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. 1.15
  2. 9.2
  3. 2.3
  4. 4.6


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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