Sample of two radioactive nuclides `A` and `B` are taken. `lambda_(A)` and `lambda_(B)` are the disintergration constants of `A` and `B` respectively. In which of the following cases, the two sample can simultaneously have the same decay rate at any time ?

Options

A. Initial rate of decay of `A` is twice the initial rate of decay of `B` and `lambda_(A)=lambda_(B)`
B. Initial rate of decay of `A` is twice the initial rate of decay of `B` and `lambda_(A) gt lambda_(B)`
C. Initial rate of decay of `B` is twice the initial rate of decay of `A` and `lambda_(A) gt lambda_(B)`
D. Initial rate of decay of `B` is same as the rate of decay of `A` at `t=2h` and `lambda_(B)=lambda_(A)`

Answer

B

Explanation

`|(dN)/(dt)|rarr`rate of decay `=lambda N`
(a) `t=0`
`(lambda)/(A) N_(0A)=2 (lambda)/(B)N_(0B) rArrN_(0A)=2N_(0B)`
`|(dN)/(dt)|_(A)=lambda_(A)N_(A)=lambdaA_(A)N_(0A)e^(-lambdat)=2 lambda_(A)N_(0A)e^(-lambda At)`
`|(dN)/(dt)|_(B)=lambda_(B)N_(B)=lambda_(B)N_(0B)e^(-lambda_(A)t)`
`:'lambda_(A)=lambda_(B)`
`|(dN)/(dt)|!=|(dN)/(dt)|_(B)`
(b) `lambda_(A)N_(0A)=2lambda_(B)N_(0B)(lambda_(A) gtlambda_(B))`
at `t=t`
`|(dN)/(dt)|=lambdaN_(0)e^(-lambdat)`
`rArr` If equal : `lambda_(A)N_(0A)e^(-lambda t)=lambda_(B)N_(0)e^(-lambda bt)`
`2=e^((lambda_(A)-lambda_(B))t)(lambda_(A) gtlambda_(B)) rArr t` can have real solution.
(c) Proceed similar to (b) t will have no real solution.
(d) `lambda_(B)N_(0B)=lambda_(A)N_(0A)e^(-2lambdaA)`
`N_(0B)=N_(0A)e^(02lambdaA)`
at some `t=t`, if equal roots occur
`lambda_(B)N_(0B)e^(-lambda_(B)t)=lambda_(A)N_(0A)^(-lambda At)`
`N_(0B)e^(-lambda Bt)=lambda_(A)N_(0A)e^(-lambda At)(lambda_(A)=lambda_(B))`
`N_(0B)e^(-lambda A t)=N_(0B)e^(2lambda A)e^(-lambda_(At))`
`rArr e^(2 lambda A)=1` not possible.