ \lambda = \frac{h}{\sqrt{2 m E_k}} \] Since both particles have the same kinetic energy \( E_k \): \[ \lambda \propto \frac{1}{\sqrt{m}} \] Because the mass of an electron is much smaller than that of a proton (\( m_e \ll m_p \)), the **electron** will have a much larger de-Broglie wavelength: \[ \lambda_e > \lambda_p \]
In simple words: Wavelength is inversely proportional to the square root of mass when kinetic energy is constant. Since the electron is much lighter than the proton, it will have a much larger wavelength.

Exam Tip: State the inverse mass relation \( \lambda \propto \frac{1}{\sqrt{m}} \) to support your qualitative conclusion.

Question 750. An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has the largest/ shortest de-Broglie wavelength ? 
Answer: At constant kinetic energy: \[ \lambda \propto \frac{1}{\sqrt{m}} \] The masses of the three particles follow the order: \[ m_e < m_p < m_\alpha \] Therefore:
* **Largest Wavelength:** The **electron** (since it has the smallest mass, \( m_e \)).
* **Shortest Wavelength:** The **alpha particle** (since it has the largest mass, \( m_\alpha \)).
In simple words: Since wavelength depends inversely on mass at equal kinetic energy, the lightest particle (electron) has the longest wavelength, and the heaviest particle (alpha) has the shortest.

Exam Tip: Arrange the masses in order clearly to justify both the "largest" and "shortest" answers.

Question 751. An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other ? 
Answer: The relationship between kinetic energy \( E_k \) and wavelength \( \lambda \) is: \[ E_k = \frac{h^2}{2 m \lambda^2} \] Since both particles share the same wavelength \( \lambda \): \[ E_k \propto \frac{1}{m} \] Because the mass of the electron is much smaller than that of the alpha particle (\( m_e \ll m_\alpha \)), the kinetic energy of the electron is much greater than that of the alpha particle: \[ E_{k,e} > E_{k,\alpha} \]
In simple words: Energy is inversely proportional to mass when wavelengths are equal. Since the electron is much lighter than the alpha particle, it must have much more kinetic energy to achieve the same wavelength.

Exam Tip: Write the inverse relation \( E_k \propto \frac{1}{m} \) to show complete mathematical understanding.

Question 752. Matter waves are associated with the material particles only if they are in motion. Why ? 
Answer: According to the de-Broglie relation, the wavelength is: \[ \lambda = \frac{h}{m v} \] If a particle is at rest (\( v = 0 \)), its wavelength becomes mathematically infinite: \[ \lambda = \infty \] An infinite wavelength has no physical meaning as a wave, proving that matter waves can only exist and be associated with particles when they are actively in motion (\( v \neq 0 \)).
In simple words: If a particle is completely still, its velocity is zero, which mathematically drives its wavelength to infinity. An infinite wavelength cannot exist as a wave, so waves only exist when particles are moving.

Exam Tip: Show the substitution of \( v = 0 \implies \lambda = \infty \) to justify your conclusion.

Question 753. State the laws of photoelectric emission. 
Answer: The fundamental laws governing photoelectric emission are:
1. For a given photosensitive surface, the emission rate of photoelectrons (photoelectric current) is directly proportional to the intensity of the incident light, provided the frequency exceeds the threshold.
2. The maximum kinetic energy of the emitted photoelectrons is independent of the light's intensity but is directly proportional to the frequency of the incident radiation.
3. There exists a minimum cut-off frequency, called the threshold frequency (\( \nu_0 \)), below which no photoelectric emission can occur, regardless of how intense the light is.
4. Photoelectric emission is an instantaneous process with no measurable time lag (less than \( 10^{-9}\text{ s} \)) between the incidence of light and the release of electrons.
In simple words: 1. Brighter light means more electrons are ejected. 2. Higher frequency (color) light means faster electrons are ejected, while brightness doesn't affect speed. 3. There is a minimum threshold frequency below which nothing happens. 4. Electrons fly out instantly when hit by light.

Exam Tip: Present these laws as a clear, numbered list of four points to match standard textbook layouts.

