A radioactive element` X` converts into another stable element `Y`. Half-life of `X` is `2h`. Initially, only `X` is present. After time `t`, the ratio of atoms of `X` and `Y` is found to be `1:4` Then `t` in hours is .

Options

A. `2`
B. `4`
C. between` 4` and `6`
D. `6`

Answer

C

Explanation

Let `N_(2)` be the number of atoms of X at time `t =0`. Then, at `t=4h` (two half-lives),
`N_x =N_0/4` and `N_y =3N_0/4`
`because (N_(x))/(N_(y)) =(1)/(3) ~~0.33`
At `t=6th` (three half-lives), `N_(x) =(N_(0))/(8)` and `N_(y) =(7N_(0))/(8)` or `(N_(x))/(N_(y)) =(1)/(7) ~~0.142`
The given ratio `(1)/(4)` lies between `(1)/(3)` and `(1)/(7)`.
Therefore , t lies between `4 h` and `6h`.