Question

A radioactive nuclei 'A' starts decaying from time t = 0 following law of radioactive decay. It decays in two ways. In first case, $A \rightarrow 2B + 3C$ where B and C are stable daughter nuclei. In second case, $A \rightarrow B + D$ where B and D are stable daughter nuclei. Initially only A is present. Decay constant for the two decays are $4hr^{-1}$ and $2hr^{-1}$ respectively. Find the ratio of total number of nuclei to the total number of radioactive nuclei at time $t = \frac{ln(2)}{3}hour$.

Answer 1

13

Answer 2

Let the initial number of A nuclei be $N_A(0)$.

1. Decay Law:
   $N_A(t) = N_A(0)e^{-(\lambda_1+\lambda_2)t} = N_A(0)e^{-6t}$.

2. Total decayed nuclei:
   $N_{\text{decayed}} = N_A(0)\left(1-e^{-6t}\right)$.

3. Branching Ratios:

   • For $A\to 2B+3C$ (mode 1): probability $= \frac{\lambda_1}{\lambda_1+\lambda_2} = \frac{4}{6} = \frac{2}{3}$.
   • For $A\to B+D$ (mode 2): probability $= \frac{2}{6} = \frac{1}{3}$.

4. Daughter nuclei produced:

   • Mode 1 gives 5 nuclei per decay (2B + 3C):
     $N_{\text{daughters,1}} = \frac{2}{3}N_A(0)\left(1-e^{-6t}\right)\times 5$.
   • Mode 2 gives 2 nuclei per decay (B + D):
     $N_{\text{daughters,2}} = \frac{1}{3}N_A(0)\left(1-e^{-6t}\right)\times 2$.

   Total daughters:

   $$N_{\text{daughters}} = N_A(0)\left(1-e^{-6t}\right)\left[\frac{2}{3}\times 5 + \frac{1}{3}\times 2\right] = N_A(0)\left(1-e^{-6t}\right)\left(\frac{10+2}{3}\right) = 4N_A(0)\left(1-e^{-6t}\right).$$

5. Total number of nuclei at time $t$:

   $$N_{\text{total}} = N_A(t) + N_{\text{daughters}} = N_A(0)e^{-6t} + 4N_A(0)\left(1-e^{-6t}\right).$$

6. Ratio (Total nuclei)/(Radioactive nuclei):

   $$\frac{N_{\text{total}}}{N_A(t)} = \frac{e^{-6t} + 4\left(1-e^{-6t}\right)}{e^{-6t}} = 4e^{6t} - 3.$$

7. Substitute $t = \frac{\ln 2}{3}$:
   Here, $6t = 6\left(\frac{\ln 2}{3}\right) = 2\ln2 = \ln 4$.
   Thus,

   $$\frac{N_{\text{total}}}{N_A(t)} = 4e^{\ln 4} - 3 = 4\times 4 - 3 = 16-3 = 13.$$

The ratio is 13.