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

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$2$

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(A) The activity of a radioactive sample is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of radioactive nuclei present at time $t$.Given $R_1 = \lambda N_1$ and $R_2 = \lambda N_2$.The number of atoms disintegrated in the time interval $(t_2 - t_1)$ is $\Delta N = N_1 - N_2$.Since $N = \frac{R}{\lambda}$,we have $\Delta N = \frac{R_1 - R_2}{\lambda}$.The decay constant $\lambda$ is related to the half-life $T$ by $\lambda = \frac{\ln 2}{T}$.Substituting $\lambda$ into the expression for $\Delta N$:$\Delta N = \frac{(R_1 - R_2)T}{\ln 2}$.We are given that $\Delta N = \frac{n(R_1 - R_2)T}{\ln 4}$.Since $\ln 4 = \ln(2^2) = 2 \ln 2$,we have:$\Delta N = \frac{n(R_1 - R_2)T}{2 \ln 2}$.Equating the two expressions for $\Delta N$:$\frac{R_1 - R_2}{\ln 2} = \frac{n(R_1 - R_2)}{2 \ln 2}$.Canceling common terms $(R_1 - R_2)$ and $\ln 2$ from both sides:$1 = \frac{n}{2}$,which implies $n = 2$.

Activity of a radioactive sample is $R_1$ at a time $t_1$ and $R_2$ at a time $t_2$. Its half-life period is $T$. The number of atoms that have disintegrated in the time interval $(t_2 - t_1)$ is equal to $\frac{n(R_1 - R_2)T}{\ln 4}$. Then '$n$' is equal to

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