The radioactivity of a sample is R_1 at time | vedclass

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Question

$R_{1}=N_{1} \lambda, R_{2}=N_{2} \lambda$

Also,

$T=\log _{e} \cdot \frac{2}{\lambda}$ or $\lambda=\log _{e} \cdot \frac{2}{T}$

$	herefore R

Vedclass · Accepted Answer

$(R_1 - R_2) T$

Vedclass · Answer

$R_{1}=N_{1} \lambda, R_{2}=N_{2} \lambda$

Also,

$T=\log _{e} \cdot \frac{2}{\lambda}$ or $\lambda=\log _{e} \cdot \frac{2}{T}$

$	herefore R_{1}-R_{2}=\left(N_{1}-N_{2}ight) \lambda$

$=\left(N_{1}-N_{2}ight) \log _{e} \cdot \frac{2}{T}$

$	herefore\left(N_{1}-N_{0}ight)=\frac{\left(R_{1}-R_{2}ight) T}{\log _{e} 2}$

i.e., $\left(N_{1}-N_{2}ight) \propto\left(R_{1}-R_{2}ight) T$

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 $T.$ Number of atoms that have disintegrated in time $(T_2 - T_1)$ is proportional to

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