CBSE Class 12 Physics Dual Nature Of Matter And Radiation Worksheet Set 02

CBSE Class 12 Physics Dual nature of radiation Atoms and Nuclei.Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects. 

1 What is the stopping potential applied to a a photocell if the maximum kinetic energy of a photoelectron is 5eV ?

2 Work functions of two metals A and B are 4eV and 10 eV respectively . Which metal has the higher threshold wavelength ?

3 Two beams ,one of red light and the other of blue light , of same intrensity incident on a metallic surface to emit Photoelectrons. Which one of them emits electrons of greater kinetic energy?

4 How does the stopping potential of a Photo cell change ,when i) the intensity of the incident radiation is halved? Ii) frequency of incident radiation increases ?

5 If the potential difference used to accelerate electrons is tripled , by what factor the de Broglie wavelength of electron beam change?

6 An electron and proton have the same kinetic energy .Which one of them has the larger de Broglie wavelength.

7 An alpha particle and a proton are accelerated from rest by the same potential. Find the ratio of their de Broglie wavelengths.

8 Show graphically the variation of de Broglie wavelength λ of an electron with i)√V ii) V where V is the potential through which an electron is accelerated from rest.

9 Name the experiment which verified the wave nature of electrons.

Which phenomenon was observed in this experiment using an electron beam?

10 Why Caesium oxide is coated on the cathode of Photo electric cell?

Important Questions for NCERT Class 12 Physics Dual Nature Of Matter And Radiation 

Question. The stopping potential doubles when the frequency of the incident light changes from n to 3v/2. Then the work function of the metal must be    
(a) hν/2
(b) hν
(c) 2hν
(d) none of the above

Answer : A

Question. The force on a hemisphere of radius 1 cm if a parallel beam of monochromatic light of wavelength 500 nm. falls on it with an intensity of 0.5 W/cm2, striking the curved surface in a direction which is perpendicular to the flat face of the hemisphere is (assume the collisions to be perfectly inelastic)    
(a) 5.2 × 10^–13 N
(b) 5.2 × 10^–12 N
(c) 5.22 × 10^–9 N
(d) zero

Answer : C

Question. A 15.0 eV photon collides with and ionizes a hydrogen atom. If the atom was originally in the ground state (ionization potential =13.6 eV), what is the kinetic energy of the ejected electron?    
(a) 1.4 eV
(b) 13.6 eV
(c) 15.0 eV
(d) 28.6 eV

Answer : A

Question. A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by   
(a)  B²/2VE²
(b) 2VB²/E²
(c) 2VE²/B²
(d) E²/2VB²
(Where V is the potential difference between cathode and anode)

Answer : D

Question. In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of
(a) collisions between the charged particles emitted from the cathode and the atoms of the gas
(b) collision between different electrons of the atoms of the gas
(c) excitation of electrons in the atoms
(d) collision between the atoms of the gas. 

Answer : A

Question. In a discharge tube ionization of enclosed gas is produced due to collisions between
(a) neutral gas atoms/molecules
(b) positive ions and neutral atoms/molecules
(c) negative electrons and neutral atoms/molecules
(d) photons and neutral atoms/molecules. 

Answer : C

Question. J.J. Thomson’s cathode-ray tube experiment demonstrated that
(a) cathode rays are streams of negatively charged ions
(b) all the mass of an atom is essentially in the nucleus
(c) the e/m of electrons is much greater than the e/m of protons
(d) the e/m ratio of the cathode-ray particles changes when a different gas is placed in the discharge tube 

Answer : C

Question. Which of the following is not the property of cathode rays ?
(a) It produces heating effect.
(b) It does not deflect in electric field.
(c) It casts shadow.
(d) It produces fluorescence.

Answer : B

Question. Who evaluated the mass of electron indirectly with help of charge?
(a) Thomson
(b) Millikan
(c) Rutherford
(d) Newton 

Answer : A

Question. In a discharge tube at 0.02 mm, there is formation of
(a) Crooke’s dark space
(b) Faraday’s dark space
(c) both space partly
(d) none of these.

