JEE Physics Nuclei MCQs Set 08

Answer: (a) \({}^{11}_{6}\mathrm{C}\) → \({}^{11}_{5}\mathrm{B}\) + \({}^{0}_{+1}e\) + neutrino

Explanation:
1. In β⁺ decay a proton changes into a neutron: Z decreases by 1 (6 → 5, boron), A stays 11
2. A neutrino is emitted together with the positron.

Teacher's Note:
1. β⁺ decay occurs in proton-rich nuclei, so n/p rises.
2. Positron emission lowers Z by 1.

Answer: (c) \({}^{97}_{43}\mathrm{Tc}\) → \({}^{97}_{42}\mathrm{Mo}\) + \({}^{0}_{+1}e\) + neutrino

Explanation:
1. β⁺ decay: Z = 43 − 1 = 42 (molybdenum), A = 97
2. A neutrino accompanies the positron.

Teacher's Note:
1. The daughter nucleus of β⁺ decay is an isobar of the parent.
2. Z decreases by one.

Answer: (a) \({}^{120}_{54}\mathrm{Xe}\) + \({}^{0}_{-1}e\) → \({}^{120}_{53}\mathrm{I}\) + X-rays + neutrino

Explanation:
1. A proton changes into a neutron by capturing the electron: Z = 54 − 1 = 53 (iodine), A = 120
2. The vacancy in the K-shell is filled by an outer electron, giving X-rays, and a neutrino is emitted.

Teacher's Note:
1. K-capture is similar to β⁺ decay: the neutron-to-proton ratio increases.
2. The characteristic X-rays are a signature of K-capture.

Answer: (a) 1 : 9

Explanation:
1. Activity = λN = (0.693/T)N, so equal activities give N₁/T₁ = N₂/T₂
2. N_x : N_y = T_x : T_y = 3 : 27 = 1 : 9.

Teacher's Note:
1. For equal activity the element with the shorter half-life has fewer atoms.
2. Activity is proportional to N / T½.

Answer: (b) 12.011

Explanation:
1. Average = (98.9 × 12 + 1.1 × 13)/100
2. = (1186.8 + 14.3)/100 = 12.011.

Teacher's Note:
1. Atomic weight is the abundance-weighted mean of isotopic masses.
2. Since C-12 is by far the most abundant, the average is only slightly above 12.

Answer: (a) 4T

Explanation:
1. 6.25% = 1/16 = (1/2)⁴
2. So 4 half-lives are needed: t = 4T.

Teacher's Note:
1. Express the final fraction as a power of 1/2 to count half-lives.
2. Activity falls in the same ratio as the number of nuclei.

Answer: (d) 26/27

Explanation:
1. Fraction left after 7 days = 1/3, so each 7-day interval multiplies the remainder by 1/3
2. After 21 days (3 intervals) the fraction left = (1/3)³ = 1/27
3. Fraction decayed = 1 − 1/27 = 26/27.

Teacher's Note:
1. The decay is exponential, so equal times give equal ratios of remaining nuclei.
2. Do not add decayed fractions; multiply remaining fractions.

Answer: (b) 3200

Explanation:
1. 25% = 1/4 = (1/2)²
2. Time = 2 half-lives = 2 × 1600 = 3200 years.

Teacher's Note:
1. Two half-lives always leave one quarter.
2. Keep the time unit the same as the half-life unit.

Answer: (c) 5T

Explanation:
1. 3.125% = 1/32 = (1/2)⁵
2. So t = 5T.

Teacher's Note:
1. 1/32 is 2⁻⁵, so five half-lives are needed.
2. 1% needs t = T log₂100 = 6.64T, a useful comparison.

Answer: (a) 6.64T

Explanation:
1. N/N₀ = (1/2)^(t/T) = 0.01
2. t/T = log₂100 = ln100/ln2 = 4.605/0.693 = 6.64
3. So t = 6.64T.

Teacher's Note:
1. When the fraction is not a power of 1/2, use logarithms.
2. 1% lies between 6 half-lives (1.56%) and 7 half-lives (0.78%).

Answer: (b) 7.1 μg

Explanation:
1. Activity = 8 × 10⁻³ × 3.7 × 10¹⁰ = 2.96 × 10⁸ decays/s
2. λ = 0.693/(5.3 × 3.156 × 10⁷ s) = 4.14 × 10⁻⁹ s⁻¹
3. N = A/λ = 7.15 × 10¹⁶ atoms
4. Mass = (7.15 × 10¹⁶ / 6.02 × 10²³) × 60 g ≈ 7.1 × 10⁻⁶ g = 7.1 μg.

Teacher's Note:
1. Use N = A/λ to find the number of atoms from the required activity.
2. Convert the half-life into seconds before finding λ.

Answer: (b) 1.2 × 10⁻⁷ g/s

Explanation:
1. Fissions per second = 10⁴ / (200 × 1.6 × 10⁻¹³) = 3.1 × 10¹⁴
2. Mass per second = (3.1 × 10¹⁴ / 6.02 × 10²³) × 235 g = 1.2 × 10⁻⁷ g/s.

