`A overset(lambda_(1)) rarr B overset (lambda_(2))rarr C`
t=0 `N_(0)` 0 0
` t N_(1) N_(2) N_(3)` In the above radioactive decay C is stable nucleus. Then:

Options

A. rate of decay of `A` will first increase and then decreases
B. number of nuclei of `B` first increase and then decreases
C. if `lambda_(2) gt lambda_(1)`, then activity of `B` will always be higher than activity of `A`
D. if `lambda gt gt lambda_(2)`, then number of nucleus of `C` will always be less than number of nucleus of `B`.

Answer

B

Explanation

Rate of decay of `A` keeps on decreasing continuosly because concentration of `A` decreases with time. `rArr A` is false
Initial rate fo production of `B` is `lambda_(1)N_(0)` and rate of decay is zero. With time, as the number of `B` atom in increase, the rate of its production decreases and its rate of decay increases. Thus the number of nuclei of `B` will first increase and then decrease. `rArr B` is the correct choice
The initial activity of `B` zero whereas initial activity of `A` is `lambda_(1)N_(0) rArrC` is false.
As time `t rarr : N_(A)=0,N_(B)=0` and `N_( C)=N_(0)rArrD` is false