Answer : A

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. Thermions are
(a) protons
(b) electrons
(c) photons
(d) positrons

Answer : B

Question. A source of light is placed at a distance of 50 cm from a photo cell and the stopping potential is found to be V0. If the distance between the light source and photo cell is made 25 cm, the new stopping potential will be :
(a) V0/2
(b) V0
(c) 4V0
(d) 2V0

Answer : B

Question. Photoelectric emission occurs only when the incident light has more than a certain minimum
(a) power
(b) wavelength
(c) intensity
(d) frequency 

Answer :  D

Question. A 200 W sodium street lamp emits yellow light of wavelength 0.6 μm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
(a) 1.5 × 10²⁰
(b) 6 × 10¹⁸
(c) 62 × 10²⁰
(d) 3 × 10¹⁹

Answer: A

Question. A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ/2 . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is :
(h = Planck's constant, c = speed of light)
(a) hc/λ
(b) 2hc/λ
(c) hc/3λ
(d) hc/2λ

Answer: D

Question. An electron of mass m and a photon have same energy E.
The ratio of de-Broglie wavelengths associated with them is :
(a) 1/c(E/2m)^1/2
(b) (E/2m)^1/2
(c) c(2mE)^1/2
(d) 1/xc(2m/E)^1/2

Answer: A

Question. The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is :-

Answer: A

Question. Photoelectric work function of a metal is 1eV. Light of wavelength λ = 3000 Å falls on it. The photo electrons come out with velocity
(a) 10 metres/sec
(b) 10² metres/sec
(c) 10⁴ metres/sec
(d) 10⁶ metres/sec

Answer: D

DIRECTIONS : Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer from the following-
(a) Statement-1 is false, Statement-2 is true
(b) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
(d) Statement-1 is true, Statement-2 is false

Question. Statement-1 : Photoelectric saturation current increases with the increase in frequency of incident light.
Statement-2 : Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.

Answer: D

Question. Statement-1 : Though light of a single frequency (monochromatic) is incident on a metal, the energies of emitted photoelectrons are different.
Statement-2 : The energy of electrons emitted from inside the metal surface, is lost in collision with the other atoms in the metal.

Answer: A

Question. Statement-1 : Photosensitivity of a metal is high if its work function is small.
Statement-2 : Work function = hf₀ where f₀ is the threshold frequency.

Answer: B

Question. Statement-1 : In process of photoelectric emission, all emitted electrons do not have same kinetic energy.
Statement-2 : If radiation falling on photosensitive surface of a metal consists of different wavelength then energy acquired by electrons absorbing photons of different
wavelengths shall be different.

Answer: B

Question. Statement-1 : The de-Broglie wavelength of a molecule (in a sample of ideal gas) varies inversely as the square root of absolute temperature.
Statement-2 : The rms velocity of a molecule (in a sample of ideal gas) depends on temperature.

Answer: A
 

Short Answer Type Questions
 

Question. How does the energy of a photon change if its wavelength is doubled?
Answer: The energy associated with a photon is given by \( E = h\nu = \frac{hc}{\lambda} \). This relationship indicates that the photon's energy is inversely proportional to its wavelength, which can be written as \( E \propto \frac{1}{\lambda} \). Consequently, if the wavelength is doubled, the energy of the photon decreases to exactly half of its initial value.
In simple words: As the wavelength of light gets longer, its energy becomes smaller. Doubling the wavelength cuts the photon's energy in half.

Exam Tip: Always state the formula \( E = \frac{hc}{\lambda} \) clearly before explaining the inverse relationship to secure full marks.

Question. Why are alkali metals highly suitable for photoelectric emission?
Answer: Alkali metals possess extremely low work functions. Because of this physical property, even low-energy visible light carries enough energy to successfully trigger the photoelectric emission of electrons from their surfaces.
In simple words: Alkali metals hold onto their electrons very loosely. This means even ordinary visible light can easily knock electrons out of them.

Exam Tip: When explaining photoelectric sensitivity, always highlight the term "low work function" as it is the key evaluation keyword.

Question. Why is ultraviolet radiation more effective than visible light for causing photoelectric emission?
Answer: Ultraviolet rays are highly effective for photoelectric emission because they have a higher frequency than visible light. According to Planck's equation, this higher frequency means ultraviolet photons carry significantly more energy.
In simple words: Ultraviolet light has a very high frequency, which makes its photons much more energetic and better at releasing electrons.

Exam Tip: Relate frequency to photon energy using \( E = h\nu \) to explain why higher frequency radiations are more effective.

