JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Modern Physics

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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

\frac{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.

(A) $3.38 \times 1 0^{24}$ ✓ Correct
(B) $6.76 \times 1 0^{24}$
(C) $6.76 \times 1 0^{23}$
(D) $1.69 \times 1 0^{24}$

Answer: (A)

Solution

Substance A starts with

\frac{320}{16} = 20

mol; after two days, which is two of its one-day half-lives,

\frac{20}{2 ^{2}} = 5

mol remain. Substance B starts with

\frac{320}{32} = 10

mol; after two days, which is four of its half-day half-lives,

\frac{10}{2 ^{4}} = 0.625

mol remain. The total is

5.625

mol, and multiplying by Avogadro's number,

5.625 \times 6.022 \times 1 0^{23} = 3.38 \times 1 0^{24}

atoms are left.

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Modern Physics

Solution

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