CBSE Class 12 Physics Atoms and Nuclei Assignment Set 03

Question. The size of the atom is proportional to
(a) A
(b) A1/3
(c) A2/3
(d) A–1/3

Answer: B

Question. The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is B = – me⁴ / 8n² ∈²₀ h² (m = electron mass).
If one decides to work in a frame of reference where the electron is at rest, the proton would be moving arround it. By similar arguments, the binding energy would be B = – me⁴ / 8n² ∈²₀ h² (M = proton mass)
This last expression is not correct because
(a) n would not be integral
(b) Bohr-quantisation applies only to electron
(c) the frame in which the electron is at rest is not inertial
(d) the motion of the proton would not be in circular orbits, even approximately

Answer: C

Question. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because
(a) of the electrons not being subject to a central force
(b) of the electrons colliding with each other
(c) of screening effects
(d) the force between the nucleus and an electron will no longer be given by Coulomb’s law

Answer: A

Question. The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is
(a) 1/2
(b) 2/237
(c) 1/137
(d) 1/237

Answer: C

Question. For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,
(a) because Bohr model gives incorrect values of angular momentum.
(b) because only one of these would have a minimum energy.
(c) angular momentum must be in the direction of spin of electron.
(d) because electrons go around only in horizontal orbits.

Answer: A

Question. O₂ molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms
(a) is not important because nuclear forces are short-ranged.
(b) is as important as electrostatic force for binding the two atoms.
(c) cancels the repulsive electrostatic force between the nuclei.
(d) is not important because oxygen nucleus have equal number of neutrons and protons.

Answer: A

Question. In the following transitions of the hydrogen atom, the one which gives an absorption line of highest frequency is
(a) n = 1 to n = 2
(b) n = 3 to n = 8
(c) n = 2 to n = 1
(d) n = 8 to n = 3

Answer: A

Question. To explain his theory, Bohr used
(a) conservation of linear momentum
(b) quantisation of angular momentum
(c) conservation of quantum
(d) none of the options

Answer: B

Question. Taking the Bohr radius as a0 = 53 pm, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr’s model, will be about
(a) 53 pm
(b) 27 pm
(c) 18 pm
(d) 13 pm

Answer: C

Question. The ratio of energies of the hydrogen atom in its first to second excited state is
(a) 1 : 4
(b) 4 : 1
(c) – 4 : – 9
(d) – (1/4) : – (1/9)

Answer: D

Question. If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be
(a) (16/25)λ
(b) (9/16)λ
(c) (20/7)λ
(d) (20/13)λ

Answer: C

Question. Hydrogen H, deuterium D, singly-ionised helium He+ and doubly-ionised lithium Li⁺⁺ all have one electron around the nucleus. Consider n = 2 to n = 1 transition. The wavelengths of the emitted radiations are λ₁, λ₂, λ₃, and λ₄ m m m m respectively. Then approximately
(a) λ₁ = 2λ₂ = 2√λ₃ = 3√λ₄
(b) λ₁ = λ₂ = 2λ₃ = 3λ₄
(c) λ₁ = 2λ₂ = 4λ₃ = 9λ₄
(d) 4λ₁ = 2λ₂ = 2λ₃ = λ₄

Answer: C

Question. The Bohr model for the spectra of a H-atom
(a) will not be applicable to hydrogen in the molecular from.
(b) will not be applicable as it is for a He-atom.
(c) is valid only at room temperature.
(d) predicts continuous as well as discrete spectral lines.

Answer: A, B

Question. Let E_n =1/8∈²₀ me4/n²h² be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency (E₂ – E₁)/h falls on it,
(a) it will not be absorbed at all.
(b) some of atoms will move to the first excited state.
(c) all atoms will be excited to the n = 2 state.
(d) no atoms will make a transition to the n = 3 state.

Answer: B, D

Question. The wavelength of the first line of Lyman series in hydrogen is 1216 Å. The wavelength of the second line of the same series will be
(a) 912 Å
(b) 1026 Å
(c) 3648 Å
(d) 6566 Å

Answer: B

Question. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is
(a) 10.20 eV
(b) 20.40 eV
(c) 13.6 eV
(d) 27.2 eV

Answer: A

Question. When an electron in an atom goes from a lower to a higher orbit, its
(a) kinetic energy (KE) increases, potential energy (PE) decreases
(b) KE increases, PE increases
(c) KE decreases, PE increases
(d) KE decreases, PE decreases

Answer: C

Question. According to Bohr’s theory, the energy of radiation in the transition from the third excited state to the first excited state for a hydrogen atom is
(a) 0.85 eV
(b) 13.6 eV
(c) 2.55 eV
(d) 3.4 eV

Answer: C

Question. Given the value of Rydberg constant is 10⁷ m^–1, the wave number of the last line of the Balmer series in hydrogen spectrum will be
(a) 0.25 × 10⁷ m^–1
(b) 2.5 × 10⁷ m^–1
(c) 0.025 × 10⁴ m^–1
(d) 0.5 × 10⁷ m^–1

Answer: A

Question. A set of atoms in an excited state decays.
(a) in general to any of the states with lower energy.
(b) into a lower state only when excited by an external electric field.
(c) all together simultaneously into a lower state.
(d) to emit photons only when they collide.

