A heavy nucleus Q of half-life 20 minutes undergoes alpha-decay with probability of 60% and beta-decay with probability of 40%. Initially, the number of Q nuclei is 1000. The number of alpha-decays of Q in the first one hour is

The half-life of nucleus $$Q$$ is given as $$T_{1/2}=20\text{ min}$$, independent of the decay mode.

First convert the half-life to the decay constant. For radioactive decay,
$$\lambda=\frac{\ln 2}{T_{1/2}}$$ $$-(1)$$

Substituting $$T_{1/2}=20\text{ min}$$ into $$(1)$$:
$$\lambda = \frac{\ln 2}{20} = \frac{0.693}{20}\,\text{min}^{-1}$$
$$\lambda \approx 0.03465\;\text{min}^{-1}$$

Let the initial number of $$Q$$ nuclei be $$N_0 = 1000$$. After time $$t$$, the number of undecayed nuclei is
$$N(t)=N_0\,e^{-\lambda t}$$ $$-(2)$$

The time interval of interest is one hour: $$t = 60\text{ min}$$. Using $$(2)$$:
$$N(60)=1000\,e^{-0.03465 \times 60}=1000\,e^{-2.079}$$
$$e^{-2.079}\approx 0.125$$
$$N(60)\approx 1000 \times 0.125 = 125$$

Therefore, the total number of nuclei that have decayed in the first hour is
$$\text{Total decays}=N_0-N(60)=1000-125=875$$

Each decay is either an $$\alpha$$-decay (probability $$0.60$$) or a $$\beta$$-decay (probability $$0.40$$). Hence, the expected number of $$\alpha$$-decays is
$$N_\alpha = 0.60 \times 875 = 525$$

Thus, the number of $$\alpha$$-decays of $$Q$$ in the first one hour is 525.

Option D which is: 525