Nuclei - Result Question 77
78. In a radioactive material the activity at time $t_1$ is $R_1$ and at a later time $t_2$, it is $R_2$. If the decay constant of the material is $\lambda$, then
[2006]
(a) $R_1=R_2 e^{\lambda(t_1-t_2)}$
(b) $R_1=R_2 e^{(t_2 / t_1)}$
(c) $R_1=R_2$
(d) $R_1=R_2 e^{-\lambda(t_1-t_2)}$
Show Answer
Answer:
Correct Answer: 78. (d)
Solution:
- (d) Let at time $t_1 & t_2$, number of particles be $N_1 & N_2$. So,
$R_1=\frac{d N_1}{d t}=-\lambda N_1 ; \quad R_2=\frac{d N_2}{d t}=-\lambda N_2$
$\frac{R_1}{R_2}=\frac{\lambda N_1}{\lambda N_2}=\frac{N_1}{N_1 e^{-\lambda(t_2-t_1)}}=e^{\lambda(t_2-t_1)}$
$R_1=R_2 e^{\lambda(t_2-t_1)}=R_2 e^{-\lambda(t_1-t_2)}$
