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Atoms and Nuclei

Section: Physics

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Question : 17 of 53

Marks: +1, -0

In a radioactive sample,

{}_{19}^{40}\mathrm{K}

nuclei either decay into stable

{}_{20}^{40}\mathrm{Ca}

nuclei with decay constant

4.5 \times 10^{-10}

per year or into stable

{}_{18}^{40}

Ar nuclei with decay constant

0.5 \times 10^{-10}

per year. Given that in this sample all the stable

{}_{20}^{40}\mathrm{Ca}

{}_{18}^{40}

Ar nuclei are produced by the

{}_{19}^{40}\mathrm{K}

nuclei only. In time

t \times 10^{9}

years, if the ratio of the sum of stable

{}_{20}^{40}\mathrm{Ca}

{}_{18}^{40}

Ar nuclei to the radioactive

{}_{19}^{40}\mathrm{K}

nuclei is 99, the value of will be : [Given

\ln 10=2.3

[JEE Adv 2019 P1]

4.6
2.3
9.2
1.15

Solution:

\lambda = \lambda_1 + \lambda_2 = 5 \times 10^{-10} \; \text{per year}\;

N = N_0 e^{-\lambda t}

N_0 - N = N_{\text{stable}}

N = N_{\text{radioactive}}

\frac{N_0}{N} - 1 = 99

\frac{N_0}{N} = 100

\frac{N}{N_0} = e^{-\lambda t} = \frac{1}{100}

\Rightarrow \lambda t = 2 \ln 10

=4.6

t = 9.2 \times 10^{9} \; \text{years}\;

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