The half- life of $\mathrm{Au}^{198}$ is 2.7 days. The activity of $1.50 \mathrm{mg}^{198}$ of $\mathrm{Au}^{198}$ if atomic weight is $198 \mathrm{~g} \mathrm{~mol}^{-1}$ is, $\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23} / \mathrm{mol}\right)$

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Modern Physics

The half- life of

Au^{198}

is 2.7 days. The activity of

1.50 mg^{198}

Au^{198}

if atomic weight is

198 g mol^{- 1}

is,

(N_{A} = 6 \times 1 0^{23} / mol)

(A)240 Ci
(B)252 Ci
(C)535 Ci
(D)357 Ci✓ Correct

Answer: (D)

Solution

The number of nuclei in

1.50 mg

Au^{198}

N = \frac{1.50 \times 1 0 ^{- 3}}{198} \times 6 \times 1 0^{23} = 4.55 \times 1 0^{18}

and the decay constant that goes with a half-life of

2.7

days is

λ = \frac{ln 2}{t _{1/2}} = \frac{0.693}{2.7 \times 86400} = 2.97 \times 1 0^{- 6} s^{- 1}

so the activity is

λ N = 2.97 \times 1 0^{- 6} \times 4.55 \times 1 0^{18} = 1.35 \times 1 0^{13} decays per second

One curie is

3.7 \times 1 0^{10}

decays per second, and dividing gives an activity close to

3.6 \times 1 0^{2}

curie, that is about

357 Ci

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

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