Question

The half-life of a radioactive sample is $T$. If the activities of the sample at time ${{t}_{1}}$​ and ${{t}_{2}}$ $\left( {{t}_{1}}<{{t}_{2}} \right)$ are ${{R}_{1}}$ and ${{R}_{2}}$, then the number of atoms disintegrated in time \[{{t}_{2}}-{{t}_{1}}\] is proportional to:
$\begin{align}
  & \text{A}\text{. }\left( {{R}_{1}}-{{R}_{2}} \right)T \\
 & \text{B}\text{. }\left( {{R}_{1}}+{{R}_{2}} \right)T \\
 & \text{C}\text{. }\left( \dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)T \\
 & \text{D}\text{. }\dfrac{{{R}_{1}}+{{R}_{2}}}{T} \\
\end{align}$

Answer 1

Hint: The process by which an unstable atomic nucleus loses energy by radiation is called radioactive decay. The radioactive decays per unit time are directly proportional to the number of nuclei of radioactive compounds in the sample. We will use the expression for radioactive decay relating the activity of a sample to the half-life of the sample and the decay constant.

Complete step-by-step solution:
Radioactive decay is described as the process by which an unstable atomic nucleus loses energy by radiation. A sample material containing radioactive nuclei is considered as radioactive.
The decay of radioactive elements occurs at a fixed constant rate. The half-life of a radioisotope is the time required for one half of the concentration of the unstable substances to degrade into a more stable material. We can say that half-life is the time required for a radioactive sample to reduce to half of its initial value. The half-life of a radioactive sample is represented by ${{T}_{(\dfrac{1}{2})}}$.
The decay constant is described as the proportionality between the size of the population of radioactive atoms in a sample and the rate at which the population decreases as a result of radioactive decay. The decay constant is represented by the symbol $\lambda $.
The number of nuclei decayed in particular time being is given as,
$N={{N}_{o}}-{{N}_{o}}{{e}^{-\lambda t}}$
Where,
$N$is the number of decayed nuclei
${{N}_{o}}$ is the number of initial nuclei
$\lambda $ is the decay constant
$t$ is the time
The activity of a radioactive sample is described as the rate at which radioactive particles are emitted. Activity is usually represented by the symbol $A$.
For a number of nuclei $N$, in a sample of a radioactive nuclide, the activity $A$ is related to the decay constant $\lambda $, which is the probability of decay per nucleus per unit time.
If, during time $\Delta t$, $N$ changes by $\Delta N$, then the activity $A$, is the rate of decay and given as,
$A=-\dfrac{\Delta N}{\Delta t}=\lambda N$
Activity of a radioactive sample is given as,
$A=\lambda N$
Where,
$\lambda $ is the decay constant
$N$ is the number of disintegrated nuclei
Activity at time ${{t}_{1}}$ is,
${{R}_{1}}=\lambda {{N}_{1}}$
Activity at time ${{t}_{2}}$ is,
${{R}_{2}}=\lambda {{N}_{2}}$
The number of disintegrated nuclei in time \[{{t}_{2}}-{{t}_{1}}\] is given as,
${{N}_{1}}-{{N}_{2}}=\dfrac{{{R}_{1}}}{\lambda }-\dfrac{{{R}_{2}}}{\lambda }$
Half-life of sample is given as,
${{T}_{(\dfrac{1}{2})}}=\dfrac{\ln 2}{\lambda }$
Therefore,
$\dfrac{1}{\lambda }=\dfrac{{{T}_{(\dfrac{1}{2})}}}{\ln 2}$
Half-life of given sample is $T$
$\dfrac{1}{\lambda }=\dfrac{T}{\ln 2}$
Thus,
${{N}_{1}}-{{N}_{2}}=\left( {{R}_{1}}-{{R}_{2}} \right)\dfrac{T}{\ln 2}$
The value of $\ln 2$ is constant.
${{N}_{1}}-{{N}_{2}}\propto \left( {{R}_{1}}-{{R}_{2}} \right)T$
The number of atoms disintegrated in time \[{{t}_{2}}-{{t}_{1}}\] is proportional to $\left( {{R}_{1}}-{{R}_{2}} \right)T$
Hence, the correct option is A.

Note: The activity of a radioactive sample is expressed by the number of disintegrations taking place at its core at any given moment of time. The activity also represents the number of radiations emitted by the sample of radioactive matter. The activity of a radioactive sample is inversely proportional to the half-life of the sample. The longer the half-life of a sample, the lower will be its activity.