Question

If the half-life of an element is 69.3 hours then how much of its decay in the 10th to 11th hours. Initial activity = $50\mu Ci$.
(A)- 1%
(B)- 2%
(C)- 3%
(D)- 4%

Answer 1

This question can be solved by the formula \text {%decay}=\left( \dfrac{{{N}_{1}}-{{N}_{2}}}{{{N}_{1}}} \right)\text{ x 100}, where ${{N}_{1}}$ is the number of active nuclei at 10 hours and ${{N}_{2}}$ is the number of active nuclei at 11 hours.

Complete step by step solution:
We know the decay equation is:
$N={{N}_{0}}{{e}^{-\lambda t}}$
N is the number of active nuclei at time t, ${{N}_{0}}$ is the number of active nuclei at the initial time, and $\lambda$ is the decay constant.
At time 10 hours, the t will be 10 and let this equation be noted as ${{N}_{1}}$. This equation for active nuclei at time 10 hours is written as:
${{N}_{1}}={{N}_{0}}{{e}^{-10\lambda }}$
At time 11 hours, the t will be 11 and let this equation be noted as ${{N}_{2}}$. This equation for active nuclei at time 11 hours is written as:
${{N}_{2}}={{N}_{0}}{{e}^{-11\lambda }}$
The percentage of decay can be calculated as:
$\% decay=\left( \dfrac{{{N}_{1}}-{{N}_{2}}}{{{N}_{1}}} \right)\text{ x 100}$
So putting the value of active nuclei at 10 hours and active nuclei at 11 hours will be:
$\%\text{ }decay=\left( \dfrac{{{N}_{0}}{{e}^{-10\lambda }}-{{N}_{0}}{{e}^{-11\lambda }}}{{{N}_{0}}{{e}^{-10\lambda }}} \right)\text{ x 100}$
So this equation becomes:
$\%\text{ }decay=\left( 1-\dfrac{1}{{{e}^{\lambda }}} \right)\text{ x 100}$
We know the formula of decay constant is:
$\lambda =\dfrac{0.693}{{{t}_{1/2}}}$
We are given the value of half-life in the question as 69.3 hrs, so putting the value in the above equation, we get
$\lambda =\dfrac{0.693}{69.3}=0.01$
Putting this value in the percentage equation,
$\%\text{ }decay=\left( 1-\dfrac{1}{{{e}^{0.01}}} \right)\text{ x 100}$
$\%\text{ }decay=0.01\text{ x 100}$
$\%\text{ }decay=1$
So, the percentage of decay is 1.

Therefore, the correct answer is an option (a)- $1\%$.

Note: The value of active nuclei at 10 hours will be more than the active nuclei present at 11 hours. Don't get confused that the initial activity has not been used in the solution.