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

Hint: 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.