The fraction of a radioactive material which remains active after time t is `9//16`. The fraction which remains active after time `t//2` will be .

Options

A. `(4)/(5)`
B. `(7)/(8)`
C. `(3)/(5)`
D. `(3)/(4)`

Answer

D

Explanation

`(9)/(16)=((1)/(2))^((t)/(T))`
`(N)/(N_(0))=((1)/(2))^((t)/(2T))`
`((N)/(N_(0)))^(2)=((1)/(2))^(t//T)`
or `(N)/(N_(0)^(2))=(9)/(16)` or `(N)/(N_(0)) =(3)/(4)`
Note the special techaniq used in the problem.