JEE Physics Nuclei MCQs Set 13

Answer: (b) They are charge dependent

Explanation:
1. Nuclear forces are the strongest forces known, short ranged (about 1 to 2 fm) and saturated, so each nucleon interacts only with its neighbours.
2. They are charge independent: the proton-proton, neutron-neutron and proton-neutron nuclear forces are nearly equal.
3. So the statement that they are charge dependent is incorrect.

Teacher's Note:
1. Charge independence of the nuclear force is one of its key properties.
2. Coulomb force, by contrast, depends on charge.

Answer: (b) 10¹⁷ kg/m³

Explanation:
1. Nuclear radius \( R=R_0A^{1/3} \) with \( R_0\approx1.2\times10^{-15}\ \text{m} \), so volume \( V=\frac43\pi R_0^3A \).
2. Mass \( =Am_p \), so density \( \rho=\frac{m_p}{\frac43\pi R_0^3}=\frac{1.67\times10^{-27}}{7.2\times10^{-45}}\approx2.3\times10^{17}\ \text{kg/m}^3 \).
3. This is of the order of 10¹⁷ kg/m³, independent of A.

Teacher's Note:
1. Nuclear density is the same for all nuclei because volume is proportional to A.
2. It is about 10¹⁴ times the density of water.

Answer: (c) only for A = 12

Explanation:
1. The atomic mass unit is defined as one-twelfth of the mass of a carbon-12 atom.
2. Hence the carbon-12 atom has mass exactly 12 u. For other nuclides the mass differs slightly from A u because of the mass defect and the proton-neutron mass difference.

Teacher's Note:
1. The unit u is defined through carbon-12.
2. Other atoms have masses close to, but not exactly equal to, A u.

Answer: (b) Q = E2 - 2E1

Explanation:
1. Energy released equals the increase in total binding energy: \( Q=B_{final}-B_{initial}=E_2-2E_1 \).

Teacher's Note:
1. Binding energy is the energy released when a nucleus is formed from its nucleons.
2. More binding energy afterwards means energy is released.

Answer: (b) 160 MeV

Explanation:
1. Binding energy of X: \( 200\times7.4=1480\ \text{MeV} \).
2. Binding energy of the products: \( 110\times8.2+90\times8.2=1640\ \text{MeV} \).
3. Energy released \( =1640-1480=160\ \text{MeV} \).

Teacher's Note:
1. Energy released is the difference of the total binding energies of products and reactant.
2. Total binding energy = (number of nucleons) x (binding energy per nucleon).

Answer: (c) 220 MeV

Explanation:
1. Parent: \( 200\times6.5=1300\ \text{MeV} \).
2. Fragments: \( 80\times7+120\times8=560+960=1520\ \text{MeV} \).
3. Energy released \( =1520-1300=220\ \text{MeV} \).

Teacher's Note:
1. Count total binding energy before and after.
2. A positive result means energy is released.

Answer: (d) 23.6 MeV

Explanation:
1. Binding energy of two deuterons: \( 2\times(2\times1.1)=4.4\ \text{MeV} \).
2. Binding energy of He-4: \( 4\times7.0=28\ \text{MeV} \).
3. Energy released \( =28-4.4=23.6\ \text{MeV} \).

Teacher's Note:
1. Each deuteron has 2 nucleons.
2. Subtract the initial total binding energy from the final.

Answer: (a) zero

Explanation:
1. The end product of a decay series is a stable nucleus that does not disintegrate.
2. A nucleus that never decays has decay constant zero.

Teacher's Note:
1. The decay constant is the probability per unit time that a nucleus decays.
2. For stable nuclei this probability is zero.

Answer: (c) \( \frac1e \)

Explanation:
1. Mean life \( \tau=\frac{1}{\lambda} \).
2. At \( t=\tau \): \( \frac{N}{N_0}=e^{-\lambda\tau}=e^{-1}=\frac1e \).

Teacher's Note:
1. Mean life is the time for the amount to fall to 1/e of its initial value.
2. It is longer than the half-life by the factor 1/ln 2 = 1.44.

Answer: (d) a given nucleus may decay at any time after t = 0

Explanation:
1. Radioactive decay is a random process: each nucleus has the same constant probability of decaying per unit time.
2. So any particular nucleus can decay at any instant after t = 0, even immediately, and some nuclei survive beyond 16 days (one quarter remain at 16 days).

Teacher's Note:
1. Half-life is a statistical property of a large sample, not a deadline for any one nucleus.
2. After n half-lives a fraction (1/2)^n of nuclei is still undecayed, never exactly zero.

