JEE Advanced Chemistry: Nuclei questions with solutions

Q1. Two radioactive substances A and B initially have amounts in the ratio 4:1 (i.e., [A]0 = 4x and [B]0 = x). After a certain time, the amounts remaining are equal. If n1 and n2 are the number of half-lives elapsed for A and B respectively during this time, what is the value of (n1 - n2)?

1. 1
2. 2
3. 3
4. 4

Answer: 2

Using the radioactive decay law: after n half-lives, the remaining amount is N0 * (1/2)ⁿ. For A: remaining = 4x * (1/2)ⁿ1 = 4x / 2ⁿ1. For B: remaining = x * (1/2)ⁿ2 = x / 2ⁿ2. Setting them equal: 4x / 2ⁿ1 = x / 2ⁿ2. Dividing both sides by x: 4 / 2ⁿ1 = 1 / 2ⁿ2. Cross-multiplying: 4 * 2ⁿ2 = 2ⁿ1, so 2² * 2ⁿ2 = 2ⁿ1, giving 2^(n1) = 2^(n2 + 2), hence n1 - n2 = 2.

Q2. Gadolinium-153, used to detect osteoporosis, has a half-life of 242 days. Which value is closest to the percentage of Gd-153 remaining in a patient after 2 years?

1. 33%
2. 26%
3. 12.0%
4. 6.25%

Answer: 12.0%

2 years = 2*365 = 730 days. Number of half-lives n = 730/242 ≈ 3.02. Fraction remaining = (1/2)³.02 ≈ (0.5)³ * (0.5)⁰.02 ≈ 0.125 * 0.986 ≈ 0.123, i.e., about 12.3%. The closest option is 12.0%.

Q3. Nuclide x(A)^y (mass number A, atomic number y) and p(B)^q are isobars. Nuclide m(C)ⁿ is an isotone with x(A)^y. Select the correct option.

1. y = n - m
2. n = q - x + m
3. x = p
4. (y - x) = (q - p)

Answer: (y - x) = (q - p)

Standard notation: ^A_y X means mass number A, atomic number y. Isobars (A)^y(X) and (B)^q(p... wait the notation xA^y is confusing. Let us interpret: nuclide 1 has atomic number x, mass number A (so neutrons = A-x), element symbol y is likely the symbol not the atomic number. Let me re-read: 'xA^y' — likely x=atomic number (Z), A=element symbol, y=mass number. Similarly pB^q: p=Z, q=mass number. mCⁿ: m=Z, n=mass number. Isobars xA^y and pB^q: same mass number -> y = q. Isotones xA^y and mCⁿ: same neutron number -> y-x = n-m. From (y-x)=(n-m) and isobar condition y=q: (q-x)=(n-m) -> n = q-x+m. That matches option B. Also (y-x)=(n-m) -> but option D says (y-x)=(q-p): since y=q (isobars), this becomes (q-x)=(q-p) -> x=p, which is option C. So from isobar: y=q. From isotone: y-x = n-m, i.e., neutrons equal: y-x = n-m. Option D: (y-x)=(q-p). Since y=q: q-x = q-p -> x=p. That would mean option C and D are equivalent and either true or false together. Since x and p are atomic numbers of different elements (A and B), x need not equal p. So D is not necessarily true. The isotone condition gives y-x = n-m, which means n = y-x+m = q-x+m (using y=q) -> option B is correct.

Q4. The price of a radioactive substance is directly proportional to its activity. If the initial price of a radioactive sample with half-life of 6 months is Rs 4096, what will be the price after 5 years?

1. Rs 4
2. Rs 8
3. Rs 16
4. Rs 4096

Answer: Rs 4

After 10 half-lives, the activity drops by a factor of 2¹⁰ = 1024. Price = 4096/1024 = Rs 4.

Q5. Select the correct option(s) regarding radioactivity of U²³⁵ and U3O8 (where all uranium is U²³⁵):

1. Specific activity of U²³⁵ is greater than specific activity of U3O8 (All 'U' is U²³⁵)
2. Activity of 1 mol U²³⁵ is greater than activity of 1 mol U3O8 (All 'U' is U²³⁵)
3. Decay constant of U²³⁵ is same whether it is pure U²³⁵ or U3O8 (All 'U' is U²³⁵)
4. Rate of radioactive disintegration increases with increase in temperature

Answer: Decay constant of U²³⁵ is same whether it is pure U²³⁵ or U3O8 (All 'U' is U²³⁵)

The decay constant is an intrinsic nuclear property and does not change with chemical environment or temperature — this statement is correct. Specific activity of U²³⁵ (pure) is higher than U3O8 because U3O8 has more mass per U atom (oxygen contributes mass but no radioactivity). For 1 mol: U3O8 contains 3 mol of U²³⁵ atoms, so its activity is 3 times that of 1 mol U²³⁵. Temperature has no effect on nuclear decay rates.

Q6. The rate of radioactive decay of a radioactive element depends on which of the following?

1. Amount of radioactive element present
2. Temperature
3. Pressure
4. Isotopic form of the radioactive element

Answer: Amount of radioactive element present

The rate of radioactive decay follows the law: dN/dt = -lambda * N, where lambda is the decay constant (unique to each isotope) and N is the number of radioactive atoms present. Temperature and pressure have no effect on nuclear decay. Different isotopes of the same element have different decay constants, but the rate still depends on the amount (N) present.

Q7. A radioactive isotope undergoes decay by emitting 7 alpha particles and 6 beta particles to eventually form a stable product. What is the difference between the atomic number of the original radioactive isotope and that of the stable product?

1. 4
2. 6
3. 8
4. 10

Answer: 8

Seven alpha emissions reduce the atomic number by 14, while six beta emissions increase it by 6, giving a net decrease of 8. The difference in atomic number between parent and daughter is 8.