JEE Physics Nuclei MCQs Set 03

Answer: (c) \( \frac{N_0}{32} \)

Explanation:
1. \( \frac{N_0}{16} = N_0\left(\frac12\right)^4 \), so 4t equals four half-lives and the half-life is t
2. At 5t there are 5 half-lives: \( N = \frac{N_0}{2^5} = \frac{N_0}{32} \). Answer: (c)

Teacher's Note:
1. Find the half-life from the given fall in the counting rate.
2. The counting rate is proportional to the number of active nuclei.

Answer: (c) \( \sqrt2 : 1 \)

Explanation:
1. After 2 h: \( N_A = \frac{N_0}{2} \) and \( N_B = \frac{N_0}{\sqrt2} \)
2. Decay constants \( \lambda_A = 2\lambda_B \)
3. \( \frac{R_A}{R_B} = \frac{\lambda_A N_A}{\lambda_B N_B} = 2\cdot\frac{1/2}{1/\sqrt2} = \sqrt2 \). Answer: (c)

Teacher's Note:
1. Activity equals decay constant times the number of nuclei left.
2. Half-life is inversely proportional to the decay constant.

Answer: (c) 63%

Explanation:
1. After one mean life \( t = \frac1\lambda \), \( N = N_0e^{-1} = 0.37N_0 \)
2. The fraction decayed \( = 1 - 0.37 = 0.63 \), that is 63%. Answer: (c)

Teacher's Note:
1. Mean life is 1/lambda.
2. 37% remain, so 63% decay.

Answer: (d) \( \frac{1}{9\lambda} \)

Explanation:
1. \( \frac{N_1}{N_2} = \frac{N_0e^{-10\lambda t}}{N_0e^{-\lambda t}} = e^{-9\lambda t} \)
2. Setting this equal to \( e^{-1} \) gives \( 9\lambda t = 1 \), so \( t = \frac{1}{9\lambda} \). Answer: (d)

Teacher's Note:
1. Divide the two exponential decay laws.
2. Equate the exponents.

Answer: (c) Curve (c)

Explanation:
1. The rate of formation of Y equals the rate of decay of X
2. \( R = \lambda N = \lambda N_0e^{-\lambda t} \)
3. This is a maximum at \( t = 0 \) and falls exponentially. Answer: (c)

Teacher's Note:
1. The product forms at the same rate as the parent decays.
2. Activity decreases exponentially with time.

Answer: (d) sometimes equal to and sometimes more than its atomic number

Explanation:
1. Mass number = number of protons + number of neutrons
2. It equals the atomic number only for ordinary hydrogen (\( {}_1^1H \)), which has no neutron
3. For all other nuclei it is larger. Answer: (d)

Teacher's Note:
1. Mass number counts protons and neutrons.
2. Hydrogen-1 is the only nucleus without a neutron.

Answer: (a) 1, 3

Explanation:
1. Alpha emission: mass number \( A-4 \), atomic number \( Z-2 \)
2. Beta emission: atomic number rises by 1, mass number unchanged
3. Daughter: \( {}_{Z-1}^{A-4}Y \)
4. Neutrons \( = (A-4) - (Z-1) = A-Z-3 \)
5. So statements (1) and (3) are correct. Answer: (a)

Teacher's Note:
1. Track mass number and atomic number separately.
2. Beta emission converts a neutron into a proton.

Answer: (d) \( \rho \)

Explanation:
1. Nuclear density is the same for all nuclei, because \( R = R_0A^{1/3} \) and so the volume is proportional to the mass number
2. The density of the smaller nucleus is \( \rho \). Answer: (d)

Teacher's Note:
1. Nuclear matter has nearly constant density, independent of the mass number.
2. Volume scales with the number of nucleons.

Answer: (c) \( M_2 < 2M_1 \)

Explanation:
1. Calcium-40 is made of exactly twice the nucleons of neon-20
2. The binding energy per nucleon is greater for Ca-40, so its mass defect is more than double that of Ne-20
3. Hence \( M_2 < 2M_1 \)
4. Also \( M_1 < 10(m_n+m_p) \) because of the mass defect, so (d) is wrong. Answer: (c)

Teacher's Note:
1. Mass of a nucleus is less than the sum of its nucleons by the mass defect.
2. Larger nuclei in this range are more tightly bound per nucleon.

Answer: (b) 100

Explanation:
1. \( K_\alpha = Q\cdot\frac{A-4}{A} \)
2. \( 48 = 50\cdot\frac{A-4}{A} \)
3. \( 48A = 50A - 200 \), so \( A = 100 \). Answer: (b)

Teacher's Note:
1. The energy is shared in the inverse ratio of the masses of the products.
2. The lighter alpha particle gets the larger share.

Answer: (c) 128 μCi

Explanation:
1. 60 hours = 10 half-lives
2. Initial activity \( = 0.125\times2^{10} = 0.125\times1024 = 128 \) μCi. Answer: (c)

Teacher's Note:
1. Divide the time by the half-life to get the number of half-lives.
2. Multiply by 2 for each half-life.