Question 754. Why photoelectric effect cannot be explained on the basis of wave nature of light ? Give reasons. 
Answer: Wave theory fails to explain the photoelectric effect for three main reasons:
1. **Kinetic Energy vs Intensity:** Wave theory predicts that more intense (brighter) light should transfer more energy, ejecting faster electrons. However, experiments show that maximum kinetic energy is completely independent of intensity.
2. **Existence of Threshold Frequency:** Wave theory predicts that any frequency of light should eventually eject electrons if the beam is made bright enough. However, experiments show that emission stops completely below the threshold frequency \( \nu_0 \).
3. **Instantaneous Emission:** Wave theory predicts a notable time delay as the wave slowly transfers energy to the electrons. However, experiments show that emission is completely instantaneous with no delay.
In simple words: Wave theory wrongly predicts that: 1. brighter light should make electrons faster, 2. any light can eject electrons if bright enough, and 3. there should be a delay as energy builds up. These incorrect predictions show wave theory cannot explain the effect.

Exam Tip: Contrast wave theory expectations directly with actual experimental results for each of the three points.

Question 755. (i) Using photon picture of light, show how Einstein’s photoelectric equation can be established.
(ii) Write three salient features observed in photoelectric effect which can be explained using this equation. 
Answer: The derivations and features are:
**(i) Derivation of Einstein's Equation:**
In the photon model, light consists of discrete packets of energy called photons. Each photon carries an energy \( E = h \nu \). When a photon strikes a metallic surface, its energy is completely absorbed by a single bound electron. This energy is split into two parts:
1. A portion is used as the work function (\( W_0 = h \nu_0 \)) to free the electron from the surface.
2. The remaining energy is carried away as the maximum kinetic energy \( K_{\text{max}} \) of the escaped electron.
By conservation of energy: \[ h \nu = W_0 + K_{\text{max}} \] \[ K_{\text{max}} = h \nu - W_0 = h(\nu - \nu_0) \] This is Einstein's photoelectric equation.

**(ii) Three Salient Features Explained:**
1. **Threshold Frequency:** If \( \nu < \nu_0 \), \( K_{\text{max}} \) becomes negative, which is physically impossible. This proves that emission can only occur when \( \nu \ge \nu_0 \).
2. **Intensity Independence of \( K_{\text{max}} \):** The equation shows \( K_{\text{max}} \) depends only on \( \nu \), explaining why making the light brighter (intensity) does not increase electron speed.
3. **Linear Relation with Frequency:** The equation shows that \( K_{\text{max}} \) increases linearly as frequency increases.
In simple words: (i) A photon hits an electron, transferring all its energy (\( h\nu \)). The electron uses a fixed amount to break free (work function \( W_0 \)) and takes the rest as kinetic energy, giving \( K_{\text{max}} = h(\nu - \nu_0) \). (ii) This explains why: 1. there is a minimum frequency limit, 2. brightness doesn't affect speed, and 3. higher frequency means faster electrons.

Exam Tip: Write down the final equation \( K_{\text{max}} = h(\nu - \nu_0) \) in a prominent box to finalize your proof.

Question 756. (i) Plot a graph showing the variation of photocurrent versus collector potential for three different intensities I1 > I2 > I3, two of which ( I1 and I2) have the same frequency \nu and the third has frequency \nu1 > \nu.
(ii) Explain the nature of curves on the basis of Einstein’s equation. 
Answer: The configurations are:
**(i) Graphical Plot:**
Curves 1 and 2 (same frequency \( \nu \)) start at the same stopping potential \( -V_{01} \), but reach different saturation currents since \( I_1 > I_2 \). Curve 3 (higher frequency \( \nu_1 \)) starts further to the left at a more negative stopping potential \( -V_{02} \).

**(ii) Explanation:**
According to Einstein's equation, the stopping potential \( V_0 \) is: \[ e V_0 = h \nu - W_0 \] Since \( I_1 \) and \( I_2 \) share the same frequency \( \nu \), they must share the same stopping potential \( V_{01} \). Since \( \nu_1 > \nu \), curve 3 requires a larger retarding stopping potential \( V_{02} \), placing its intercept further to the left.
In simple words: (i) Curves 1 and 2 start at the same stopping point because they share the same color (frequency), but curve 1 climbs higher because it is brighter. Curve 3 starts further left because it has a higher frequency. (ii) This perfectly matches Einstein's equation, which says stopping voltage depends solely on frequency.