Teacher's Note:
1. Rate of fission × mass of one atom gives the mass consumed per second.
2. Mass of one U-235 atom = 235/6.02 × 10²³ g.

Answer: (a) 2.5 × 10⁴ years

Explanation:
1. Number of atoms in 1 kg = (1000/235) × 6.02 × 10²³ = 2.56 × 10²⁴
2. Total energy = 2.56 × 10²⁴ × 3.2 × 10⁻¹¹ = 8.2 × 10¹³ J
3. Time = 8.2 × 10¹³/100 = 8.2 × 10¹¹ s ≈ 2.6 × 10⁴ years.

Teacher's Note:
1. Total energy divided by power gives the time.
2. One year is about 3.15 × 10⁷ s.

Answer: (d) 10³³ eV

Explanation:
1. Number of deuterons in 2 kg = (2000/2) × 6.02 × 10²³ = 6 × 10²⁶
2. Each reaction uses 2 deuterons: 3 × 10²⁶ reactions
3. Energy = 3 × 10²⁶ × 3.2 MeV ≈ 10²⁷ MeV = 10³³ eV.

Teacher's Note:
1. Remember that one reaction consumes two nuclei.
2. 1 MeV = 10⁶ eV.

Answer: (d) 5 × 10²⁶ MeV

Explanation:
1. Number of nuclei = (1000/235) × 6.02 × 10²³ = 2.56 × 10²⁴
2. Energy = 2.56 × 10²⁴ × 200 MeV = 5.1 × 10²⁶ MeV (about 8 × 10¹³ J, or 2.3 × 10⁷ kWh).

Teacher's Note:
1. Check the units of each option before comparing.
2. 8 × 10¹³ J is also 5 × 10³² eV, so the other options do not match.

Answer: (b) 1.8 MeV

Explanation:
1. Photon energy = rest energy of pair + kinetic energy of pair
2. = 2 × 0.51 + 0.78 = 1.80 MeV.

Teacher's Note:
1. Energy conservation: hν = 2m₀c² + KE.
2. The threshold energy is 1.02 MeV.

Answer: (d) 0.0024

Explanation:
1. Δm = 2.23/931 = 0.0024 u
2. Mass defect is positive by definition.

Teacher's Note:
1. Divide the binding energy in MeV by 931 to get u.
2. Mass defect cannot be negative.

Answer: (c) 3.125 × 10¹⁶

Explanation:
1. Energy per fission = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J
2. Number per second = 10⁶/3.2 × 10⁻¹¹ = 3.125 × 10¹⁶.

Teacher's Note:
1. Power in watts divided by joules per fission.
2. 1000 kW = 10⁶ W.

Answer: (b) 3.045 MeV

Explanation:
1. KE = total energy − rest energy = 3.555 − 0.511 = 3.044 MeV ≈ 3.045 MeV.

Teacher's Note:
1. Relativistic energy: E = KE + m₀c².
2. The rest energy of an electron is 0.511 MeV.

Answer: (a) 7 × 10⁻³⁸ cc

Explanation:
1. R = 1.3 × 8^(1/3) = 2.6 fermi = 2.6 × 10⁻¹³ cm
2. V = (4/3)πR³ = 4.19 × 1.76 × 10⁻³⁸ ≈ 7 × 10⁻³⁸ cc.

Teacher's Note:
1. 1 fermi = 10⁻¹⁵ m = 10⁻¹³ cm.
2. Cube the radius carefully.

Answer: (b) 4.25 MeV

Explanation:
1. Δm = 238.05079 − 234.04363 − 4.00260 = 0.00456 u
2. Q = 0.00456 × 931 = 4.25 MeV.

Teacher's Note:
1. Atomic masses can be used directly because the electron count balances.
2. Most of the energy goes to the α-particle.

Answer: (d) 2.5 × 10¹⁶

Explanation:
1. Energy per fission = 250 × 1.6 × 10⁻¹³ = 4 × 10⁻¹¹ J
2. Rate = 10⁶/4 × 10⁻¹¹ = 2.5 × 10¹⁶ per second.

Teacher's Note:
1. Fissions per second = power / energy per fission.
2. Use 1 MW = 10⁶ W.

Answer: (b) 10.837 MeV

Explanation:
1. Mass of 3 α-particles = 3 × 4.00388 = 12.01164 u
2. Δm = 12.01164 − 12.0000 = 0.01164 u
3. Energy required = 0.01164 × 931 = 10.84 MeV.

Teacher's Note:
1. Energy is needed because the products have more mass than the reactant.
2. Use the mass values given in the problem.

Answer: (a) 27

Explanation:
1. R = R₀A^(1/3) so A^(1/3) = 3.6/1.2 = 3
2. A = 3³ = 27.

Teacher's Note:
1. Cube the ratio R/R₀ to get A.
2. A = 27 is aluminium.

Answer: (c) 23.8 MeV

Explanation:
1. B.E. of 2 deuterons = 2 × (2 × 1.15) = 4.6 MeV
2. B.E. of He-4 = 4 × 7.1 = 28.4 MeV
3. Q = 28.4 − 4.6 = 23.8 MeV.

Teacher's Note:
1. Multiply by the mass number to get total binding energy.
2. Q is the difference between product and reactant binding energies.