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(D) The activity of a radioactive sample is given by $R = N \lambda$,where $N$ is the number of radioactive nuclei and $\lambda$ is the decay constant

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$(R_1 - R_2)$

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(D) The activity of a radioactive sample is given by $R = N \lambda$,where $N$ is the number of radioactive nuclei and $\lambda$ is the decay constant.The decay constant $\lambda$ is related to the half-life $T$ by $\lambda = \frac{\ln 2}{T}$.At time $T_1$,the activity is $R_1 = N_1 \lambda$,so $N_1 = \frac{R_1}{\lambda} = \frac{R_1 T}{\ln 2}$.At time $T_2$,the activity is $R_2 = N_2 \lambda$,so $N_2 = \frac{R_2}{\lambda} = \frac{R_2 T}{\ln 2}$.The number of atoms that have disintegrated in the time interval $(T_2 - T_1)$ is the difference in the number of nuclei present at these times: $\Delta N = N_1 - N_2$.Substituting the expressions for $N_1$ and $N_2$:$\Delta N = \frac{R_1 T}{\ln 2} - \frac{R_2 T}{\ln 2} = \frac{T}{\ln 2} (R_1 - R_2)$.Since $T$ and $\ln 2$ are constants,the number of disintegrated atoms is proportional to $(R_1 - R_2)$.

The radioactivity of a sample is $R_1$ at a time $T_1$ and $R_2$ at a time $T_2$. If the half-life of the specimen is $T$,the number of atoms that have disintegrated in the time $(T_2 - T_1)$ is proportional to

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