JEE Physics Nuclei MCQs Set 14

Answer: (c) 210 and 80

Explanation:
1. Two alpha decays reduce the mass number by 8 and the atomic number by 4: 218 - 8 = 210, 84 - 4 = 80.
2. Beta minus raises Z by 1 and beta plus lowers Z by 1, so these cancel; gamma emission changes neither number.
3. So A6 has mass number 210 and atomic number 80.

Teacher's Note:
1. Alpha decay: A falls by 4 and Z falls by 2. Beta-minus: Z rises by 1. Beta-plus: Z falls by 1. Gamma: no change.
2. Add up the changes step by step.

Answer: (a) 11

Explanation:
1. Nuclear density is independent of A: \( \rho=\frac{m_{nucleon}}{\frac43\pi R_0^3}=\frac{1.6\times10^{-27}}{\frac43\pi(1.5\times10^{-15})^3}\approx1.1\times10^{17}\ \text{kg m}^{-3} \).
2. Density of water is 10³ kg/m³, so the ratio is about \( 1.1\times10^{14}=11\times10^{13} \), giving n = 11.

Teacher's Note:
1. Mass is A nucleons and volume is proportional to A, so A cancels from the density.
2. Nuclear density is about 10¹⁷ kg/m³, enormously larger than that of ordinary matter.

Answer: (a) 6

Explanation:
1. Number of atoms in 120 g: \( \frac{120}{240}\times6\times10^{23}=3\times10^{23} \).
2. Total energy: \( 3\times10^{23}\times200=6\times10^{25}\ \text{MeV} \)

Teacher's Note:
1. Moles = mass / molar mass, and each mole contains N_A atoms.
2. Multiply the number of atoms by the energy per fission.

Answer: (c) 1 : 1

Explanation:
1. Nuclear radius \( R=R_0A^{1/3} \), so the volume is proportional to A.
2. Mass is proportional to A as well, hence the density \( \rho=\frac{A\,m}{\frac43\pi R_0^3A} \) is the same for all nuclei.
3. Ratio = 1 : 1.

Teacher's Note:
1. Nuclear matter has almost constant density for all nuclei.
2. This follows from the R proportional to A^(1/3) law.

Answer: (a) 2

Explanation:
1. Momentum conservation: \( m_1v_1=m_2v_2\Rightarrow\frac{m_1}{m_2}=\frac{v_2}{v_1}=\frac23 \).
2. Nuclear radius is proportional to \( A^{1/3} \), hence to \( m^{1/3} \): \( \frac{R_1}{R_2}=\left(\frac23\right)^{1/3} \). So x = 2.

Teacher's Note:
1. The parent nucleus is at rest, so the two fragments have equal and opposite momenta.
2. The lighter fragment moves faster.

Answer: (a) \( \frac18 \)

Explanation:
1. Number of half-lives: n = 90/30 = 3.
2. \( \frac{N}{N_0}=\left(\frac12\right)^3=\frac18 \)

Teacher's Note:
1. After each half-life the undecayed amount halves.
2. The type of decay does not affect the calculation.

Answer: (a) 87

Explanation:
1. Mass number check: 242 - 2 x 4 = 234, as given.
2. Change in atomic number: two alphas give -4, one electron (beta-minus) gives +1, two positrons give -2.
3. \( P=92-4+1-2=87 \)

Teacher's Note:
1. A beta-minus decay raises Z by 1, and a positron emission lowers it by 1.
2. Electrons and positrons do not change the mass number.

Answer: (a) \( 3.38\times10^{24} \)

Explanation:
1. A: 2 days = 2 half-lives, so the mass left is 320/4 = 80 g, i.e. 80/16 = 5 mol = \( 3.01\times10^{24} \) atoms.
2. B: 2 days = 4 half-lives, so the mass left is 320/16 = 20 g, i.e. 20/32 = 0.625 mol = \( 3.76\times10^{23} \) atoms.
3. Total: \( 3.01\times10^{24}+0.376\times10^{24}\approx3.38\times10^{24} \)

Teacher's Note:
1. Find the mass left from the number of half-lives, then convert to moles and atoms.
2. The mass number gives the molar mass in grams.

Answer: (b) A is false but R is true

Explanation:
1. Reason R is true.
2. Because R = R0 A^(1/3), the volume is proportional to A and the density is the same for every nuclide, so the Assertion, which ranks the densities, is false.

Teacher's Note:
1. The same R0 A^(1/3) law that is quoted in R actually shows that the nuclear density is constant.
2. So the assertion fails even though the reason is correct.

Answer: (a) 300

Explanation:
1. The decay constants add: \( \lambda=\lambda_1+\lambda_2 \), so \( \frac1{T}=\frac1{T_1}+\frac1{T_2}=\frac1{300}+\frac1{30}=\frac{11}{300} \) (in s).
2. \( T=\frac{300}{11}\ \text{s} \), so alpha = 300.

Teacher's Note:
1. Convert 5 minutes to 300 s before adding.
2. When two decay modes act together, the reciprocals of the half-lives add.

Answer: (d) a neutron has a larger rest mass than a proton

Explanation:
1. Free decay needs the products to have less total rest energy than the parent.
2. The neutron is slightly heavier than the proton (by about 1.29 MeV/c²), so neutron to proton + electron + antineutrino is allowed.
3. The reverse process would need energy to be supplied, so it does not occur for a free proton.

Teacher's Note:
1. A free particle can decay only into products of smaller total mass.
2. A neutron is not a proton plus an electron; it is a distinct particle.

Answer: (b) 28.4 MeV

Explanation:
1. Mass defect: \( \Delta m=2(1.0073)+2(1.0087)-4.0015=0.0305\ \text{u} \).
2. \( E=0.0305\times931.5\approx28.4\ \text{MeV} \)

Teacher's Note:
1. Binding energy = (mass defect) x 931.5 MeV per u.
2. Helium-4 contains 2 protons and 2 neutrons.

Answer: (a) 6

Explanation:
1. Nucleus A has 17 protons and 17 neutrons, so A = 34 and its binding energy is 34 x 1.2 = 40.8 MeV.
2. Nucleus B: 26 x 1.8 = 46.8 MeV.
3. Difference = 46.8 - 40.8 = 6 MeV.

Teacher's Note:
1. Total binding energy = (binding energy per nucleon) x (number of nucleons).
2. Find the mass number of A from the condition that protons equal neutrons.