The half life of a radioactive sample is $T$ . If the activities of the sample at time t $_{1}$ and t $_{2}$ ( $t_{1} < t_{2}$ ) are $R_{1}$ and $R_{2}$ respectively, then the number of atoms disintegrated in time $t_{2} - t_{1}$ is proportional to

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Answer 1

Activity of a radioactive sample is given as $a = λ N$

Activity at time $t_{1}$ $R_{1} = λ N_{1}$

Activity at time $t_{2}$ $R_{2} = λ N_{2}$

Then, the number of atoms disintegrated in time $t_{2} - t_{1}$ is $N_{1} - N_{2} = \frac{R _{1}}{λ} - \frac{R _{2}}{λ}$

Half life time period is: $T_{\frac{1}{2}} = \frac{ln 2}{λ}$

$N_{1} - N_{2} = (R_{1} - R_{2}) \frac{T}{ln 2}$

Answer 2

Activity of a radioactive sample is given as

a = λ N

Activity at time

t_{1}

R_{1} = λ N_{1}

Activity at time

t_{2}

R_{2} = λ N_{2}

Then, the number of atoms disintegrated in time

t_{2} - t_{1}

is

N_{1} - N_{2} = \frac{R _{1}}{λ} - \frac{R _{2}}{λ}

Half life time period is:

T_{\frac{1}{2}} = \frac{ln 2}{λ}

N_{1} - N_{2} = (R_{1} - R_{2}) \frac{T}{ln 2}