JEE Physics Nuclei MCQs Set 25

Answer: (d) X and Z are isotopes

Explanation:
1. Alpha emission lowers Z by 2 and A by 4. Two beta-minus emissions raise Z by 2 and leave A unchanged.
2. So Z has the same atomic number as X but mass number 4 less: X and Z are isotopes.

Teacher's Note:
1. Isotopes share atomic number, isobars share mass number. Track Z and A separately for each step.

Answer: (d) 8 and 6

Explanation:
1. Mass number falls by 32, and each alpha removes 4: \( 32/4=8 \) alpha particles.
2. Eight alphas lower Z by 16 but Z falls only by 10, so 6 beta-minus particles (each raises Z by 1) are emitted.

Teacher's Note:
1. First find alphas from the change in mass number. Betas then correct the atomic number.

Answer: (d) Neutrons

Explanation:
1. Natural radioactivity emits alpha (helium nuclei), beta (electrons) and gamma (electromagnetic) radiation.
2. Neutrons are not emitted in natural radioactive decay.

Teacher's Note:
1. Memorise the three natural emissions: alpha, beta and gamma. Neutron emission occurs in nuclear reactions.

Answer: (c) energy is released due to de-excitation of a nucleus

Explanation:
1. A nucleus left in an excited state after alpha or beta decay drops to a lower level.
2. The energy difference leaves as a gamma photon.

Teacher's Note:
1. Gamma rays come from nuclear energy level transitions, just as visible light comes from atomic ones. The proton number and mass number do not change.

Answer: (d) \( {}^{1}_{0}n \)

Explanation:
1. Mass number: 27 + 4 = 30 + A, so A = 1.
2. Charge: 13 + 2 = 15 + Z, so Z = 0.
3. A particle with A = 1 and Z = 0 is a neutron.

Teacher's Note:
1. Balance both the mass number and the charge. This was the reaction in which artificial radioactivity was discovered.

Answer: (c) \( 7\times10^5 \)

Explanation:
1. \( N=N_0\left(\tfrac12\right)^{t/T}=10^6\times2^{-1/2} \).
2. \( =10^6\times0.707\approx7\times10^5 \).

Teacher's Note:
1. Half the time of a half-life leaves about 71 percent. It is the reciprocal of root 2.

Answer: (a) 0.5

Explanation:
1. In 30 minutes P completes 3 half-lives: fraction \( (1/2)^3=1/8 \).
2. Q completes 2 half-lives: fraction \( (1/2)^2=1/4 \).
3. Ratio = (1/8)/(1/4) = 0.5.

Teacher's Note:
1. Count the number of half-lives for each isotope separately. Then compare the surviving fractions.

Answer: (c) \( \dfrac{e-1}{e} \)

Explanation:
1. After one mean life \( t=1/\lambda \), \( N=N_0e^{-1} \).
2. Fraction decayed = \( 1-\dfrac1e=\dfrac{e-1}{e} \) (about 63 percent).

Teacher's Note:
1. In one mean life the sample falls to 1/e of its original size. About 63 percent has decayed by then.

Answer: (d) \( \dfrac1{9\lambda} \)

Explanation:
1. \( \dfrac{N_1}{N_2}=\dfrac{e^{-10\lambda t}}{e^{-\lambda t}}=e^{-9\lambda t} \).
2. Setting this equal to \( e^{-1} \) gives \( t=\dfrac1{9\lambda} \).

Teacher's Note:
1. Divide the two exponentials and the decay constants subtract. Equate the remaining exponent to minus one.

Answer: (a) 16.425

Explanation:
1. \( \dfrac{N}{N_0}=\dfrac1{20}=2^{-t/T}\Rightarrow t=T\log_2 20 \).
2. \( \log_220=4.322\Rightarrow t=3.8\times4.322=16.4 \) days.

Teacher's Note:
1. Write the surviving fraction as a power of two. log base 2 of 20 is a little over 4.3.

Answer: (b) 32 min

Explanation:
1. The rate halves between 4 min and 36 min.
2. So one half-life = 36 - 4 = 32 min.

Teacher's Note:
1. When the count rate halves, exactly one half-life has passed. The elapsed time between the two readings is the half-life.

Answer: (a) 20 min

Explanation:
1. At 33 percent decay, 67 percent remains. At 67 percent decay, 33 percent remains.
2. The amount has halved (67 to 33.5 is nearly a factor of 2), so the time is about one half-life, 20 min.

Teacher's Note:
1. The two remaining fractions are about 2 : 1. Whenever the ratio is 2, one half-life has elapsed.

Answer: (b) 200

Explanation:
1. 1600 to 100 is a factor of 16 = \( 2^4 \), so 4 half-lives in 8 s and \( T=2 \) s.
2. At 6 s, 3 half-lives have passed: \( 1600/2^3=200 \) counts/s.