Answer: (b) fusion

Explanation:
1. Two light nuclei (carbon-12 and helium-4) combine into a heavier nucleus (oxygen-16) with release of energy.
2. This is nuclear fusion.

Teacher's Note:
1. Fusion joins light nuclei; fission splits heavy ones.
2. Alpha and beta decay involve the emission of a particle from a single nucleus.

Answer: (d) captures a gamma-ray photon of energy 3 MeV

Explanation:
1. Rest energy of the separated nucleons \( =939+940=1879\ \text{MeV} \), which is 3 MeV more than the deuteron.
2. The deuteron must therefore absorb (capture) energy of at least 3 MeV, for example from a gamma-ray photon of 3 MeV.

Teacher's Note:
1. Breaking a bound system costs energy, so the deuteron has to absorb a photon.
2. The binding energy of the deuteron here is 3 MeV.

Answer: (c) 5 alpha and 4 beta

Explanation:
1. Fall in mass number: \( 226-206=20 \), so \( \frac{20}{4}=5 \) alpha particles.
2. These lower Z by 10, from 88 to 78. The final Z is 82, so 4 beta-minus emissions raise it by 4.

Teacher's Note:
1. Mass number changes only through alpha emission.
2. Then use the atomic number to find the number of beta emissions.

Answer: (a) The alpha particle has more kinetic energy than the daughter nucleus

Explanation:
1. Momentum conservation: the two products have equal and opposite momenta p.
2. Kinetic energy \( =\frac{p^2}{2m} \) is inversely proportional to mass.
3. The alpha particle is much lighter, so it carries the larger kinetic energy.

Teacher's Note:
1. Equal momentum with unequal masses means unequal kinetic energies.
2. The lighter product always takes most of the energy.

Answer: (c) M2 < 2M1

Explanation:
1. Two neon-20 nuclei contain the same 40 nucleons (20 protons and 20 neutrons) as one calcium-40 nucleus.
2. Calcium-40 has a larger binding energy per nucleon (about 8.55 MeV against 8.03 MeV for neon-20), so it has a larger total binding energy.
3. A larger binding energy means a smaller mass: M2 < 2M1.

Teacher's Note:
1. Mass of a nucleus = mass of its separate nucleons minus the binding energy divided by c².
2. Compare the total binding energies of the two systems with the same number of nucleons.

Answer: (a) 5.34 MeV

Explanation:
1. Total binding energy of C-13 \( =13\times7.5=97.5\ \text{MeV} \).
2. Total binding energy of C-12 \( =12\times7.68=92.16\ \text{MeV} \).
3. Energy to remove one neutron \( =97.5-92.16=5.34\ \text{MeV} \).

Teacher's Note:
1. Removing a neutron from C-13 leaves C-12.
2. The energy needed is the difference between the two total binding energies.

Answer: (a) 2 : 1

Explanation:
1. The rate of emission is the activity \( R=\lambda N \).
2. For equal rates, \( \lambda_AN_A=\lambda_BN_B \).
3. \( \frac{N_A}{N_B}=\frac{\lambda_B}{\lambda_A}=\frac21 \).

Teacher's Note:
1. Probability of getting a particle per unit time is proportional to the activity.
2. The nucleus with the smaller decay constant must be present in greater numbers.

Answer: (a) λ = λ1 + λ2

Explanation:
1. The decay probabilities per unit time for the two independent processes add.
2. Hence \( \lambda=\lambda_1+\lambda_2 \), and the mean life satisfies \( \frac1\tau=\frac1{\tau_1}+\frac1{\tau_2} \).

Teacher's Note:
1. Rates of independent decay modes add.
2. The effective half-life is shorter than either partial half-life.

Answer: (b) 8, 6

Explanation:
1. \( \frac{238-206}{4}=8 \) alpha particles; they reduce Z by 16 to 76.
2. To reach Z = 82 we need 6 beta-minus particles.

Teacher's Note:
1. Mass number falls only via alpha emission.
2. Each beta-minus raises Z by one.

Answer: (a) 172 and 69

Explanation:
1. Two alpha decays reduce the mass number by 8: 180 gives 172, and the atomic number by 4: 72 gives 68.
2. One beta-minus decay raises the atomic number by 1: 69.
3. Gamma emission changes neither. So A4 has mass number 172 and atomic number 69.

Teacher's Note:
1. Track mass number and atomic number step by step.
2. Beta-minus increases Z by 1; gamma changes nothing.