Answer: A

Fill in the Blanks

Question. The angle of scattering θ for zero value of impact parameter b is _________________.
Answer: 180°

Question. The frequency spectrum of radiation emitted as per Rutherford’s model of atom is _____________.
Answer: continuous

Question. The force responsible for scattering of alpha particle with target nucleus is _______________.
Answer: electrostatic force

Question. According to de Broglie a stationary orbit is that which contains an _______________ number of de Broglie waves associated with the revolting electron.
Answer: integral

Question. _______________ is a physical quantity whose dimensions are the same as that of Plank’s constant.
Answer: Angular momentum

Question. _______________ series of hydrogen spectrum lies in the visible region electromagnetic spectrum.
Answer: Balmer

Question. ______________ is the ionisation potential of hydrogen atom.
Answer: 13.6 eV

Question. Total energy of electron in a stationary orbit is _________________, which means the electron is bound to the nucleus and is not free to leave it
Answer: negative

Question. The value of Rydberg constant is _________________.
Answer: 1.09 × 10⁷ m^–1

Question. When an electron jumps from 2nd stationary orbit of hydrogen atom to 1st stationary orbit, the energy emitted is _________________.
Answer: 10.2 eV
 

Short Answer Type Questions
 

Question. What main conclusion did Rutherford draw from Geiger-Marsden's alpha-particle scattering experiment regarding where the positive charge and mass of an atom are concentrated?
Answer: Rutherford concluded that almost all the positive charge and the entire mass of an atom are concentrated in a tiny region at the center, which is called the nucleus.
In simple words: The experiment proved that nearly all of an atom's mass and positive charge are packed into a single, tiny central spot called the nucleus.

Exam Tip: Be sure to use the word "nucleus" and state that both mass and charge are concentrated in this tiny region to get full credit.

Question. Which observation in the alpha-particle scattering experiment led to the conclusion that most of the space inside an atom is empty?
Answer: This conclusion was based on the observation that the vast majority of the alpha particles passed straight through the thin gold foil without experiencing any deflection.
In simple words: Since almost all of the alpha particles went straight through the foil without being pushed aside, it shows that the atom is mostly empty space.

Exam Tip: Highlighting that the particles went "undeflected" is the key phrase examiners look for to justify a hollow atomic structure.

Question. Two nuclei have mass numbers in the ratio 1:27. What is the ratio of their nuclear radii?
Answer: The nuclear radius is related to the mass number by the formula:
\( R = R_0 A^{1/3} \)
Therefore, the ratio of the radii is:
\( \frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} \)
Substituting the given ratio \( \frac{A_1}{A_2} = \frac{1}{27} \):
\( \frac{R_1}{R_2} = \left(\frac{1}{27}\right)^{1/3} = \frac{1}{3} \)
Thus, the ratio of their nuclear radii is 1:3.
In simple words: Since the radius of a nucleus grows with the cube root of its mass, a mass ratio of 1 to 27 gives a size ratio of 1 to 3.

Exam Tip: Always state the formula \( R = R_0 A^{1/3} \) explicitly before performing the calculation to secure step marks.

Question. Two nuclei have mass numbers in the ratio 1:8. Find the ratio of their nuclear radii.
Answer: Using the empirical formula for nuclear radius:
\( R = R_0 A^{1/3} \)
Taking the ratio of the radii for both nuclei:
\( \frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} \)
Substituting the mass number ratio of \( \frac{1}{8} \):
\( \frac{R_1}{R_2} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2} \)
This gives the ratio \( R_1:R_2 = 1:2 \).
In simple words: The nuclear radius is proportional to the cube root of the mass number. For a mass ratio of 1 to 8, the radius ratio is the cube root, which is 1 to 2.

Exam Tip: Do not leave the final answer as a fraction; express it as a ratio \( 1:2 \) to match the question's format.

Question. Compare the ionizing power of alpha particles with that of beta particles.
Answer: Alpha particles possess a significantly higher ionizing power compared to beta particles.
In simple words: Alpha particles can knock electrons out of nearby atoms much more strongly than beta particles can.

