The radioactivity of a sample is R_1 at time | vedclass

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(B) The activity of a radioactive sample is given by $R = N \lambda,$ where $N$ is the number of radioactive atoms present and $\lambda$ is the decay

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$(R_1 - R_2) T$

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(B) The activity of a radioactive sample is given by $R = N \lambda,$ where $N$ is the number of radioactive atoms present and $\lambda$ is the decay constant.At time $T_1,$ $R_1 = N_1 \lambda \implies N_1 = R_1 / \lambda.$At time $T_2,$ $R_2 = N_2 \lambda \implies N_2 = R_2 / \lambda.$The number of atoms that have disintegrated in the time interval $(T_2 - T_1)$ is $\Delta N = N_1 - N_2.$Substituting the values of $N_1$ and $N_2,$ we get $\Delta N = (R_1 - R_2) / \lambda.$Since the half-life $T = \ln(2) / \lambda,$ we have $\lambda = \ln(2) / T.$Substituting $\lambda$ in the expression for $\Delta N,$ we get $\Delta N = (R_1 - R_2) \cdot T / \ln(2).$Since $\ln(2)$ is a constant,$\Delta N \propto (R_1 - R_2) T.$

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

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