JEE Physics Nuclei MCQs Set 09

Answer: (a) −4.03 MeV

Explanation:
1. Mass of reactants = 1.007825 + 3.016049 = 4.023874 u
2. Mass of products = 2 × 2.014102 = 4.028204 u
3. Q = (4.023874 − 4.028204) × 931 = −4.03 MeV.

Teacher's Note:
1. Q is negative when products are heavier than reactants.
2. The reaction is endothermic.

Answer: (b) 5.4 MeV

Explanation:
1. Momentum conservation: KE_α ∝ 1/m, so KE_α = Q × (A − 4)/A
2. = 5.5 × 216/220 = 5.4 MeV.

Teacher's Note:
1. The α-particle gets the larger share because it is lighter.
2. The remaining 0.1 MeV goes to the recoiling nucleus.

Answer: (a) 1.35 g

Explanation:
1. Mass of radioactive material remaining: m = m₀e^(−λt) with t = 2/λ
2. m = 10 × e⁻² = 1.35 g.

Teacher's Note:
1. Mean life τ = 1/λ.
2. The decay products are different elements and are not counted.

Answer: (d) \(\frac{A-Z-11}{Z-1}\)

Explanation:
1. Mass number of the product = A − 12
2. Atomic number = Z − 6 + 5 = Z − 1
3. Neutrons = (A − 12) − (Z − 1) = A − Z − 11
4. Ratio n/p = (A − Z − 11)/(Z − 1).

Teacher's Note:
1. Each α removes 2 protons and 2 neutrons; each β⁻ converts a neutron into a proton.
2. Neutron number is always A minus Z of the product.

Answer: (c) 0.461 min⁻¹

Explanation:
1. 975/9750 = 1/10 = e^(−5λ)
2. λ = ln10/5 = 2.303/5 = 0.461 min⁻¹.

Teacher's Note:
1. The activity falls by a factor of 10 in 5 minutes.
2. ln10 = 2.303.

Answer: (c) 1 : 4

Explanation:
1. Number left: N₁ = 2 × (1/2)⁶ = 2/64 and N₂ = 3 × (1/2)⁴ = 3/16 = 12/64
2. Activity = λN with λ ∝ 1/T: A₁ : A₂ = (1/2)(2/64) : (1/3)(12/64) = 1 : 4.

Teacher's Note:
1. Activity depends on both N and λ, so include the half-lives again.
2. λ is inversely proportional to the half-life.

Answer: (a) 34.39%

Explanation:
1. Remaining each month = 0.9, so after 4 months remaining = 0.9⁴ = 0.6561
2. Disintegrated = 1 − 0.6561 = 0.3439 = 34.39%.

Teacher's Note:
1. Do not simply multiply 10% by 4; the decay is exponential.
2. Multiply the remaining fractions.

Answer: (b) 4t/3

Explanation:
1. 7/8 decayed leaves 1/8 = (1/2)³, so t = 3T
2. 15/16 decayed leaves 1/16 = (1/2)⁴, so the time is 4T = 4t/3.

Teacher's Note:
1. Express the fraction left as a power of 1/2.
2. T = t/3.

Answer: (a) 1080

Explanation:
1. λ = λ₁ + λ₂, so 1/T = 1/1620 + 1/810 = 3/1620, giving T = 540 years
2. One-fourth remains after 2 half-lives: 2 × 540 = 1080 years.

Teacher's Note:
1. Decay constants add; half-lives do not.
2. The effective half-life is shorter than either.

Answer: (c) N₀[e^(−λt₁) − e^(−λt₂)]

Explanation:
1. Number present at t₁ = N₀e^(−λt₁), at t₂ = N₀e^(−λt₂)
2. Number decayed = N(t₁) − N(t₂) = N₀[e^(−λt₁) − e^(−λt₂)].

Teacher's Note:
1. The earlier number minus the later number gives the decayed atoms.
2. Option 2 would be negative.

Answer: (c) \(\frac{1}{\sqrt{2}}\)

Explanation:
1. N/N₀ = (1/2)^((T/2)/T) = (1/2)^(1/2) = 1/√2 ≈ 0.707.

Teacher's Note:
1. Half a half-life leaves 70.7%, not 75%.
2. Decay is exponential, not linear.

Answer: (c) 87.5%

Explanation:
1. 90 days = 3 half-lives, remaining = 1/8 = 12.5%
2. Disintegrated = 87.5%.

Teacher's Note:
1. Decayed = 100% − remaining.
2. After 3 half-lives, 7/8 has decayed.

Answer: (c) 33 days

Explanation:
1. (1/2)^(t/10) = 1/10, so t/10 = log₂10 = 3.32
2. t = 33.2 days.