Exam Tip: A direct, one-sentence statement is sufficient for this comparison-style question in 1-mark sections.

Question. A radioactive substance has a half-life of 30 days. How much time will it take for the quantity of the substance to decay to one-fourth of its initial value?
Answer: The remaining fraction of a radioactive substance is given by:
\( \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \)
Where \( n \) is the number of half-lives. Here, we require:
\( \frac{N}{N_0} = \frac{1}{4} \implies \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^2 \)
This gives \( n = 2 \) half-lives.
The total decay time \( t \) is:
\( t = n T_{1/2} = 2 \times 30 = 60 \text{ days} \)
Thus, it will take 60 days to reduce to one-fourth of its original mass.
In simple words: To decay to a quarter of its original amount, the substance must go through two half-lives. Since one half-life is 30 days, two of them will take 60 days.

Exam Tip: Clearly show the calculation for the number of half-lives \( n \) before calculating the total time \( t \).

Question. Why are neutrons considered highly effective projectiles for inducing nuclear reactions inside a target nucleus?
Answer: Because neutrons carry no electrical charge, they do not experience any electrostatic repulsion from the positively charged nucleus. This allows them to easily penetrate deep into the target nucleus and get absorbed, thereby altering the neutron-to-proton ratio and initiating a reaction.
In simple words: Since neutrons have no charge, they can enter a nucleus without being pushed away by electrical forces, which easily changes the balance of the nucleus.

Exam Tip: Emphasize the phrase "no electrostatic repulsion" to explain why neutral particles are ideal for nuclear bombardment.

Question. How does the neutron-to-proton ratio of a heavy nucleus change after it undergoes alpha decay?
Answer: The ratio of neutrons to protons inside the nucleus increases after the emission of an alpha particle.
In simple words: When a nucleus emits an alpha particle, the proportion of neutrons compared to protons in the remaining nucleus goes up.

Exam Tip: Be ready to explain this by showing that an alpha particle removes an equal number of protons and neutrons (2 of each), which naturally increases the ratio in heavy nuclei where neutrons outnumber protons.

Question. Why do alpha particles have high ionizing power when passing through a medium?
Answer: Due to their relatively large mass and double positive charge, alpha particles easily attract and pull out electrons from the atoms and molecules they collide with along their path.
In simple words: Alpha particles are heavy and carry a strong positive charge, which allows them to easily tear electrons away from other atoms they run into.

Exam Tip: Mention both "larger mass" and "greater charge" as the dual physical causes behind their high ionization ability.

Question. The half-life of a radioactive isotope is 20 minutes. What fraction of the original mass of this substance will remain undecayed after one hour?
Answer: Given the half-life \( T_{1/2} = 20 \text{ minutes} \) and the total decay time \( t = 1 \text{ hour} = 60 \text{ minutes} \).
The number of half-lives is:
\( n = \frac{t}{T_{1/2}} = \frac{60}{20} = 3 \)
Using the decay formula:
\( \frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \)
Thus, \( \frac{1}{8} \) of the initial mass will remain undecayed after one hour.
In simple words: An hour has three 20-minute periods. Halving the starting mass three times leaves you with exactly one-eighth of what you began with.

Exam Tip: Ensure that both time values are in the same units (minutes) before calculating the value of \( n \).

Question. What changes occur in the mass number and atomic number of a nucleus when it undergoes gamma decay?
Answer: During gamma decay, the nucleus merely loses excess energy. It does not undergo any change in its atomic number or mass number, meaning it remains the exact same isotope.
In simple words: When a nucleus releases a gamma ray, it only releases energy. It does not turn into a different element or change its weight.

Exam Tip: Explicitly state that both the mass number \( A \) and the atomic number \( Z \) remain completely unchanged.

Question. Can a radioactive nucleus emit both an alpha particle and a beta particle simultaneously?
Answer: No, a single nucleus cannot emit an alpha particle and a beta particle at the same time. It will decay by emitting either one or the other. However, if the nucleus is left in an excited state after the decay, it can also emit a gamma ray.
In simple words: A nucleus has to choose to throw out either an alpha or a beta particle, never both at once. But it can release a gamma ray right after doing either of those if it still has leftover energy.

Exam Tip: Emphasize that alpha and beta decays are mutually exclusive for a single disintegration event, while gamma emission is a secondary process.

Question. What is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1:3?
Answer: The ratio of their nuclear densities is 1:1, because nuclear density is a constant value that does not depend on the mass number \( A \).
In simple words: All atomic nuclei are packed with the same tightness, so their density ratio is always 1 to 1 regardless of how heavy they are.

Exam Tip: State clearly that nuclear density is "independent of mass number" to earn full marks on this common conceptual question.