Nuclei - Result Question 71
72. 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 $(t_1-t_2)$ :
[2010]
(a) $\lambda(A_1-A_2)$
(b) $A_1 t_1-A_2 t_2$
(c) $A_1-A_2$
(d) $(A_1-A_2) / \lambda$
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Answer:
Correct Answer: 72. (d)
Solution:
- (d) Activity is given by
$A=\frac{d N}{d t}=-\lambda N$
Activity at time $t_1$ is
$A_1=-\lambda N_1$
and activity at time $t_2$ is $A_2=-1 N_2$ As $t_1>t_2$, therefore, number of atoms remained after time $t_1$ is less than that remained after time $t_2$. That is, $N_1<N_2$.
$\therefore$ number of nuclei decayed in $(t_1-t_2)$
$=N_2-N_1=\frac{(A_1-A_2)}{\lambda}$
If a radioactive element $A$ disintegrates to form another radioactive element $B$, which inturn disintegrates to form another element $C$,
Rate of disintegration of $A=\frac{d N_1}{d t}=-\lambda_1 N_1$
Rate of disintegration of $B=\frac{d N_2}{d t}=-\lambda_2 N_2$
Net rate of formation of $B=$ Rate of disintegration of A - Rate of disiutegration of $B=\lambda_1 N_1-\lambda_2 N_2$
