The decay constant of a radio isotope is lamb | vedclass

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Question

$A_{1}=\lambda N_{1}$ at time $t_1$

$A_{2}=\lambda N_{2}$ at time $t_2$

Therefore, number of nuclei decayed during time interval $\left(t_{

Vedclass · Accepted Answer

$(A_1-A_2)/&lambda;$

Vedclass · Answer

$A_{1}=\lambda N_{1}$ at time $t_1$

$A_{2}=\lambda N_{2}$ at time $t_2$

Therefore, number of nuclei decayed during time interval $\left(t_{1}\,-\,t_{2}\right)$ is

$N_{1}-N_{2}=\frac{\left[A_{1}-A_{2}\right]}{\lambda}$

The decay constant of a radio isotope is $\lambda$. If $A_1$ and $A_2$ are its activities at times $t_1$ and $t_2$ respectively, the number of nuclei which have decayed during the time $(t_1 - t_2)$

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