The half-life of ${}^{198}\mathrm{Au}$ is 3 days. If atomic weight of ${}^{198}\mathrm{Au}$ is $198\mathrm{~g}/\mathrm{mol}$ then the activity of 2 mg of ${}^{198}\mathrm{Au}$ is [ in disintegration/ second ] :

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Modern Physics

The half-life of

^{198} Au

is 3 days. If atomic weight of

^{198} Au

198 g / mol

then the activity of 2 mg of

^{198} Au

is [ in disintegration/ second ] :

(A) $6.06 \times 1 0^{18}$
(B) $2.67 \times 1 0^{12}$
(C) $16.18 \times 1 0^{12}$ ✓ Correct
(D) $32.36 \times 1 0^{12}$

Answer: (C)

Solution

The decay constant follows from the half-life:

λ = \frac{ln 2}{T _{1/2}} = \frac{0.693}{3 \times 24 \times 3600} = 2.67 \times 1 0^{- 6} s^{- 1}

and

2

mg of the sample holds

N = \frac{2 \times 1 0 ^{- 3}}{198} \times 6.02 \times 1 0^{23} = 6.08 \times 1 0^{18} nuclei

The activity is the product of the two:

A = λ N = 2.67 \times 1 0^{- 6} \times 6.08 \times 1 0^{18} \approx 16.2 \times 1 0^{12}

so the sample decays at about

16.2 \times 1 0^{12}

disintegrations per second.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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