Question. Can X-rays induce the photoelectric effect in metals like sodium, zinc, and copper?
Answer: Yes, X-rays can cause the photoelectric effect in metals such as sodium, zinc, and copper. Since X-rays have extremely high frequencies, their photons carry more than enough energy to overcome the work functions of these metals.
In simple words: Yes, X-rays are very energetic and can easily knock electrons out of metals like sodium, zinc, and copper.

Exam Tip: Start with a clear "Yes" or "No" when answering direct questions before providing the scientific justification.

Question. How does the maximum kinetic energy of emitted photoelectrons change if the intensity of the incident light is increased?
Answer: The maximum kinetic energy of emitted photoelectrons remains completely unchanged. This is because the kinetic energy depends solely on the frequency of the incident light and is entirely independent of its intensity.
In simple words: Making the light brighter does not make the ejected electrons move any faster. Their speed only depends on the color or frequency of the light.

Exam Tip: Clearly state that intensity only affects the number of photoelectrons emitted per second, while frequency determines their kinetic energy.

Question. If the maximum kinetic energy of emitted photoelectrons is \( 5\text{ eV} \), calculate the stopping potential.
Answer: The relation between stopping potential \( V_0 \) and maximum kinetic energy is given by \( V_0 = \frac{K_{\text{max}}}{e} \). Substituting the given value, we get \( V_0 = \frac{5\text{ eV}}{e} \)
\( \implies V_0 = 5\text{ V} \). Thus, the stopping potential is \( 5\text{ V} \).
In simple words: To stop electrons that have 5 electron-volts of energy, you need a stopping voltage of exactly 5 volts.

Exam Tip: Always show the division by the elementary charge \( e \) to demonstrate how electron-volts (\text{eV}) convert directly to volts (\text{V}) for stopping potential.

Question. How does the threshold wavelength of a metal relate to its work function? Compare sodium and copper in terms of electron emission and threshold wavelength.
Answer: The work function is defined as \( W_0 = h\nu_0 = \frac{hc}{\lambda_0} \), which means the threshold wavelength is inversely proportional to the work function, \( \lambda_0 \propto \frac{1}{W_0} \). Because sodium has a lower work function than copper, it requires less energy to release electrons. Consequently, sodium has a higher threshold wavelength than copper.
In simple words: A metal with a lower work function holds its electrons less tightly, making them easier to release. This also means it can work with longer, lower-energy wavelengths of light.

Exam Tip: Remember that a lower work function always corresponds to a longer (higher) threshold wavelength because of their inverse relationship.

Question. State one common application of a photocell.
Answer: Photocells find an important application in the reproduction of sound from motion picture films.
In simple words: Photocells are used in movie theaters to help turn the soundtrack printed on film reels into actual sound we can hear.

Exam Tip: Mentioning "reproduction of sound in cinema" or "light meters in photography" are excellent, high-scoring examples of photocell uses.

Question. An electron and a proton have the same kinetic energy. Which particle has a longer de Broglie wavelength and why?
Answer: The kinetic energy \( K \) of a particle is related to its momentum \( p \) by \( K = \frac{p^2}{2m} \), which gives \( p = \sqrt{2mK} \). The de Broglie wavelength is expressed as \( \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \). For an electron and a proton with equal kinetic energy, the ratio of their wavelengths is \( \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \). Since the mass of an electron is much smaller than that of a proton (\( m_e < m_p \)), it follows that \( \lambda_e > \lambda_p \). Thus, the electron has a longer de Broglie wavelength.
In simple words: Lighter particles move faster when they have the same kinetic energy as heavier ones. Because the electron is much lighter than a proton, it has a longer wavelength.

Exam Tip: Clearly derive \( \lambda = \frac{h}{\sqrt{2mK}} \) to show how wavelength is inversely proportional to the square root of mass for a constant kinetic energy.

Question. In a Davisson-Germer experiment, if the scattering angle \( \phi \) is \( 52^\circ \), find the glancing angle \( \theta \) with the crystal planes.
Answer: The relationship between the glancing angle \( \theta \) and the scattering angle \( \phi \) is given by \( \theta = 90^\circ - \frac{\phi}{2} \). Substituting the scattering angle:
\( \implies \theta = 90^\circ - \frac{52^\circ}{2} \)
\( \implies \theta = 90^\circ - 26^\circ = 64^\circ \). Therefore, the glancing angle is \( 64^\circ \).
In simple words: Using the geometry of the experiment, we find the glancing angle by subtracting half of the scattering angle from 90 degrees, giving 64 degrees.