Teacher's Note:
1. 1/10 lies between 3 half-lives (1/8) and 4 half-lives (1/16).
2. So the answer lies between 30 and 40 days.

Answer: (c) 0.693 N/n seconds

Explanation:
1. Activity = n = λN, so λ = n/N
2. T½ = 0.693/λ = 0.693 N/n seconds.

Teacher's Note:
1. Activity is the number of decays per second.
2. T½ = 0.693/λ.

Answer: (b) 6.03 g

Explanation:
1. Remaining = 10 × 2^(−20/15) = 10 × 0.397 = 3.97 g
2. Disintegrated = 10 − 3.97 = 6.03 g.

Teacher's Note:
1. The question asks for the disintegrated mass, not the remaining mass.
2. 20/15 = 4/3 half-lives.

Answer: (b) A₂ = A₁e^((t₁ − t₂)/T)

Explanation:
1. A = A₀e^(−t/T) where T = 1/λ is the mean life
2. A₂/A₁ = e^(−(t₂ − t₁)/T) = e^((t₁ − t₂)/T).

Teacher's Note:
1. Divide the two activity equations to eliminate A₀.
2. Mean life T replaces 1/λ.

Answer: (d) 2.3 × 10⁸ atoms/s

Explanation:
1. λ = 0.693/1.2 × 10⁷ = 5.78 × 10⁻⁸ s⁻¹
2. Decay rate = λN = 5.78 × 10⁻⁸ × 4 × 10¹⁵ = 2.3 × 10⁸ atoms/s.

Teacher's Note:
1. Decay rate = λN.
2. Use T½ in seconds.

Answer: (a) 2 : 1

Explanation:
1. Equal probabilities of emission per second means equal activities: λ_AN_A = λ_BN_B
2. N_A/N_B = λ_B/λ_A = 2/1.

Teacher's Note:
1. The nucleus with the smaller λ must be present in greater number for the same activity.
2. λ_A : λ_B = 1 : 2.

Answer: (c) 62/41

Explanation:
1. Mass number = 238 − 32 = 206; atomic number = 92 − 16 + 6 = 82 (lead)
2. Neutrons = 206 − 82 = 124
3. n/p = 124/82 = 62/41.

Teacher's Note:
1. The end product of the uranium series is Pb-206.
2. Reduce the fraction to its simplest form.

Answer: (b) +1.7 MeV

Explanation:
1. For β⁻ decay, Q = [m(P) − m(S)]c² using atomic masses
2. = (31.97391 − 31.97207) × 931 = 0.00184 × 931 = 1.71 MeV.

Teacher's Note:
1. With atomic masses, the electron masses cancel in β⁻ decay.
2. A positive Q means the decay is energetically allowed.

Answer: (b) 2.13 Ci

Explanation:
1. N = (15 × 10⁻³/90) × 6.02 × 10²³ = 1.0 × 10²⁰
2. Activity = λN = 7.88 × 10⁻¹⁰ × 1.0 × 10²⁰ = 7.9 × 10¹⁰ Bq
3. In curie = 7.9 × 10¹⁰/3.7 × 10¹⁰ = 2.13 Ci.

Teacher's Note:
1. Find N from the mass and molar mass, then use A = λN.
2. Divide by 3.7 × 10¹⁰ to convert Bq to curie.

Answer: (c) \(\frac{2}{\ln 2}\) s

Explanation:
1. Let the fraction left after 2 s be r: first interval decays N(1 − r) = n, next interval decays Nr(1 − r) = 0.5n
2. Dividing, r = 0.5, so 2 s is one half-life: T½ = 2 s
3. Mean life = T½/ln2 = 2/ln2 s.

Teacher's Note:
1. Equal time intervals give equal ratios of remaining nuclei.
2. Mean life = T½/0.693.

Answer: (b) \(\frac{1}{3\lambda}\)

Explanation:
1. N₁/N₂ = e^(−6λt)/e^(−3λt) = e^(−3λt)
2. Setting e^(−3λt) = e⁻¹ gives t = 1/(3λ).

Teacher's Note:
1. The relevant exponent is the difference of the decay constants.
2. Initial numbers are equal and cancel.

Answer: (a) 20

Explanation:
1. 450 decayed leaves 150 = 600/4, i.e. two half-lives
2. t = 2 × 10 = 20 minutes.

Teacher's Note:
1. Find the remaining number, not the decayed number, to count half-lives.
2. 600 → 300 → 150.

Answer: (a) 11.6 days

Explanation:
1. 0.2 = (1/2)^(t/5), so t/5 = log₂5 = 2.32
2. t = 5 × 2.32 = 11.6 days.

Teacher's Note:
1. 20% = 1/5, between 2 half-lives (25%) and 3 half-lives (12.5%).
2. So the answer lies between 10 and 15 days.