For a radioactive material, its activity `A` and rate of change of its activity of `R` are defined as `A=(-dN)/(dt)` and `R=(-dA)/(dt)`, where `N(t)` is the number of nuclei at time `t`. Two radioactive source `P` (mean life `tau`) and `Q` (mean life `2 tau`) have the same activity at `t=0`. Their rates of activities at `t=2 tau` are `R_(p)` and `R_(Q)`, respectively. If `(R_(P))/(R_(Q))=(n)/(e )`, then the value of `n` is:

Answer

B

Explanation

`N=N_(0)e^(-lambdat)`
`A = -(dN)/(dt)`
`rArr A=lambdaN_(0)e^(-lambdat)`
P and Q have same activity at t = 0. If `N_(PO)` and `N_(QO)` are the number of nuclei at t = 0, then for t = 0 we can write the following:
`A = lambda_(P).N_(PO)=lambda_(Q).N_(QO)`
`rArr N_(PO)/N_(QO) =lambda_(Q)/lambda_(P)`
Further we can write the following:
`lambda_(P) =1/tau` and `lambda_(Q)=1/(2tau)`
`rArr lambda_(Q)/lambda_(P)=1/2`
`rArr N_(PO)/N_(QO)=1/2`
Activity can be written as follows:
`A=-(dN)/(dt)= lambdaN rArr A=lambdaN`
Rate of change of activity can be written as follows:
`R=-(dA)/(dt) =-lambda.(dN)/(dt) =lambda^(2)N`
`rArr R_(P)/R_(Q)=lambda_(P)^(2)/lambda_(Q)^(2).N_(P)/N_(Q)`
At `t=2tau`,
`N_(P) =N_(PO)/e^(2)` and `N_(Q)=N_(QO)/e`
`N_(P)/N_(Q) =N_(PO)/e^(2)` and `N_(Q)=N_(QO)/e`
`R_(P)/R_(Q)=4 xx 1/(2e) =2/e rArr n=2`