Exam Tip: Clearly label both stopping potentials on the negative potential axis to show how they relate to the two frequencies.

Question 757. The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which of the two has higher value of threshold frequency ? Justify your answer. 
Answer: **Metal A** has the higher threshold frequency.
**Justification:**
The intercept of the line on the frequency axis represents the threshold frequency \( \nu_0 \) of the metal. From the graph, the line for metal A starts further to the right, meaning its threshold frequency is higher: \[ \nu_{0,A} > \nu_{0,B} \]
In simple words: Metal A has its starting point further to the right on the frequency axis, meaning it has a higher threshold frequency. Since it needs higher-frequency light to start emitting, it has a larger work function.

Exam Tip: Draw a quick reference arrow showing that threshold frequency increases towards the right on the horizontal axis.

Question 758. In a photoelectric effect experiment, the graph between the stopping potential (V0) and frequency ( \nu ) of the incident radiation on two different metal plates P & Q are shown in figure. Explain.
(i) Which of the metal plates P & Q has greater value of work function ?
(ii) What does the slope of lines depict ?
Answer: The details are:
**(i) Greater Work Function:** **Metal Q** has the greater work function.
* **Reason:** The threshold frequency of metal Q (\( \nu_{0,Q} \)) is larger than that of P (\( \nu_{0,P} \)) because its line starts further to the right on the frequency axis. Since \( W_0 \propto \nu_0 \), metal Q has a higher work function.
**(ii) Significance of the Slope:** The slope of both lines represents the ratio of Planck's constant to the elementary charge: \[ \text{slope} = \frac{h}{e} \] Since \( h \) and \( e \) are universal constants, the slope is identical for both lines, making them parallel.
In simple words: (i) Metal Q has its threshold point further to the right, meaning it requires higher-energy light to free electrons, so its work function is larger. (ii) The slope of both lines is the ratio \( h/e \), which is a universal constant, explaining why the lines are parallel.

Exam Tip: State clearly that the slope \( h/e \) is independent of the nature of the metal plate.

Question 759. The following graph shows the variation of stopping potential (V0) with frequency ( \nu ) of the incident radiation for two photosensitive surfaces X and Y. 
(i) Which of the metals has larger threshold wavelength ? Give reason. X Y
(ii) Explain giving reason, which metal gives out electrons having larger kinetic energy, for the same wavelength of incident radiation ?
(iii) If the distance between the light source and metal X is halved, how will the kinetic energy of emitted from it change ? Give reason.
Answer: The detailed analysis is:
**(i) Larger Threshold Wavelength:** **Metal X** has the larger threshold wavelength.
* **Reason:** From the graph, metal X has a lower threshold frequency than metal Y (\( \nu_{0,X} < \nu_{0,Y} \)). Since wavelength is inversely proportional to frequency (\( \lambda_0 = \frac{c}{\nu_0} \)), the lower frequency of X corresponds to a larger threshold wavelength: \[ \lambda_{0,X} > \lambda_{0,Y} \]
**(ii) Larger Kinetic Energy:** **Metal X** will emit photoelectrons with larger kinetic energy.
* **Reason:** Since both metals are exposed to the same wavelength (same photon energy \( E \)), and metal X has a smaller work function than Y (\( W_{0,X} < W_{0,Y} \)), less energy is spent freeing the electrons from X, leaving more leftover energy as kinetic energy: \[ K_{\text{max},X} > K_{\text{max},Y} \]
**(iii) Effect of halving the distance:** The kinetic energy of the emitted electrons will **remain completely unchanged**.
* **Reason:** Halving the distance doubles the light intensity, but kinetic energy depends solely on the frequency of the light and is independent of intensity or distance.
In simple words: (i) Metal X has a lower threshold frequency, meaning its threshold wavelength is longer (larger). (ii) Metal X has a lower work function, so it wastes less of the photon's energy, releasing faster electrons. (iii) Moving the light closer makes the light brighter, but has no effect on the energy of individual photons, so electron speed remains the same.

