The count rate observed from a radioactive source at `t` sound was `N_0` and at `4t` second it was `(N_0)/(16)`. The count rate observed at `(11/2)t` second will be

Options

A. `(N_(0))/(128)`
B. `(N_(0))/(64)`
C. `(N_(0))/(32)`
D. None of these

Answer

B

Explanation

Let initially substance have `N_(i)` nuclei then `N=N_(i)`
`e^(-lambda t), (dN)/(dt)= -lambdaN_(i)e^(-lambdat)`
it is given that, At `t=t, (dN)/(dt)= -lambdaN_(i)e^(-lambda t)=N_(0)`
and at `t=4t, (dN)/(dt)= -lambdaN_(i)e^(-4lambdat)=(N_(0))/(16)`
Dividing both, we get `e^(3lambda t)=16`, At `t=11t//2`,
`(dN)/(dt)= -lambdaN_(i)e^(-(11)/(2)lambda t)= - lambdaN_(i)e^(-(8)/(2)lambda t)xx e^((3)/(2)lambda t)=(N_(0))/(64)`