In a radioactive sample, nuclei either decay into stable nuclei with decay constant per year or into stable nuclei with decay constant per year. Given that in this sample all the stable and nuclei are produced by the nuclei only. In time years, i

Question

In a radioactive sample, ${}_{19}^{40}K$ nuclei either decay into stable ${}_{20}^{40}{Ca}$ 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 ${}_{20}^{40}{Ca}$ and ${}_{18}^{40}{Ar}$ nuclei are produced by the ${}_{19}^{40}K$ nuclei only. In time $t \times 10^{9}$ years, if the ratio of the sum of stable ${}_{20}^{40}{Ca}$ and ${}_{18}^{40}{Ar}$ nuclei to the radioactive ${}_{19}^{40}K$ nuclei is 99, the value of $t$ will be [Given: $l n 10 = 2.3$ ] [2019]

Smart Achievers · Accepted Answer

(1)

Here $- \frac{d N}{d t} = λ_{1} N + λ_{2} N$

In integrating on both sides,

$t = \frac{2.303}{λ_{1} + λ_{2}} \log_{10} (\frac{N_{0}}{N})$ $\Rightarrow t = \frac{2.303}{5 \times 10^{- 10}} \log_{10} (\frac{100}{1})$

$[∵ λ_{1} + λ_{2} = 4.5 \times 10^{- 10} + 0.5 \times 10^{- 10} and \frac{N_{0}}{N} = 100]$

$∴$ $t = 9.2 \times 10^{9} year$

Solution

1 9.2

2 4.6

3 1.15

4 2.3

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