Exam Tip: For part (iii), distinguish clearly between "kinetic energy" (independent of distance) and "photoelectric current" (which increases as distance decreases).

Question 760. An electron is accelerated from rest through a potential V. Obtain the expression for the de-Broglie wavelength. 
Answer: When an electron of mass \( m \) and charge \( e \) is accelerated through a potential difference \( V \), its kinetic energy \( E_k \) is: \[ E_k = e V \] The momentum \( p \) is related to kinetic energy by: \[ p = \sqrt{2 m E_k} = \sqrt{2 m e V} \] According to de-Broglie's relation, the associated wavelength \( \lambda \) is: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}} \] Substituting the constant values (\( h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s} \), \( m = 9.1 \times 10^{-31}\text{ kg} \), and \( e = 1.6 \times 10^{-19}\text{ C} \)): \[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times V}} \] \[ \lambda = \frac{1.227 \times 10^{-9}}{\sqrt{V}}\text{ m} = \frac{12.27}{\sqrt{V}}\text{ \AA} \]
In simple words: Accelerating an electron through a voltage V gives it kinetic energy eV. Linking this to its momentum and substituting the physical constants gives a simple final wavelength formula of \( \frac{12.27}{\sqrt{V}} \) Angstroms.

Exam Tip: Show the transition from the general formula \( \lambda = \frac{h}{\sqrt{2meV}} \) to the simplified numeric value of \( \frac{12.27}{\sqrt{V}}\text{ \AA} \) to secure full marks.

Question 761. Describe briefly how Davisson-Germer experiment demonstrated the wave nature of electrons.
Answer: **Principle:**
The experiment is based on the principle that a beam of moving electrons behaves as a wave and undergoes diffraction when scattered by a nickel crystal, satisfying Bragg's law: \[ 2d \sin\theta = n \lambda \]
**Working:**
An electron gun shoots electrons at a nickel crystal. The scattered electrons are collected at various angles, and their intensity is measured. A prominent peak in intensity is observed at an accelerating voltage of \( 54\text{ V} \) and a scattering angle of \( \phi = 50^\circ \). This peak is due to constructive wave interference.
The angle of inclination \( \theta \) is: \[ \theta = 180^\circ - \phi = \frac{180^\circ - 50^\circ}{2} = 65^\circ \] Using Bragg's equation with \( d = 0.91\text{ \AA} \) and \( n=1 \): \[ \lambda = 2d \sin\theta = 2 \times 0.91 \times \sin(65^\circ) \approx 1.65\text{ \AA} \]
Using de-Broglie's theoretical formula for \( V = 54\text{ V} \): \[ \lambda = \frac{12.27}{\sqrt{54}} \approx 1.67\text{ \AA} \] The experimental and theoretical values are in excellent agreement, confirming the wave nature of moving electrons.
In simple words: The experiment measured how electrons scattered off a nickel crystal. They saw a strong peak in intensity at 54 Volts and 50 degrees, which is a classic diffraction pattern. Calculating the wavelength from this pattern matched de-Broglie's theoretical formula perfectly, proving electrons act as waves.

Exam Tip: Highlight the comparison of the two calculated wavelengths (\( 1.65\text{ \AA} \) and \( 1.67\text{ \AA} \)) to show how the experiment verified de-Broglie's hypothesis.

Question 762. The wavelength \lambda of a photon and the de-Broglie wavelength of an electron have the same value. Show that the energy of a photon is \frac{2\lambda mc}{h} times the kinetic energy of electron. Where m, c and h have their usual meaning.
Answer: Let the common wavelength be \( \lambda \).
1. **Energy of a Photon (\( E_p \)):** \[ E_p = \frac{h c}{\lambda} \] ----- (1)
2. **Kinetic Energy of an Electron (\( E_k \)):**
The de-Broglie wavelength of the electron is \( \lambda = \frac{h}{p} \implies p = \frac{h}{\lambda} \). The kinetic energy \( E_k \) is: \[ E_k = \frac{p^2}{2m} = \frac{h^2}{2 m \lambda^2} \] ----- (2)
3. **Finding the Ratio:**
Divide equation (1) by equation (2): \[ \frac{E_p}{E_k} = \frac{\left(\frac{h c}{\lambda}\right)}{\left(\frac{h^2}{2 m \lambda^2}\right)} = \frac{h c}{\lambda} \times \frac{2 m \lambda^2}{h^2} = \frac{2 \lambda m c}{h} \] Rearranging this gives: \[ E_p = \left( \frac{2 \lambda m c}{h} \right) E_k \] This mathematically proves the required relation.
In simple words: By writing down the photon energy formula and the electron kinetic energy formula using the same wavelength, and then dividing the photon energy by the electron energy, the terms simplify to exactly \( \frac{2\lambda mc}{h} \).

