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(B) The decay process is $X \xrightarrow{\alpha} Y \xrightarrow{\beta} Z$.Let $N_X$ and $N_Y$ be the number of nuclei of $X$ and $Y$ present at any ti

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$10^{7}$

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(B) The decay process is $X \xrightarrow{\alpha} Y \xrightarrow{\beta} Z$.Let $N_X$ and $N_Y$ be the number of nuclei of $X$ and $Y$ present at any time $t$.The rate of change of the number of daughter nuclei $Y$ is given by: $\frac{dN_Y}{dt} = \lambda_X N_X - \lambda_Y N_Y$.Since the rate of emission of $\beta$-particles $(\lambda_Y N_Y)$ is constant over several months, we have $\frac{dN_Y}{dt} \approx 0$.This implies $\lambda_X N_X = \lambda_Y N_Y$.The rate of emission of $\alpha$-particles is $\lambda_X N_X$, and the rate of emission of $\beta$-particles is $\lambda_Y N_Y$.Given that the rate of $\beta$-emission is $10^{7} / 	ext{h}$, the rate of $\alpha$-emission must also be $10^{7} / 	ext{h}$.Therefore, the number of $\alpha$-particles emitted in one hour is $10^{7}$.

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