Answer: (a) A, B and D only

Explanation:
1. Alpha decay lowers Z by 2 and A by 4, so (A) is correct.
2. Beta-plus decay turns a proton into a neutron: Z decreases by 1 and A is unchanged, so (B) is correct.
3. Beta-minus decay turns a neutron into a proton: Z increases by 1, so (C) is incorrect.
4. Gamma emission only removes excitation energy; Z and A are unchanged, so (D) is correct.

Teacher's Note:
1. Beta-minus raises Z, beta-plus lowers Z.
2. Gamma decay does not change the composition of the nucleus.

Answer: (a) A, B and C only

Explanation:
1. The energy E is shared between the beta particle and the antineutrino (the recoil of the daughter is negligible), so each carries between zero and E: (A) and (B) are correct.
2. A neutron changes into a proton (or vice versa), so N/Z changes: (C) is correct.
3. Mass number is unchanged in beta decay, so (D) is incorrect.

Teacher's Note:
1. Beta energy has a continuous spectrum from zero up to the Q-value.
2. Beta decay converts one type of nucleon into the other, leaving A fixed.

Answer: (a) 492.6 MeV

Explanation:
1. Fe-56 has 26 protons and 30 neutrons. Mass of the separate constituents \( =26(1.00783)+30(1.00867)=56.46368\ \text{u} \).
2. Mass defect \( =56.46368-55.9349=0.5288\ \text{u} \).
3. Binding energy \( =0.5288\times931.5=492.6\ \text{MeV} \) (about 8.8 MeV per nucleon).

Teacher's Note:
1. Using atomic masses with hydrogen atoms for the protons keeps electron masses balanced.
2. The 8.79 MeV in the options is the binding energy per nucleon.

Answer: (a) 152 min

Explanation:
1. \( R=R_0e^{-\lambda t}\Rightarrow\lambda=\frac{1}{40}\ln\frac{600}{500}=\frac{0.1823}{40}=4.56\times10^{-3}\ \text{min}^{-1} \).
2. \( T_{1/2}=\frac{0.693}{4.56\times10^{-3}}\approx152\ \text{min} \).

Teacher's Note:
1. Use the activity ratio in the exponential decay law to find λ.
2. The half-life is 0.693 divided by λ.

Answer: (a) 2.32 τ

Explanation:
1. \( \frac{dN}{dt}=R-\lambda N\Rightarrow N=\frac{R}{\lambda}\left(1-e^{-\lambda t}\right) \), and the equilibrium amount is \( \frac R\lambda \).
2. For 80%: \( 1-e^{-\lambda t}=0.8\Rightarrow e^{-\lambda t}=0.2 \).
3. \( t=\frac{\ln5}{\lambda}=\frac{\ln5}{\ln2}\tau=2.32\tau \).

Teacher's Note:
1. The nuclide builds up until its decay rate equals the production rate.
2. Write the differential equation for N and solve it.

Answer: (a) 5.21 MeV

Explanation:
1. Q-value to the ground state of Ra-224: \( (228.028726-224.020196-4.00260)\times931.5=5.52\ \text{MeV} \).
2. Because 0.217 MeV goes to the gamma ray, the Q-value for the alpha decay to the excited state is \( 5.52-0.217=5.31\ \text{MeV} \).
3. The alpha particle gets the fraction \( \frac{224}{228} \) of this: \( 5.31\times\frac{224}{228}=5.21\ \text{MeV} \).

Teacher's Note:
1. The alpha particle and the daughter share Q in the inverse ratio of their masses.
2. Subtract the gamma energy because the daughter is left excited.

Answer: (b) 5.49 MeV

Explanation:
1. Mass defect \( =238.04955-234.04095-4.002603=0.005997\ \text{u} \), so \( Q=0.005997\times931.5=5.59\ \text{MeV} \).
2. Strictly, momentum conservation gives the alpha particle the share \( \frac{234}{238} \) of Q: \( K_\alpha=5.59\times\frac{234}{238}=5.49\ \text{MeV} \).

Teacher's Note:
1. When the recoil of the daughter is not negligible, the alpha gets (A - 4)/A of Q.
2. Always compute Q from the atomic mass difference first.

Answer: (a) 1.86 MeV

Explanation:
1. For beta-minus decay the atomic masses can be used directly: \( Q=(m_P-m_S)c^2=(31.974-31.972)\times931.5=1.86\ \text{MeV} \).
2. This energy is shared between the beta particle and the antineutrino.

Teacher's Note:
1. In beta-minus decay the electron mass is already accounted for in the atomic masses.
2. The total energy released equals the sum of the two particles' kinetic energies.