Exam Tip: Be careful not to confuse the scattering angle \( \phi \) with the glancing angle \( \theta \); use the formula \( 2\theta + \phi = 180^\circ \) to double-check your work.

Question. Find the energy of a photon of wavelength \( 6 \times 10^{-7}\text{ m} \).
Answer: The energy of a photon can be calculated using the formula \( E = \frac{hc}{\lambda} \). Substituting the given values:
\( \implies E = \frac{6.6 \times 10^{-34}\text{ J s} \times 3 \times 10^8\text{ m/s}}{6 \times 10^{-7}\text{ m}} \)
\( \implies E = 3.3 \times 10^{-19}\text{ J} \).
In simple words: By multiplying Planck's constant and the speed of light, and then dividing by the light's wavelength, we find that each photon carries \( 3.3 \times 10^{-19} \) Joules of energy.

Exam Tip: Always include standard SI units like Joules (\text{J}) in your final energy calculations to avoid losing marks.

Question. State Einstein's photoelectric equation and explain how the maximum velocity of photoelectrons changes as the wavelength of the incident light decreases.
Answer: According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by \( E_k = \frac{1}{2}mv^2 = h\nu - W_0 = \frac{hc}{\lambda} - W_0 \). This formula indicates that the velocity of the ejected electrons is related to wavelength such that \( v \propto \frac{1}{\sqrt{\lambda}} \) approximately. Consequently, as the wavelength of the incident radiation decreases, the maximum velocity of the emitted photoelectrons increases.
In simple words: Shorter wavelengths of light have more energy. This extra energy goes into kicking the electrons out with a higher speed.

Exam Tip: Remember that decreasing the wavelength increases the frequency, which provides more kinetic energy and thus higher speed to the photoelectrons.

Question. Draw a graph showing how the stopping potential varies with the frequency of incident radiation for a photoelectric material.
Answer: The graph below displays the variation of stopping potential with frequency. The intercept on the frequency axis represents the threshold frequency \( \nu_0 \) below which no photoelectric emission occurs.
In simple words: The graph shows that as you increase the frequency of the light past a certain starting point, the stopping voltage needed to halt the electrons goes up in a straight line.

Exam Tip: Always mark the threshold frequency \( \nu_0 \) on the frequency axis where the straight line starts. The slope of this line equals \( \frac{h}{e} \).

Question. Light of wavelength \( 4 \times 10^{-7}\text{ m} \) is incident on a metal X having a work function of \( 2.5\text{ eV} \). Will this metal exhibit photoelectric emission?
Answer: First, we calculate the energy of the incident photons:
\( E = \frac{hc}{\lambda} \)
\( \implies E = \frac{6.6 \times 10^{-34}\text{ J s} \times 3 \times 10^8\text{ m/s}}{4 \times 10^{-7}\text{ m}} \)
\( \implies E = 4.95 \times 10^{-19}\text{ J} \) Converting this energy to electron-volts (\text{eV}):
\( E = \frac{4.95 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ J/eV}} \approx 3.1\text{ eV} \) Since the energy of the incident photons (\( 3.1\text{ eV} \)) is greater than the work function of metal X (\( 2.5\text{ eV} \)), photoelectric emission will occur. Hence, metal X will emit photoelectrons.
In simple words: The energy of the light is about 3.1 electron-volts. Since this is higher than the metal's work function of 2.5 eV, the light has enough energy to knock out electrons.

Exam Tip: Always convert energy from Joules to electron-volts by dividing by \( 1.6 \times 10^{-19} \) so you can directly compare it to the work function of the metal.

Question. A photon and an electron have the same wavelength \( \lambda \). Prove that the total energy of the electron is greater than the energy of the photon.
Answer: For a photon, the energy is given by:
\( E_1 = \frac{hc}{\lambda} \) For an electron, the de Broglie wavelength is \( \lambda = \frac{h}{mv} \), which gives the relativistic mass \( m = \frac{h}{\lambda v} \). The total energy of the electron is:
\( E_2 = mc^2 = \left(\frac{h}{\lambda v}\right)c^2 = \frac{hc^2}{\lambda v} \) Taking the ratio of the two energies:
\( \frac{E_2}{E_1} = \frac{\left(\frac{hc^2}{\lambda v}\right)}{\left(\frac{hc}{\lambda}\right)} = \frac{c}{v} \) Since the velocity of the electron \( v \) is always less than the speed of light \( c \) (\( v < c \)), the ratio \( \frac{c}{v} > 1 \). Therefore,
\( \implies E_2 > E_1 \), which proves that the total energy of the electron is greater than that of the photon.
In simple words: Because an electron travels slower than the speed of light, its total energy relative to its wavelength is always greater than that of a photon of the same wavelength.