Teacher's Note:
1. Find the half-life from the total fall first. Then count the half-lives up to the time asked.

Answer: (a) \( \dfrac{N_0}{2^8} \)

Explanation:
1. 0 to 0.5 hr (A only): one half-life, factor 1/2.
2. 0.5 to 1.5 hr (B only): 1 hr = 4 half-lives of 1/4 hr, factor 1/16.
3. 1.5 to 2 hr (both): decay constants add, so effective half-life \( T=\dfrac{(1/2)(1/4)}{1/2+1/4}=\dfrac16 \) hr; 0.5 hr = 3 half-lives, factor 1/8.
4. Total: \( \tfrac12\cdot\tfrac1{16}\cdot\tfrac18=\dfrac1{256}=2^{-8} \).

Teacher's Note:
1. When two processes act together their decay constants add. Handle each time interval separately and multiply the factors.

Answer: (a) 1 : 2 : 3

Explanation:
1. All lumps are the same substance, so they have the same decay constant.
2. Activity \( A=A_0e^{-\lambda t} \) falls by the same factor for all, so the ratio stays 1 : 2 : 3.

Teacher's Note:
1. A common decay constant means every sample shrinks by the same factor. Ratios are therefore preserved.

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

Explanation:
1. Radioactive decay is random; half-life is a statistical property of a large number of nuclei.
2. Any single nucleus may decay at any instant, and some will survive beyond 16 days.

Teacher's Note:
1. Half-life describes the behaviour of a large population, not of one nucleus. A single nucleus has no fixed lifetime.

Answer: (a) 4

Explanation:
1. \( A=\lambda N \). With \( N_1=2N_2 \): \( \lambda_1=\dfrac{A_1}{2N_2},\ \lambda_2=\dfrac{2A_1}{N_2} \).
2. \( \lambda_1/\lambda_2=1/4 \), and since \( T\propto1/\lambda \), \( T_1/T_2=4 \).

Teacher's Note:
1. Activity equals decay constant times number of nuclei. Half-life varies inversely with decay constant.

Answer: (b) 1.50

Explanation:
1. \( R=R_0A^{1/3} \), so \( R_1/R_2=(135/40)^{1/3}=(3.375)^{1/3}=1.5 \).

Teacher's Note:
1. Nuclear radius varies as the cube root of mass number. 135/40 is 27/8, a perfect cube ratio.

Answer: (b) \( 1:2^{1/3} \)

Explanation:
1. Momentum conservation: \( m_1v_1=m_2v_2 \), so \( m_1:m_2=1:2 \) (the faster part is lighter).
2. Radius \( \propto A^{1/3}\propto m^{1/3} \), so \( R_1:R_2=1:2^{1/3} \).

Teacher's Note:
1. Equal and opposite momenta mean the speeds are inversely proportional to the masses. Mass number is proportional to mass.

Answer: (d) All of these

Explanation:
1. Nuclear (strong) force acts between nucleons only: proton-proton, neutron-neutron and proton-neutron.
2. An electron is a lepton and does not feel the strong force, so none of the pairs listed experiences it.

Teacher's Note:
1. The strong force acts on nucleons, not on electrons. Electrons take part in electromagnetic and weak interactions.

Answer: (a) \( {}^4\text{He} \)

Explanation:
1. Stability depends on binding energy per nucleon.
2. He-4: 27.37/4 = 6.84 MeV. Li-7: 39.3/7 = 5.61 MeV.
3. He-4 has the larger value, so it is more stable.

Teacher's Note:
1. Never compare total binding energies of nuclei of different size. Divide by the number of nucleons first.

Answer: (d) 8 MeV

Explanation:
1. For most nuclei binding energy per nucleon lies between 7 and 9 MeV, with a maximum near 8.7 MeV for iron.

Teacher's Note:
1. Nuclear energies are millions of times larger than chemical ones. A few eV per atom in chemistry, about 8 MeV per nucleon in nuclei.

Answer: (d) \( 4(x_2-x_1) \)

Explanation:
1. Binding energy of He-4 = \( 4x_2 \); of two deuterons = \( 2\times2x_1=4x_1 \).
2. Q = BE of products - BE of reactants = \( 4x_2-4x_1=4(x_2-x_1) \).

Teacher's Note:
1. Energy released equals the gain in total binding energy. Multiply binding energy per nucleon by the number of nucleons.

Answer: (a) \( E_a-3E_b=Q \)

Explanation:
1. Energy released = binding energy of the product - binding energy of the reactants.
2. \( Q=E_a-3E_b \).

Teacher's Note:
1. Q equals binding energy after minus binding energy before. Each of the three B nuclei brings its own binding energy.