Question

In the above radioactive decay $C$ is stable nucleus. Then :

• A. rate of decay of $A$ will first increase and then decrease
• B. number of nuclei of $B$ will first increase and then decrease
• C. if $\lambda_{2} >\lambda_{1}$, then activity of $B$ will always be higher than activity of $A$
• D. if $\lambda_{1} \gg \lambda_{2}$, then number of nucleus of $C$ will always be less than number of nucleus of $B$.

Answer 1

Answer: (B)

Rate of decay of $A$ keeps on decreasing continuously because concentration of $A$ decreases with time $\Rightarrow(A)$ is false.
Initial rate of production of $B$ is $\lambda_{1} N_{0}$ and rate of decay is zero. With time, as the number of $B$ atom increase, the rate of its production decrease and its rate of decay increases. Thus the number of nuclei of $B$ will first increase and then decrease.
$\Rightarrow$ (B) is the correct choice
The initial activity of $B$ is zero whereas initial activity of $A$ is $\hat{\lambda}_{1} N_{0} \Rightarrow$ (C) is false.
As time $t \rightarrow \infty: N_{A}=0, N_{B}=0$ and $N_{C}=N_{0} \Rightarrow$ (D) is false