Exam Tip: Make sure to explicitly state that the electron's velocity \( v \) is less than the speed of light \( c \) to justify why the ratio \( \frac{c}{v} \) is greater than 1.

Question. An electron and a photon have the same wavelength \( \lambda \). Show that the energy of the photon \( E_{ph} \) is related to the kinetic energy of the electron \( E_e \) by the relation \( E_{ph} = E_e \left(\frac{2mc\lambda}{h}\right) \).
Answer: Let the common wavelength be \( \lambda \). For the photon, the energy is:
\( E_{ph} = \frac{hc}{\lambda} \) For the electron, the de Broglie wavelength is \( \lambda = \frac{h}{mv} \), which gives the velocity \( v = \frac{h}{m\lambda} \). The kinetic energy of the electron is:
\( E_e = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{h}{m\lambda}\right)^2 = \frac{h^2}{2m\lambda^2} \) Now, comparing the two energies:
\( \frac{E_{ph}}{E_e} = \frac{\left(\frac{hc}{\lambda}\right)}{\left(\frac{h^2}{2m\lambda^2}\right)} = \frac{hc}{\lambda} \times \frac{2m\lambda^2}{h^2} = \frac{2mc\lambda}{h} \)
\( \implies E_{ph} = E_e \left(\frac{2mc\lambda}{h}\right) \).
In simple words: By writing down the energy equations for both particles in terms of wavelength, we can divide one by the other to find the exact ratio between them.

Exam Tip: Perform step-by-step algebraic simplification of the fraction division to avoid calculation errors.

Question. Derive an expression for the de Broglie wavelength of a gas molecule of mass \( m \) in thermal equilibrium at temperature \( T \).
Answer: The kinetic energy \( E \) of a particle of mass \( m \) is given by \( E = \frac{1}{2}mv^2 = \frac{p^2}{2m} \), where \( p \) is its momentum. Thus, the momentum can be expressed as:
\( p = \sqrt{2mE} \) The de Broglie wavelength is:
\( \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \) From the kinetic theory of gases, the average kinetic energy of a gas molecule at absolute temperature \( T \) is \( E = \frac{3}{2}k_B T \) (where \( k_B \) is Boltzmann's constant). Substituting this value:
\( \lambda = \frac{h}{\sqrt{2m \left(\frac{3}{2} k_B T\right)}} \)
\( \implies \lambda = \frac{h}{\sqrt{3m k_B T}} \).
In simple words: Since heat makes gas molecules move, we can use their temperature to find their average energy, which then lets us calculate their wave-like wavelength.

Exam Tip: Always define \( k_B \) as the Boltzmann constant and \( T \) as the absolute temperature in Kelvin when writing this derivation.

Question. Why is there no photoelectric emission when light from a bulb falls on a wooden block?
Answer: Wood has a very high work function. The energy of the photons from a standard bulb is far too low to overcome this barrier. Since the photon energy is less than the work function of wood, no photoelectrons are emitted.
In simple words: A wooden block holds its electrons very tightly. The light from a regular bulb is too weak to free them, so no electrons are released.

Exam Tip: Explain that photoelectric emission can only happen when the incident photon's energy exceeds the work function of the target material.

Question. If light with a photon energy of \( 4\text{ eV} \) falls on a molybdenum (Mo) surface, will photoelectrons be emitted?
Answer: No, molybdenum will not emit photoelectrons. This is because the work function of molybdenum is greater than \( 4\text{ eV} \) (approximately \( 4.2\text{ eV} \)), meaning the incident photon energy is insufficient to eject any electrons.
In simple words: No electrons are released because the incoming light carries only 4 electron-volts of energy, which is less than what molybdenum needs to let go of its electrons.

Exam Tip: For emission to occur, the condition \( E \ge W_0 \) must be met. If the photon energy is smaller than the work function, write a definitive "No emission".

