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 $_{18}^{40} \mathrm{Ar}$ nuclei with decay constant $0.5 \times 10^{-10}$ per year. Given that in this sample all the stable $_{20}^{40} \mathrm{Ca}$ and $_{18}^{40} \mathrm{Ar}$ nuclei are produced by the $_{19}^{40} \mathrm{~K}$ nuclei only. In time $\mathrm{t} \times 10^{9}$ years, if the ratio of the sum of stable $_{20}^{40} \mathrm{Ca}$ and $_{18}^{40} \mathrm{Ar}$ nuclei to the radioactive $_{19}^{40} \mathrm{~K}$ nuclei is 99, the value of t will be : [Given $\ln 10=2.3$]

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Modern Physics

In a radioactive sample,

_{19}^{40} K

nuclei either decay into stable

_{20}^{40} Ca

nuclei with decay constant

4.5 \times 1 0^{- 10}

per year or into stable

_{18}^{40} Ar

nuclei with decay constant

0.5 \times 1 0^{- 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 1 0^{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

ln 10 = 2.3

]

Answer: (A)

This is a parallel radioactive decay with effective decay constant:

λ = λ_{1} + λ_{2} = 4.5 \times 1 0^{- 10} + 0.5 \times 1 0^{- 10} = 5 \times 1 0^{- 10} per year

The number of radioactive nuclei:

N = N_{0} e^{- λ t}

The number of stable nuclei:

N_{s t ab l e} = N_{0} - N

Given ratio:

\frac{N _{s t ab l e}}{N} = \frac{N _{0} - N}{N} = \frac{N _{0}}{N} - 1 = 99

\frac{N _{0}}{N} = 100

e^{λ t} = 100

λ t = ln 100 = 2 ln 10 = 2 \times 2.3 = 4.6

t = \frac{4.6}{5 \times 1 0 ^{- 10}} = 9.2 \times 1 0^{9} years

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Single correct option