The decay constant of a radio isotope is (\lambda). If (A\_1... | Physics | SolveeIt | Solveeit

Today, August 24

Question

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 $(t_1-t_2)$

A

$A_1t_1-A_2t_2$

B

$A_1-A_2$

C

$(A_1-A_2)/ \lambda$

D

$\lambda (A_1-A_2)$

Answer

$(A_1-A_2)/ \lambda$

Explanation

Solution

$A_1=\lambda N_1 \, at \, time \, t_1$
$A_2=\lambda N_2$ at time $t_2$
Therefore, number of nuclei decayed during time interval $(t_1-t_2)$ is
$N_1-N_2 =\frac{[A_1-A_2]}{\lambda}$