Calculation of Alpha Decays for Heavy Nucleus in One Hour

To determine the number of $\alpha$-decays of nucleus $Q$ in the first one hour, we follow these steps:

1. Given Data:

• Initial number of $Q$ nuclei, $N_0 = 1000$
• Half-life of nucleus $Q$, $T_{1/2} = 20\text{ minutes}$
• Time duration, $t = 1\text{ hour} = 60\text{ minutes}$
• Probability of $\alpha$-decay, $P_\alpha = 60\% = 0.6$
• Probability of $\beta$-decay, $P_\beta = 40\% = 0.4$

2. Number of half-lives ($n$): $n = \frac{t}{T_{1/2}} = \frac{60\text{ min}}{20\text{ min}} = 3$

3. Number of remaining $Q$ nuclei after 1 hour ($N$): Using the radioactive decay law: $N = N_0 \left(\frac{1}{2}\right)^n = 1000 \times \left(\frac{1}{2}\right)^3 = \frac{1000}{8} = 125$

4. Total number of decayed $Q$ nuclei ($\Delta N$): $\Delta N = N_0 - N = 1000 - 125 = 875$

5. Calculation of $\alpha$-decays: Since $60\%$ of the total decays occur via $\alpha$-decay, the total number of $\alpha$-decays is given by: $N_\alpha = P_\alpha \times \Delta N = 0.6 \times 875 = 525$

Thus, the number of $\alpha$-decays during the first hour is 525.

Correct Option: (D)