JEE Physics Nuclei MCQs Set 12

Answer: (a) \( \dfrac{E}{4}(A+4) \)

Explanation:
1. Momentum conservation gives equal and opposite momenta: \( p_\alpha = p_X \).
2. Kinetic energy is \( \dfrac{p^2}{2m} \propto \dfrac{1}{m} \), so \( \dfrac{K_\alpha}{E} = \dfrac{A}{4} \) and \( K_\alpha = \dfrac{AE}{4} \).
3. Total energy released \( = E + \dfrac{AE}{4} = \dfrac{E}{4}(A+4) \).

Teacher's Note:
1. Lighter fragment takes the larger share of kinetic energy.
2. Mass numbers can stand in for masses in this ratio.

Answer: (a)

Explanation:
1. In β decay the energy is shared between the electron and the antineutrino, so electrons are emitted with all energies from zero up to a maximum \( E_0 \).
2. The number of electrons is small at E = 0, rises to a peak, and falls smoothly to zero at \( E_0 \).
3. This continuous spectrum is graph (a).

Teacher's Note:
1. A continuous β spectrum led Pauli to propose the neutrino.
2. An α spectrum, in contrast, has discrete lines.

Answer: (c) 23.6 MeV

Explanation:
1. Total binding energy of two deuterons \( = 2\times2.2 = 4.4\ \text{MeV} \).
2. Binding energy of He-4 is 28 MeV.
3. Energy released \( = 28 - 4.4 = 23.6\ \text{MeV} \).

Teacher's Note:
1. Energy released in fusion equals the gain in total binding energy.
2. Remember to count the binding energy of both deuterons.

Answer: (a) \( \dfrac{N_0}{8} \)

Explanation:
1. Number of half-lives \( = \dfrac{15}{5} = 3 \).
2. Amount left \( = N_0\left(\dfrac12\right)^3 = \dfrac{N_0}{8} \).

Teacher's Note:
1. Divide the elapsed time by the half-life to get the number of halvings.
2. Each half-life halves the remaining amount.

Answer: (b) 10¹⁰ years

Explanation:
1. The γ emission follows the decay of uranium, so its rate falls with the half-life \( 5\times10^{9} \) yr.
2. The rate drops from 20 to 5, that is by a factor of 4 = 2², which takes two half-lives.
3. Time \( = 2\times5\times10^{9} = 10^{10} \) years.

Teacher's Note:
1. A fall by a factor of 4 means two half-lives.
2. The excited state of thorium lives for a very short time, so the rate is controlled by the uranium decay.

Answer: (d) V ∝ A

Explanation:
1. The nuclear radius is \( R = R_0A^{1/3} \).
2. So \( V = \dfrac43\pi R^3 = \dfrac43\pi R_0^3A \), which is proportional to A.

Teacher's Note:
1. Nuclear density is nearly the same for all nuclei, so volume follows mass number.
2. Radius, not volume, goes as the cube root of A.

Answer: (c) \( {}^{m-4}_{n}Z \)

Explanation:
1. An α particle lowers the mass number by 4 and the atomic number by 2.
2. Each β⁻ particle raises the atomic number by 1 and leaves the mass number unchanged.
3. Final mass number \( m-4 \); final atomic number \( n-2+2 = n \).

Teacher's Note:
1. Two β⁻ emissions undo the drop in atomic number caused by one α emission, giving an isotope of the original element.
2. Track A and Z separately.

Answer: (b) TTF

Explanation:
1. (i) True. Fission fragments are neutron rich, so they undergo β⁻ decay.
2. (ii) True. The daughter nucleus is usually left in an excited state and de-excites by emitting γ rays.
3. (iii) False. Binding energy per nucleon rises up to about A = 56 (iron) and then slowly falls for heavier nuclei.

Teacher's Note:
1. The binding energy per nucleon curve has its peak near iron.
2. Both fission of heavy nuclei and fusion of light ones release energy for this reason.

