JEE Physics Nuclei MCQs Set 16

Answer: (b) Its kinetic energy increases and its potential and total energies decrease.

Explanation:
1. In the Bohr model \( E_n=-\frac{13.6}{n^2} \) eV, \( K=+\frac{13.6}{n^2} \) eV and \( U=-\frac{27.2}{n^2} \) eV.
2. Moving to a lower n, K increases, U becomes more negative (decreases) and the total energy E = −K decreases.

Teacher's Note:
1. Total energy = −kinetic energy for bound orbits.
2. A smaller orbit means a faster electron.

Answer: (a) 1 mA

Explanation:
1. Frequency of revolution \( f=\frac{v}{2\pi r}=\frac{2\times10^6}{2\pi\times0.5\times10^{-10}}\approx6.4\times10^{15}\ \text{Hz} \).
2. \( I=ef=1.6\times10^{-19}\times6.4\times10^{15}\approx1\times10^{-3}\ \text{A}=1\ \text{mA} \).

Teacher's Note:
1. A revolving charge constitutes a current I = qf.
2. Convert Å to metres.

Answer: (d) 3

Explanation:
1. Energy needed for the ground state to the nth level: \( 13.6\left(1-\frac1{n^2}\right)=12.09\Rightarrow n=3 \).
2. From n = 3 the possible transitions are 3→2, 3→1 and 2→1: three lines.

Teacher's Note:
1. Find the level reached from the energy absorbed.
2. From n there are n(n − 1)/2 possible lines.

Answer: (c) doubly ionised lithium

Explanation:
1. For hydrogen-like species \( \frac1\lambda=RZ^2\left(1-\frac14\right) \).
2. The shortest wavelength corresponds to the largest Z²: Z = 3 for Li²⁺ (H and D have Z = 1, He⁺ has Z = 2).

Teacher's Note:
1. Transition energy is proportional to Z².
2. The slight difference between H and D (reduced mass) is negligible compared with Z².

Answer: (b) \( E_2\to E_3 \)

Explanation:
1. Absorption of light moves an electron up to a higher level, so the transition must go upwards.
2. Red light has less energy than blue light, so the gap must be smaller than \( E_4-E_2 \).
3. The upward transition with a smaller gap is \( E_2\to E_3 \).

Teacher's Note:
1. Absorption raises the electron; emission lowers it.
2. Red photons carry less energy than blue photons.

Answer: (a) \( \frac{128}{27}\lambda \)

Explanation:
1. \( \frac1\lambda=R\left(1-\frac19\right)=\frac{8R}{9} \).
2. \( \frac1{\lambda'}=R\left(\frac14-\frac1{16}\right)=\frac{3R}{16} \).
3. \( \lambda'=\frac{16}{3R}=\frac{16}{3}\times\frac{8}{9}\lambda=\frac{128}{27}\lambda \).

Teacher's Note:
1. Use the Rydberg formula for each transition.
2. Express λ′ in terms of λ by taking the ratio.

Answer: (a) the first three Lyman lines, the first two Balmer lines and the first Paschen line

Explanation:
1. A bombarding electron of 12.75 eV can raise the atom from n = 1 to n = 4 (\( 13.6\left(1-\frac1{16}\right)=12.75\ \text{eV} \)).
2. From n = 4 the transitions are 4→3, 4→2, 4→1, 3→2, 3→1, 2→1.
3. These are the first three Lyman lines (to n = 1), the first two Balmer lines (to n = 2) and the first Paschen line (4→3).

Teacher's Note:
1. An electron can transfer any part of its energy, but the atom needs exactly the level energy.
2. Lyman: to n = 1, Balmer: to n = 2, Paschen: to n = 3.

Answer: (d) either two atoms or three atoms

Explanation:
1. 12.1 eV is the 3→1 transition, 10.2 eV is 2→1, and 1.9 eV is 3→2.
2. One atom going 3→2→1 can emit 1.9 eV and 10.2 eV, but the 12.1 eV photon (3→1) must come from another atom.
3. So at least two atoms are needed; the photons could also have come from three different atoms.

Teacher's Note:
1. An atom can only complete a path from n to 1 once.
2. 3→1 is a different path from 3→2→1.

Answer: (c) the neutron-proton ratio increases as the mass number of the element increases

Explanation:
1. Stable heavy nuclei need relatively more neutrons (\( N/Z \) rises with A).
2. When the nucleus splits, each lighter fragment needs a smaller N/Z ratio, so the extra neutrons are released.