Exam Tip: Express the electron's momentum as \( p = h/\lambda \) first to show how you derived its kinetic energy.

Question 763. X-rays of wavelength ‘ \lambda ’ fall on a photo sensitive surface, emitting electrons. Assuming that the work function of surface can be neglected, prove that the de-Broglie wavelength of electrons emitted will be \sqrt{\frac{h\lambda}{2mc}} 
OR
An electromagnetic wave of wavelength \lambda is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength \lambda_1, Prove that, \lambda_1 = \sqrt{\frac{h\lambda}{2mc}}
Answer: Since the work function is negligible (\( W_0 \approx 0 \)), the maximum kinetic energy \( E_k \) of the emitted electrons equals the full energy of the incident X-ray photons: \[ E_k = E_{\text{photon}} = \frac{h c}{\lambda} \] The de-Broglie wavelength \( \lambda_1 \) of the emitted electrons is: \[ \lambda_1 = \frac{h}{\sqrt{2 m E_k}} \] Substituting the value of \( E_k = \frac{hc}{\lambda} \): \[ \lambda_1 = \frac{h}{\sqrt{2 m \left(\frac{h c}{\lambda}\right)}} = \frac{h}{\sqrt{\frac{2 m h c}{\lambda}}} = \sqrt{\frac{h^2 \lambda}{2 m h c}} = \sqrt{\frac{h \lambda}{2 m c}} \] This mathematically proves the expression.
In simple words: Since the work function is neglected, all of the photon's energy (\( \frac{hc}{\lambda} \)) turns into the electron's kinetic energy. Plugging this energy value into the electron's de-Broglie formula simplifies directly to \( \sqrt{\frac{h\lambda}{2mc}} \).

Exam Tip: Show the algebraic simplification where \( h \) inside the square root is squared to merge with the fraction.

Question 764. A proton and an \alpha - particle are accelerated through the same potential difference. Which on of the two has
(i) greater de-Broglie wavelength, and 
(ii) less kinetic energy ? Justify your answer.
Answer: The variables are related as:
**(i) Greater Wavelength:** The **proton** has the greater wavelength.
* **Reason:** For a constant potential difference, \( \lambda \propto \frac{1}{\sqrt{mq}} \). Comparing the proton (\( p \)) and the alpha particle (\( \alpha \)): \[ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{4 m_p \times 2 q_p}{m_p q_p}} = \sqrt{8} = 2\sqrt{2} > 1 \] Since \( \lambda_p = 2\sqrt{2} \lambda_\alpha \), the proton has the larger wavelength.
**(ii) Less Kinetic Energy:** The **proton** has less kinetic energy.
* **Reason:** The kinetic energy gained is \( E_k = q V \). Since both are accelerated through the same voltage \( V \), the kinetic energy is directly proportional to the charge (\( E_k \propto q \)). Because the proton has less charge than the alpha particle (\( q_p = \frac{1}{2} q_\alpha \)), it gains less kinetic energy: \[ E_{k,p} < E_{k,\alpha} \]
In simple words: (i) Because the proton is much lighter and has less charge, its wavelength is much larger (by \( 2\sqrt{2} \) times) than the alpha particle's. (ii) Since kinetic energy depends only on charge when voltage is equal, the proton (charge 1) gains half as much kinetic energy as the alpha particle (charge 2).

Exam Tip: Address both parts (i and ii) with their respective proportionalities to show clear, structured reasoning.

