Question

Decay constants of two radio-active samples A and B are 15x and 3x respectively. They have equal numbers of initial nuclei. The nation of the number of nuclei left in A and B after a time $ \dfrac{1}{6\text{x}} $ is
(A) e
(B) $ {{\text{e}}^{2}} $
(C) $ {{\text{e}}^{-1}} $
(D) $ {{\text{e}}^{-2}} $

Answer 1

Hint: According to the radioactive decay law, when a radioactive material undergoes either $ \alpha $ or $ \beta $ and $ \text{ }\!\!\lambda\!\!\text{ } $ decay, the no of nuclear undergoing the decay per unit time, is proportional to the total number of nuclei in the given sample material.
If N=number of nuclei in a sample and $ \vartriangle $ N=number of radioactive decays per unit time $ \vartriangle $ t then,
 $ \begin{align}
 & \vartriangle \text{N}\vartriangle \text{t }\!\!\alpha\!\!\text{ N} \\
 & \vartriangle \text{N}\vartriangle \text{t}=\text{ }\!\!\lambda\!\!\text{ N} \\
\end{align} $
 $ \text{ }\!\!\lambda\!\!\text{ } $ =constant of proportionality
After doing integration and putting limits this formula becomes
$ {{\text{N}}_{\text{t}}}={{\text{N}}_{\text{o}}}{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} $
This type of decay is exponential.

Complete step by step solution
We have two samples A and B. Decay constant of thses two samples are 15x and 3x respectively
i.e.
 $ \begin{align}
  & {{\text{ }\!\!\lambda\!\!\text{ }}_{\text{A}}}=15\text{x} \\
 & {{\text{ }\!\!\lambda\!\!\text{ }}_{\text{B}}}=3\text{x} \\
\end{align} $
We have to find out the ratio of the number of nuclei left in A and B after a time $ \dfrac{1}{6\text{x}} $
To find out no. of nuclei of a sample left we will use. Radioactive decay law
$ \text{N}={{\text{N}}_{\text{o}}}{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} $
Where $ {{\text{N}}_{\text{o}}} $ =initial no of nuclei
For sample A
$ {{\text{N}}_{\text{A}}}={{\text{N}}_{\text{o}}}{{\text{e}}^{-{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{A}}}\text{t}}} $
For sample B
 $ {{\text{N}}_{\text{B}}}={{\text{N}}_{\text{o}}}{{\text{e}}^{-{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{B}}}\text{t}}} $
The ratio of two sample A and sample B
 $ \dfrac{{{\text{N}}_{\text{A}}}}{{{\text{N}}_{\text{B}}}}=\dfrac{{{\text{N}}_{\text{o}}}{{\text{e}}^{-{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{A}}}\text{t}}}}{{{\text{N}}_{\text{o}}}{{\text{e}}^{-{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{b}}}\text{t}}}}={{\text{e}}^{-}}\left( {{\text{ }\!\!\lambda\!\!\text{ }}_{\text{A}}}-{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{B}}} \right)\text{t} $
Where $ {{\text{ }\!\!\lambda\!\!\text{ }}_{\text{A}}}-{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{B}}}=15\text{x}-3\text{x}=12\text{x} $
 $ \dfrac{{{\text{N}}_{\text{A}}}}{{{\text{N}}_{\text{B}}}}={{\text{e}}^{-12\text{x t}}} $
And t=6x
Put the value of ‘t’
 $ \begin{align}
  & \dfrac{{{\text{N}}_{\text{A}}}}{{{\text{N}}_{\text{B}}}}={{\text{e}}^{-}}\dfrac{12\text{x}}{\text{6x}} \\
 & ={{\text{e}}^{-2}} \\
\end{align} $
Option (D) is correct.

Note
Radioactive atoms have unstable nuclei, and when the nuclei emit radial, they become more stable. It is spontaneous breakdown of an atomic nucleus radioactive sources are based to study living organisms, to diagnose and treat diseases, to sterilize medical instruments and food, to produce energy heat and electric power etc.