Teacher's Note:
1. Heavy nuclei have a larger neutron excess.
2. The fragments must shed neutrons to become stable.

Answer: (b) \( E_a-e=3E_b \)

Explanation:
1. Energy released in a reaction = (binding energy of products) − (binding energy of reactants).
2. \( e=E_a-3E_b\Rightarrow E_a-e=3E_b \).

Teacher's Note:
1. Release of energy means the product is more tightly bound.
2. Binding energy of products minus that of reactants.

Answer: (a) \( \frac{\lambda mL}{M} \)

Explanation:
1. Number of atoms \( N=\frac mML \).
2. Activity \( A=\lambda N=\frac{\lambda mL}{M} \).

Teacher's Note:
1. Activity = decay constant × number of nuclei.
2. Number of atoms = (mass/molar mass) × Avogadro number.

Answer: (d) \( \frac{h}{2mc} \)

Explanation:
1. The photon energy must be at least the rest energy of the pair: \( \frac{hc}{\lambda}\ge2mc^2 \).
2. \( \lambda\le\frac{hc}{2mc^2}=\frac{h}{2mc} \).

Teacher's Note:
1. Threshold: photon energy = 2mc².
2. A longer wavelength means less energy.

Answer: (c) \( t \)

Explanation:
1. At time T the numbers of X and Y atoms are equal: \( N_0e^{-\lambda T}=N_0\left(1-e^{-\lambda T}\right) \).
2. \( e^{-\lambda T}=\frac12 \), so exactly half of X has decayed.
3. That is one half-life: \( T=t \).

Teacher's Note:
1. Y atoms = N0 minus remaining X atoms.
2. Equal numbers means 50% of the sample has decayed.

Answer: (b) λδt gives the fraction of atoms present which will decay in the next small time interval δt

Explanation:
1. \( \frac{dN}{N}=-\lambda\,dt \): the fraction decaying in a short interval δt is \( \lambda\,\delta t \).
2. The time for N to fall to N0/e is the mean life \( 1/\lambda \), not λ, and the other statements are not correct.

Teacher's Note:
1. λ is the probability of decay per unit time.
2. Mean life τ = 1/λ.

Answer: (a) 6.25%

Explanation:
1. The decays are simultaneous, so the decay constants add: \( \frac1T=\frac14+\frac1{12}=\frac13\Rightarrow T=3\ \text{years} \).
2. After 12 years there are \( \frac{12}{3}=4 \) half-lives: \( \left(\frac12\right)^4=\frac1{16}=6.25\% \).

Teacher's Note:
1. For parallel decay modes, 1/T = 1/T1 + 1/T2.
2. Number of half-lives = time / effective half-life.

Answer: (d) reaction of high-energy neutrons on nitrogen

Explanation:
1. Cosmic rays produce neutrons in the upper atmosphere.
2. These neutrons strike nitrogen: \( {}^{14}_{7}\text{N}+{}^1_0n\to{}^{14}_{6}\text{C}+{}^1_1p \).

Teacher's Note:
1. Carbon-14 dating relies on this reaction.
2. The target is nitrogen, the most abundant atmospheric gas.

Answer: (c) \( \frac{Z-1}{A-Z-7} \)

Explanation:
1. Two α-particles reduce Z by 4 and A by 8. Three β-particles increase Z by 3.
2. Final protons \( Z-4+3=Z-1 \); final mass number \( A-8 \).
3. Neutrons \( (A-8)-(Z-1)=A-Z-7 \).
4. Ratio protons : neutrons \( =\frac{Z-1}{A-Z-7} \).

Teacher's Note:
1. β-emission turns a neutron into a proton (Z + 1, A unchanged).
2. α-emission removes 2 protons and 2 neutrons.

Answer: (b) a continuous curve rising to a maximum and falling to zero at the maximum energy

Explanation:
1. The energy of the decay is shared between the β-particle and the antineutrino in all proportions.
2. So β-particles have a continuous distribution of energies from zero up to a maximum energy, with a peak at an intermediate value.

Teacher's Note:
1. A continuous β-spectrum led to the neutrino hypothesis.
2. Unlike α-particles, β-particles do not have discrete energies.

Answer: (c) \( \beta,\ \alpha,\ \gamma \)

Explanation:
1. First step: Z increases by 1 with A unchanged, so it is β-emission.
2. Second step: Z decreases by 2 and A by 4, so it is α-emission.
3. Third step: neither Z nor A changes, so it is γ-emission.

Teacher's Note:
1. β: Z + 1, α: Z − 2 and A − 4, γ: no change.
2. Compare Z and A before and after each step.