Question 765. A deuteron and an \alpha - particle are accelerated with the same accelerating potential. Which one of the two has -
(i) greater value of de-Broglie wavelength associated with it, it, and
(ii) less kinetic energy ? Explain.
Answer: The details are:
**(i) Greater Wavelength:** The **deuteron** has the greater wavelength.
* **Reason:** At constant potential \( V \), \( \lambda \propto \frac{1}{\sqrt{mq}} \). Comparing the deuteron (\( d \), mass \( m_d = 2m_p \), charge \( q_d = q_p \)) and the alpha particle (\( \alpha \), mass \( m_\alpha = 4m_p \), charge \( q_\alpha = 2q_p \)): \[ \frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_d q_d}} = \sqrt{\frac{4 m_p \times 2 q_p}{2 m_p \times q_p}} = \sqrt{4} = 2 \] Since \( \lambda_d = 2 \lambda_\alpha \), the deuteron has a larger wavelength.
**(ii) Less Kinetic Energy:** The **deuteron** has less kinetic energy.
* **Reason:** Gained kinetic energy is \( E_k = q V \). Since both share the same potential \( V \), the kinetic energy is proportional to the charge. Because the deuteron has less charge than the alpha particle (\( q_d = 1e < q_\alpha = 2e \)), it gains less kinetic energy: \[ E_{k,d} = \frac{1}{2} E_{k,\alpha} \]
In simple words: (i) The deuteron is lighter and has less charge, so its wavelength is exactly twice as large as the alpha particle's. (ii) Since kinetic energy depends on charge, the deuteron (charge 1) ends up with half the kinetic energy of the alpha particle (charge 2).

Exam Tip: Note the properties: \( m_d = 2m_p, q_d = q_p \) for deuteron, and \( m_\alpha = 4m_p, q_\alpha = 2q_p \) for alpha particle.

Question 766. A proton and an \alpha - particle have the same de-Broglie wavelength. Determine the ratio of-
(i) their accelerating potentials, and (ii) their speeds. 
Answer: Let the common wavelength be \( \lambda \).
**(i) Ratio of accelerating potentials (\( V_p : V_\alpha \)):**
Using the relation \( \lambda = \frac{h}{\sqrt{2mqV}} \), we get: \[ V = \frac{h^2}{2 m q \lambda^2} \implies V \propto \frac{1}{m q} \] Taking the ratio of the potential of the proton to that of the alpha particle: \[ \frac{V_p}{V_\alpha} = \frac{m_\alpha q_\alpha}{m_p q_p} = \frac{4 m_p \times 2 q_p}{m_p q_p} = \frac{8}{1} \] The ratio of their accelerating potentials is **\( 8 : 1 \)**.
**(ii) Ratio of their speeds (\( v_p : v_\alpha \)):**
Using the relation \( \lambda = \frac{h}{m v} \implies v = \frac{h}{m \lambda} \), we get: \[ v \propto \frac{1}{m} \] Taking the ratio of their speeds: \[ \frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} = \frac{4 m_p}{m_p} = \frac{4}{1} \] The ratio of their speeds is **\( 4 : 1 \)**.
In simple words: (i) To have the same wavelength, the proton needs a much higher voltage because it is lighter and has less charge, giving a voltage ratio of 8 to 1. (ii) Since speed is inversely proportional to mass at equal wavelengths, the lighter proton travels 4 times faster than the alpha particle, giving a speed ratio of 4 to 1.

Exam Tip: Write down both ratio results clearly as "8 : 1" and "4 : 1" to ensure they are easy to grade.

