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 $40000$ yr and $20$ yr, 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)

Decay occurs as
$X \xrightarrow{40000 y r} Y \xrightarrow{20 y r} Z$
As number of $Y$ nuclei does not changes with time,
this means decay rate of $X = $ decay rate of $Y.$
$\lambda_{X} N_{X}=\lambda_{Y}N_{Y}$
$\Rightarrow \frac{N_{X}}{T_{X}}=\frac{N_{Y}}{T_{Y}}$
$\Rightarrow N_{Y}=\frac{T_{Y}}{T_{X}}N_{X}$
$=\frac{20}{40000}\times4\times10^{20}$
$=2\times10^{20} $
$=2\times10^{17}$ nuclei