Question

Two radioactive substances A and B have decay constant $5\lambda$ and $\lambda$ respectively. At t=0, they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be $\left( \dfrac{1}{{{e}^{2}}} \right)$ after a time:
A. $4\lambda$
B. $2\lambda$
C. $\dfrac{1}{2\lambda }$
D. $\dfrac{1}{4\lambda }$

Answer 1

Radioactive Decays follow first order chemical kinetics and have a reaction constant or decay constant that remains fixed throughout the decay and is represented by $\lambda$. If the initial number of nuclei is given then the number of nuclei remaining at any time ‘t’ can be found using the formula given below.
Formula Used:
$N={{N}_{o}}{{e}^{-\lambda t}}$

Complete answer:
The decay of any radioactive substance follows first-order chemical kinetics. It has a reaction constant or decay constant that remains fixed throughout the decay and is represented by $\lambda$. Based on the decay constant and the initial number of nuclei in a radioactive decay we can find the number of nuclei remaining. The number of nuclei remaining can be found using the formula
$N={{N}_{o}}{{e}^{-\lambda t}}$
Here, N is the number of remaining nuclei
${{N}_{o}}$ is the number of nuclei taken initially and t is the time .
In the question it is mentioned that both the radioactive substances have the same number of nuclei at t=0, let that number be ${{N}_{o}}$. Their rate constants have also been given so using the above-mentioned formula we can find the number of nuclei remaining of each substance at any time t .
For A the number of nuclei remaining will be
${{N}_{A}}={{N}_{o}}{{e}^{-5\lambda t}}$
For B the number of nuclei remaining will be
${{N}_{B}}={{N}_{o}}{{e}^{-\lambda t}}$
The Ratio of number of nuclei of the A and B at any time t will be
$\dfrac{{{N}_{A}}}{{{N}_{B}}}=\dfrac{{{N}_{o}}{{e}^{-5\lambda t}}}{{{N}_{o}}{{e}^{-\lambda t}}}$
$\Rightarrow \dfrac{{{N}_{A}}}{{{N}_{B}}}={{e}^{-4\lambda t}}$…….(1)
We have to find the time at which
$\dfrac{{{N}_{A}}}{{{N}_{B}}}={{e}^{-2}}$…………(2)
Comparing (1) and (2) we get
$-4\lambda t=-2$
$\Rightarrow t=\dfrac{1}{2\lambda }$
Therefore, the time at which the ratio of the number of nuclei of A to those of B will be $\left( \dfrac{1}{{{e}^{2}}} \right)$ is $t=\dfrac{1}{2\lambda }$

So, Option C is correct.

Note:
The radioactive decay follows first-order kinetics. So, the time taken for the number of nuclei to halve remains constant throughout the reaction or it has a constant half-live, which is an important property of any radioactive decay. This information can be used to solve a number of problems where the number of remaining nuclei or their ratios is concerned.