Question

A parent nucleus $X$ is decaying into daughter nucleus $Y$ which in turn decays to $Z$. The half lives of $X$ and $Y$ are $40000$ years and $20$ years respectively. In a certain sample, it is found that the number of $Y$ nuclei hardly changes with time. If the number of $X$ nuclei in the sample is $4\times {10}^{20}$, the number of $Y$ nuclei present in it is

• A. $2\times {10}^{17}$
• B. $2\times {10}^{20}$
• C. $4\times {10}^{23}$
• D. $4\times {10}^{20}$

Answer 1

Answer: (A)

$2\times {10}^{17}$

For radioactive equilibrium

Rate of decay of y=rate of decay of x

so $\frac{N_x}{T_x}=\frac{N_y}{T_y}$

So $\frac{4\times {10}^{20}}{40000}=\frac{N_y}{20}$

$N_y=2\times {10}^{17}$