Question

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
A. (a) `(N_0)/(128)`
B. (b) `(N_0)/(64)`
C. (c) `(N_0)/(32)`
D. (d) None of these

Answer 1

Correct Answer - B
`(N_0)/(16)=N_0(1/2)^n` `:.` `n=4`
So, 3t times is equivalent to four half-lives. Hence, one half-life is equal to `(3t)/(4)`.
The given time `11/2t-t=9/2t` is equivalent to 6 half-lives.
`:.` `N=N_0(1/2)^6=(N_0)/(64)`