JEE Main 2023 · 29 January · evening · Single correct · +4 / −1

Substance A has atomic mass number 16 and half life of 1 day. Another substance B has atomic mass number 32 and half life of 1/2 day. If both A and B simultaneously start undergo radio activity at the same time with initial mass 320 g each, how many total atoms of A and B combined would be left after 2 days.

  1. (a) 1.69×10²⁴
  2. (b) 3.38×10²⁴
  3. (c) 6.76×10²³
  4. (d) 6.76×10²⁴
Answer and solution

Answer (b) 3.38×10²⁴

Solution

Physics sir · 3 steps

Turn grams into moles first, then count half-lives: A goes through 2 of them, B goes through 4.

  1. Grams to moles. Mass number gives the molar mass in grams. So 320 g of A is 320/16 = 20 mol, and 320 g of B is 320/32 = 10 mol.

    N = N₀(1/2)ⁿ, n = t/T_(1/2)

    SubstanceMoles at startHalf-lives in 2 daysFraction leftMoles left
    A202/1 = 21/45
    B102/(1/2) = 41/160.625
  2. Moles to atoms. Total left is 5 + 0.625 = 5.625 mol. Multiply by N_A = 6.022 × 10²³ to get 5.625 × 6.022 × 10²³ ≈ 3.39 × 10²⁴ atoms.
  3. So. About 3.4 × 10²⁴ atoms are left, which matches option (b), 3.38 × 10²⁴, within rounding of N_A.

The idea to keep. Equal masses do not mean equal numbers of atoms. Divide by the molar mass first, then apply (1/2)^(t/T_(1/2)) to each substance with its own half-life.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 110 times in JEE Main since 2020. This question is from 2023.