JEE Main 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.