`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. `6.62yr`
B. 5yr
C. `3.2yr`
D. 7yr

Answer

A

Explanation

We have `N=N_(0)((1)/(2))^(t//T)`
`thereforeN_(A)=10((1)/(2))^(t//1)impliesN_(B)=1((1)/(2))^(t//2)`
Given, `N_(A)=N_(B)`
So, `10((1)/(2))=((1)/(2))^(t//2)`
So, `10((1)/(2))^(-t//2)or 10=2^(t//2)`
`therefore "log"_(10)=t/2log_(10)2`
`1=t/2xx0.3010`
`therefore` Time, t = 0.62 yr