Question. If an electron, a proton, and an alpha particle all have the exact same kinetic energy, which one will have the shortest de Broglie wavelength?
Answer: The de Broglie wavelength of a particle with kinetic energy \( K \) is given by \( \lambda = \frac{h}{\sqrt{2mK}} \). Since the kinetic energy is constant, the wavelength is inversely proportional to the square root of the mass, \( \lambda \propto \frac{1}{\sqrt{m}} \). Because the alpha particle has the largest mass among the three, it will have the shortest de Broglie wavelength.
In simple words: For particles with the same energy, the heaviest one has the shortest wavelength. Since the alpha particle is the heaviest, it has the smallest wavelength.

Exam Tip: State the relation \( \lambda \propto \frac{1}{\sqrt{m}} \) clearly to show that a larger mass leads directly to a smaller de Broglie wavelength.

Question. How does the radius of the circular path of a charged particle in a uniform magnetic field depend on its charge, assuming its momentum remains constant?
Answer: The radius \( R \) of a circular path for a charged particle moving in a uniform magnetic field is given by \( R = \frac{mv}{qB} = \frac{p}{qB} \). When momentum \( p \) and magnetic field \( B \) are constant, the radius is inversely proportional to the charge of the particle:
\( \implies R \propto \frac{1}{q} \).
In simple words: The stronger the charge on a particle, the more it gets bent by a magnetic field, resulting in a smaller circular path.

Exam Tip: Remember the formula \( R = \frac{p}{qB} \) to easily determine how changes in momentum, magnetic field, or charge affect the radius of the path.

Question. An electron of mass \( m \) and charge \( e \) moves in a circular path of radius \( r \) under the influence of a uniform radial electric field \( E \). Write the equation of motion for this electron.
Answer: The electrostatic force acting on the electron due to the electric field is \( F_e = eE \). This force acts as the centripetal force required to keep the electron in a circular orbit. Therefore:
\( \implies eE = \frac{mv^2}{r} \).
In simple words: The electrical pull on the electron is what supplies the centripetal force needed to keep it revolving in a circle.

Exam Tip: Equate the electrostatic force \( qE \) to the centripetal force \( \frac{mv^2}{r} \) to solve orbital motion problems involving electric fields.

Question. If the accelerating potential of an electron is doubled, how will its de Broglie wavelength change?
Answer: The de Broglie wavelength of an electron accelerated through a potential difference \( V \) is given by \( \lambda = \frac{h}{\sqrt{2mqV}} \). If the potential is doubled to \( 2V \), the new wavelength becomes:
\( \implies \lambda' = \frac{h}{\sqrt{2mq(2V)}} = \frac{\lambda}{\sqrt{2}} \). Thus, the wavelength decreases to \( \frac{1}{\sqrt{2}} \) of its original value.
In simple words: When you double the voltage pushing the electron, it speeds up, causing its quantum wavelength to shrink by a factor of square root of 2.

Exam Tip: Be prepared to show that wavelength is inversely proportional to the square root of the potential difference, \( \lambda \propto \frac{1}{\sqrt{V}} \).

Question. How does the de Broglie wavelength of a particle change if its momentum is doubled?
Answer: The de Broglie wavelength is inversely proportional to its momentum, \( \lambda = \frac{h}{p} \). If the momentum is doubled to \( 2p \), the new wavelength is:
\( \implies \lambda' = \frac{h}{2p} = \frac{\lambda}{2} \). Therefore, the wavelength is halved.
In simple words: Doubling how hard a particle is moving (its momentum) cuts its wave-like wavelength exactly in half.

Exam Tip: Use the direct relationship \( \lambda = \frac{h}{p} \) to quickly calculate changes in wavelength when momentum is scaled.

Question. Find the threshold wavelength of a metal if its work function is \( 4.4\text{ eV} \).
Answer: The threshold wavelength is given by the relation \( \lambda_0 = \frac{hc}{\Phi} \). Substituting the values \( hc \approx 12420\text{ eV \AA} \) and \( \Phi = 4.4\text{ eV} \):
\( \implies \lambda_0 = \frac{12420\text{ eV \AA}}{4.4\text{ eV}} \approx 2823\text{ \AA} \). Therefore, the threshold wavelength is \( 2823\text{ \AA} \).
In simple words: To find the longest wavelength of light that can release electrons, we divide the constant of light energy by the metal's work function, which gives 2823 Angstroms.

Exam Tip: Using the approximation \( hc \approx 12400\text{ eV \AA} \) is a very convenient shortcut for calculating wavelengths directly in Angstroms from energy in electron-volts.

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