Question

A parent nucleus X is decaying into daughter nucleus Y which in turn decays into 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 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
$\begin{align}
  & \text{A}\text{. 2}\times \text{1}{{\text{0}}^{17}} \\
 & \text{B}\text{. 2}\times \text{1}{{\text{0}}^{20}} \\
 & \text{C}\text{. 4}\times \text{1}{{\text{0}}^{23}} \\
 & \text{D}\text{. 4}\times \text{1}{{\text{0}}^{20}} \\
\end{align}$

Answer 1

Hint: The elements which have unstable nuclei and radiate energy to become stable are called radioactive nuclei. The decay of a radioactive nuclei depends on the decay constant. Obtain the expression for activity and half-life of a radioactive nuclei.

Complete step by step answer:
The parent nucleus X decays to daughter nucleus Y and the nucleus Y again decays to nucleus Z.
$X\to Y\to Z$
The half life of nucleus X is given as, ${{T}_{\dfrac{1}{2}x}}=40000\text{years}$
Again, the half-life of nucleus Y is given as, ${{T}_{\dfrac{1}{2}y}}=20\text{years}$
Now, the activity of a radioactive nuclei can be defined as the rate of decay per unit time.
We can write activity as,
$A=-\dfrac{dN}{dt}$
Where, A is the activity of the radioactive nuclei and N is the number of undecayed nuclei present in the nucleus.
We can also express the activity of a radioactive nuclei as,
$A=\lambda N$
Where, $\lambda $ is the decay constant.
Now, let ${{\lambda }_{x}}$ be the decay constant for the decay of nucleus X to Y.
Again, let ${{\lambda }_{y}}$ be the decay constant for the decay of nucleus Y to Z.
Since, the activity of a radioactive element is same, we can write that,
$\begin{align}
  & {{\lambda }_{x}}{{N}_{x}}={{\lambda }_{y}}{{N}_{y}} \\
 & {{N}_{y}}=\dfrac{{{\lambda }_{x}}}{{{\lambda }_{y}}}{{N}_{x}} \\
\end{align}$
Now, the half life of a nucleus can be written as,
${{T}_{\dfrac{1}{2}}}=\dfrac{0.6932}{\lambda }$
Using this we can write that,
$\dfrac{{{\lambda }_{x}}}{{{\lambda }_{y}}}=\dfrac{{{T}_{\dfrac{1}{2}y}}}{{{T}_{\dfrac{1}{2}x}}}=\dfrac{20}{40000}=\dfrac{1}{2000}$
Again, number of X nuclei is,
${{N}_{x}}=4\times {{10}^{20}}$ substituting these values on the above equation,
$\begin{align}
  & {{N}_{y}}=\dfrac{1}{2000}\times 4\times {{10}^{20}} \\
 & {{N}_{y}}=2\times {{10}^{17}} \\
\end{align}$
So, the number of Y nuclei present will be $2\times {{10}^{17}}$
The correct option is (A).

Note: The decay rate or the activity of a radioactive nuclei depends on the decay constant. The larger the decay constant, the smaller will be the half-life and the smaller will be the decay constant, the larger will be the half-life. If the half-life of the element is small then the number of daughter nuclei will be more and if the half-life is large, the number of daughter nuclei is very low.