Answer: (b) 12 h

Explanation:
1. The intensity must fall by a factor of 64 = 2⁶.
2. This needs 6 half-lives.
3. Time \( = 6\times2 = 12 \) h.

Teacher's Note:
1. Write the ratio as a power of 2 to find the number of half-lives.
2. Each half-life halves the intensity.

Answer: (c) 931 MeV

Explanation:
1. \( 1\ \text{u} = 1.66\times10^{-27}\ \text{kg} \).
2. \( E = mc^2 = 1.66\times10^{-27}\times(3\times10^8)^2 = 1.49\times10^{-10}\ \text{J} \).
3. Dividing by \( 1.6\times10^{-13}\ \text{J/MeV} \) gives about 931 MeV.

Teacher's Note:
1. Remember 1 u = 931 MeV/c².
2. This is the standard conversion for mass defect problems.

Answer: (c) absorbing neutrons

Explanation:
1. Cadmium has a very large cross-section for neutron capture.
2. Inserting or withdrawing the rods controls the number of neutrons available for further fission, so they control the rate of the chain reaction.

Teacher's Note:
1. Control rods absorb neutrons; moderators slow them down.
2. Boron is another common control-rod material.

Answer: (c) α = 8, β = 6

Explanation:
1. The mass number falls by \( 235-203 = 32 \), and each α reduces it by 4, so \( \alpha = 8 \).
2. Eight α particles reduce Z by 16 to 76, but the end product has Z = 82.
3. Each β⁻ increases Z by 1, so \( \beta = 82-76 = 6 \).

Teacher's Note:
1. Use the mass number to count α particles, then the atomic number to count β particles.
2. This method always works for decay chains.

Answer: (c) M < Zmₚ + Nmₙ

Explanation:
1. When nucleons combine to form a nucleus, energy (the binding energy) is released.
2. By \( E = \Delta mc^2 \), the nucleus has less mass than its separate nucleons.
3. So \( M < Zm_p + Nm_n \) for every stable nucleus.

Teacher's Note:
1. The difference \( Zm_p + Nm_n - M \) is the mass defect.
2. A bound system always has less mass than its free parts.

Answer: (c) 650 years

Explanation:
1. The decay constants add: \( \lambda = \dfrac{1}{1620}+\dfrac{1}{660} \), so the effective mean life is \( \tau = \dfrac{1620\times660}{1620+660} = 469\ \text{years} \).
2. The material left is \( \dfrac14 \) when \( e^{-t/\tau} = \dfrac14 \), so \( t = \tau\ln4 = 469\times1.386 \).
3. \( t \approx 650\ \text{years} \).

Teacher's Note:
1. When a nucleus can decay in two ways, the decay constants add.
2. Mean life is the reciprocal of the decay constant.

Answer: (b) \( \dfrac{A_0}{9} \)

Explanation:
1. Activity falls exponentially, so equal time intervals give the same factor.
2. Each hour the activity is divided by \( \sqrt3 \).
3. After 3 more hours it is divided by \( (\sqrt3)^3 \) again: \( \dfrac{A_0}{\sqrt3}\cdot\dfrac{1}{3\sqrt3} = \dfrac{A_0}{9} \).

Teacher's Note:
1. In exponential decay the fractional change in equal time intervals is constant.
2. Count the total of 4 hours: \( (\sqrt3)^4 = 9 \).

Answer: (c) 12.2 h

Explanation:
1. If 99% has decayed, 1% remains, so \( \dfrac{N}{N_0} = \dfrac{1}{100} \).
2. \( t = \dfrac{T_{1/2}}{\ln2}\ln100 = \dfrac{110\times2\times2.303}{0.693} \approx 731\ \text{min} \).
3. \( 731\ \text{min} \approx 12.2\ \text{h} \).

Teacher's Note:
1. Write the remaining fraction first and use \( N = N_0e^{-\lambda t} \).
2. Convert minutes to hours at the end.