Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t = 0, the number of

Question

Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t = 0, the number of P species are 4N_0 and that of Q is N_0. Half-life of P (for conversion to R) is 1 min whereas that of Q is 2 min. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be:

Accepted Answer

1. **Decay Equation:** The number of nuclei remaining $N$ after time $t$ is given by $N = N_{initial} \left(\frac{1}{2}ight)^{t/T_{1/2}}$ .
2. **Find Time ($t$) for Equal Nuclei:**
   - For P: $N_P = 4N_0 \left(\frac{1}{2}ight)^{t/1}$
   - For Q: $N_Q = N_0 \left(\frac{1}{2}ight)^{t/2}$
   - Equating $N_P = N_Q$:
     $$ 4N_0 \left(\frac{1}{2}ight)^{t} = N_0 \left(\frac{1}{2}ight)^{t/2} $$
     $$ 4 = \frac{(1/2)^{t/2}}{(1/2)^t} = \left(\frac{1}{2}ight)^{-t/2} = 2^{t/2} $$
     $$ 2^2 = 2^{t/2} \implies \frac{t}{2} = 2 \implies t = 4 	ext{ min} $$
3. **Calculate Nuclei of R Formed:**
   - Nuclei of R formed = (Decayed P) + (Decayed Q).
   - Remaining P at 4 min: $N_P(4) = 4N_0 (1/2)^4 = 4N_0/16 = N_0/4$.
     - Decayed P = $4N_0 - N_0/4 = 15N_0/4$.
   - Remaining Q at 4 min: $N_Q(4) = N_0 (1/2)^2 = N_0/4$.
     - Decayed Q = $N_0 - N_0/4 = 3N_0/4$.
   - Total R = $\frac{15N_0}{4} + \frac{3N_0}{4} = \frac{18N_0}{4} = \frac{9N_0}{2}$.

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