Question

A radioactive element `A` with a half-value period of `2` hours decays giving a stable element `Y`. After a time `t` the ratio of `X` and `Y` atoms is `1 : 7` then `t` is :
A. 6 hours
B. 4 hours
C. between 4 and 5 hours
D. 14 hours

Answer 1

Correct Answer - A
(a) After lapse of time `t`, let the number of atoms of element and `Y` element be respectivity `n_x` and `n_Y`.
Then `(n_y)/(n_x) = 7` or `(n_y)/(n_x) + 1 = 7 +1`
or `(n_y + n_y)/(n_x) = 8`
or `(n_x)/(n_y +n_x) = (1)/(8)`
Death of an atom of mother element means the birth of an atom of daughter element.
`:. (n_x)/(n_y + n_x)(N)/(N_0) =(1)/(8) = ((1)/(2))^n`
`:. n = 3 = (t)/(T)`
But `T = 2 hours`, hence `t = 6 hours`.