Question 767. A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has –
(i) greater value of de-Broglie wavelength associated with it, it, and 
(ii) less momentum ? Give reasons to justify your answer.
Answer: The details are:
**(i) Greater Wavelength:** The **proton** has the greater wavelength.
* **Reason:** For identical potential \( V \), \( \lambda \propto \frac{1}{\sqrt{mq}} \). Both have the same charge (\( q_p = q_d = 1e \)), so the wavelength depends only on mass: \[ \lambda \propto \frac{1}{\sqrt{m}} \] Since the proton is lighter than the deuteron (\( m_p < m_d \)), its wavelength is larger: \[ \lambda_p > \lambda_d \]
**(ii) Less Momentum:** The **proton** has less momentum.
* **Reason:** The momentum is \( p = \sqrt{2 m q V} \). Since they share the same charge and potential, the momentum is directly proportional to the square root of the mass (\( p \propto \sqrt{m} \)). Because the proton is lighter: \[ p_p < p_d \]
In simple words: (i) Since they have the same charge and voltage, the lighter proton ends up with a longer wavelength. (ii) For the same reason, the lighter proton is easier to push but carries less momentum than the heavier deuteron.

Exam Tip: Write down both proportionalities (\( \lambda \propto 1/\sqrt{m} \) and \( p \propto \sqrt{m} \)) to back up both parts.

Question 768. Two metals X and Y have work functions 2 eV & 5 eV respectively. Which metal will emit electrons, when it is radiated with light of wavelength 400 nm & why ? 
Answer: The energy \( E \) of an incident photon of wavelength \( \lambda = 400\text{ nm} = 400 \times 10^{-9}\text{ m} \) is: \[ E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}}\text{ eV} \approx 3.1\text{ eV} \] Comparing with the work functions of the metals:
* For Metal X: \( W_{0,X} = 2\text{ eV} \). Since \( E > W_{0,X} \) (\( 3.1\text{ eV} > 2\text{ eV} \)), **Metal X will emit electrons**.
* For Metal Y: \( W_{0,Y} = 5\text{ eV} \). Since \( E < W_{0,Y} \) (\( 3.1\text{ eV} < 5\text{ eV} \)), Metal Y will not emit electrons.
In simple words: The incoming light carries an energy of 3.1 eV. This is enough to free electrons from Metal X (which only needs 2 eV), but not from Metal Y (which needs 5 eV). So, only Metal X will show emission.

Exam Tip: Show the photon energy calculation step-by-step to prove why the value is approximately 3.1 eV.

Question 769. Monochromatic light of frequency 6.0 X 10^{14} Hz is produced by a laser. The power emitted is 2.0 X 10^{-3} W.
(a) What is the energy of a photon in the light beam ? 
(b) Estimate the number of photons emitted per second on an average by the source.
Answer: Given:
Frequency, \( \nu = 6.0 \times 10^{14}\text{ Hz} \)
Power, \( P = 2.0 \times 10^{-3}\text{ W} = 2.0 \times 10^{-3}\text{ J/s} \)

**(a) Energy of a Photon (\( E \)):** \[ E = h \nu = (6.63 \times 10^{-34}\text{ J}\cdot\text{s}) \times (6.0 \times 10^{14}\text{ s}^{-1}) = 3.98 \times 10^{-19}\text{ J} \]
**(b) Number of photons emitted per second (\( n \)):**
The total power \( P \) is the energy of one photon multiplied by the number of photons emitted per second \( n \): \[ P = n \cdot E \implies n = \frac{P}{E} \] \[ n = \frac{2.0 \times 10^{-3}}{3.98 \times 10^{-19}} \approx 5.0 \times 10^{15}\text{ photons/second} \]
In simple words: (a) Each photon carries \( 3.98 \times 10^{-19} \) Joules of energy. (b) Dividing the total power by the energy of a single photon shows the laser shoots out \( 5 \times 10^{15} \) photons every second.

Exam Tip: Write down both results with their correct physical units (Joules for energy, and photons/second for emission rate).

Question 770. The work function for the following metals is given : 
Na : 2.75 eV and Mo : 4.175 eV
(i) Which of these will not give photoelectron emission from a radiation of wavelength 3300 A0 from a laser beam ?
(ii) What happens if the source of laser beam is brought closer ?
Answer: Given wavelength \( \lambda = 3300\text{ \AA} = 3300 \times 10^{-10}\text{ m} \).
The energy \( E \) of each incident photon is: \[ E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3300 \times 10^{-10} \times 1.6 \times 10^{-19}}\text{ eV} \approx 3.75\text{ eV} \]
**(i) Which metal will not emit electrons:**
Comparing the photon energy (\( E = 3.75\text{ eV} \)) with the work functions: