Radioactive decay of Potassium

JEE Advanced 2019 Paper 1, Question 4

In a radioactive sample, ${}^{40}_{19}{\rm K}$ nuclei decay into stable ${}^{40}_{20}{\rm Ca}$ nuclei with decay constant $4.5 \times 10^{-10}$ per year or into stable ${}^{40}_{18}{\rm Ar}$ nuclei with decay constant $0.5 \times 10^{-10}$ per year. In this sample all the stable ${}^{40}_{20}{\rm Ca}$ and ${}^{40}_{18}{\rm Ar}$ nuclei are produced by the ${}^{40}_{19}{\rm K}$ nuclei only. In time $t_1 \times 10^9$ years, if the ratio of the sum of stable ${}^{40}_{20}{\rm Ca}$ and ${}^{40}_{18}{\rm Ar}$ nuclei to the radioactive ${}^{40}_{19}{\rm K}$ nuclei is 99, the value of $t_1$ will be [Given $\ln 10 = 2.3$ ],

1.15
9.2
2.3
4.6

Solution

Let us first consider the situation where a sample of $X$ nuclei decays into a single type of nuclei $Y$ . The number $N$ of $X$ nuclei changes with time according to the equation

(1) $\begin{equation*} \frac{d N}{d t} = - \lambda_{\rm X \to Y} \cdot N \end{equation*}$

where $\lambda_{\rm X \to Y}$ is the decay constant. It is the inverse of the average lifetime of the nuclei $X$ before it decays into $Y$ . So, $\lambda_{\rm X \to Y} \cdot dt$ is the probability that a single nuclei decays in the small time interval $dt$ . When we multiply this with the total number of $X$ nuclei available we get $dN$ , the change in the number of $X$ nuclei in that that time. The negative sign in the RHS accounts for the fact that $N$ decreases over time.

Coming to the question at hand, we’re told that the ${}^{40}_{19}{\rm K}$ nuclei can decay into ${}^{40}_{20}{\rm Ca}$ or ${}^{40}_{18}{\rm Ar}$ . Let us denote the number of ${}^{40}_{19}{\rm K}$ nuclei as $N$ . Reasoning as we did above we can see that the probability for the two decays are $\lambda_{\rm K \to Ca} \cdot dt$ and $\lambda_{\rm K \to Ar} \cdot dt$ . So the total probability that a ${}^{40}_{19}{\rm K}$ nuclei decays in time $dt$ is obtained by adding these two probabilities. Therefore, the change in $N$ over a small time interval $dt$ must be

(2) $\begin{equation*} dN = - (\lambda_{\rm K \to Ca} + \lambda_{\rm K \to Ar}) dt \times N . \end{equation*}$

That is,

(3) $\begin{equation*} \frac{d N}{d t} = - (\lambda_{\rm K \to Ca} + \lambda_{\rm K \to Ar}) \cdot N \end{equation*}$

which has the solution

(4) $\begin{equation*} N(t) = N_0 e^{- \lambda_{\rm eff} t} \end{equation*}$

where $\lambda_{\rm eff} = \lambda_{\rm K \to Ca} + \lambda_{\rm K \to Ar} = 5 \times 10^{-10}$ per year and $N_0$ is the initial number of ${}^{40}_{19}{\rm K}$ nuclei at time $t=0$ . In the last line of the problem we’re told that

$\frac{N_0 - N}{N} = 99 \quad \implies \quad \frac{N_0}{N} = 100$

at time $t = t_1 \times 10^9$ years. Substituting this into (4) and simplifying,

(5) $\begin{align*} \lambda_{\rm eff} t &= \ln \left( \frac{N_0}{N} \right) \\[1em] 5 \times 10^{-10} {\rm yr}^{-1} \cdot t_1 \times 10^9 {\rm yrs} &= \ln( 100 ) \\[1em] \implies \boxed{t_1 = 9.2} \end{align*}$

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