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A parent nucleus $X$ undergoes $\alpha$ -decay with a half-life of 75000 yrs. The daughter nucleus $Y$ undergoes $\beta$ -decay with a half-life of 9 months. In a particular sample, it is found that the rate of emission of $\beta$ -particles is nearly constant (over several months) at $10^{7} / \mathrm{h} .$ What will be the number of $\alpha$ -particles emitted in an hour?
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The correct answer is:
$10^{7}$
According to the question,
$X \stackrel{\text { a-particle }}{\longrightarrow} Y \longrightarrow Z(\text { say) }$
According to Rutherford Soddy law of radioactive decay, the rate of decay of radioactive along at any instant is proportional to the number of atoms present at that instant
$\frac{d N}{d t}=-\lambda N$
since, $\lambda=$ constant (decay constant) and $N=$ number of atoms
Rate of decay of $Y$ particle is given as,
$\lambda_{X} N_{X}-\lambda_{\gamma} N_{Y}=\frac{d N_{Y}}{d t}=0$
[ $\because$ Decay rate for $\beta$ -particle become constant after some time]
Given, rate of emission of $\beta$ -particle $=10^{7} \mathrm{h}$ $\lambda_{X} N_{X}=\lambda_{\gamma} N_{Y}=10^{7} / \mathrm{h}$
$X \stackrel{\text { a-particle }}{\longrightarrow} Y \longrightarrow Z(\text { say) }$
According to Rutherford Soddy law of radioactive decay, the rate of decay of radioactive along at any instant is proportional to the number of atoms present at that instant
$\frac{d N}{d t}=-\lambda N$
since, $\lambda=$ constant (decay constant) and $N=$ number of atoms
Rate of decay of $Y$ particle is given as,
$\lambda_{X} N_{X}-\lambda_{\gamma} N_{Y}=\frac{d N_{Y}}{d t}=0$
[ $\because$ Decay rate for $\beta$ -particle become constant after some time]
Given, rate of emission of $\beta$ -particle $=10^{7} \mathrm{h}$ $\lambda_{X} N_{X}=\lambda_{\gamma} N_{Y}=10^{7} / \mathrm{h}$
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