`A` and `B` are two radioactive substances whose half lives are `1` and `2` years respectively. Initially `10 gm` of `A` and `1 gm` of `B` is taken. The time (approximate) after which they will have same quantity remaining is.

Options

A. `6.62 years`
B. 5 years
C. `3.2 years`
D. 7 years

Answer

A

Explanation

(a) `N = N_0 ((1)/(2))^(t//T_(1//2))`
`rArr N_A = 10 ((1)/(2))^(t//1)`
and `N_B = 1((1)/(2))^(t//2)`
Given`N_A = N_B rArr 10 ((1)/(2))^t = ((1)/(2))^(t//2)`
`rArr 10 = ((1)/(2))^(-t//2) rArr 10 = 2^(t//2)`. Taking log both the sides.
`log_10 = (t)/(2) log_10 2 rArr 1 = (t)/(2) xx 0.3010`
`rArr t = 6.62 years`.