An alloy is composed of two radiactive materials A and B having equal weight. The half life of A and B are 10 yrs and 20 yrs respectively. After time t, the alloy was found to consist of `(1/e)` kg of A and 1 kg of B. If the atomic weight of A and B are same, then the value of t is (Assume, `ln 2=0.7`)

Options

A. `(200/7)` yrs
B. `(10/7)` yrs
C. 7 yrs
D. 70 yrs

Answer

A

Explanation

Given, initial amount of radiactive element, A = initial amount of radioactive element, B i.e., `(N_(0))_(A)=(N_(0))_(B)` ...(i)
half life of element `A, (t_(1//2))_(A)=10` yr
and half life of element `B, (t_(1//2))_(B)=20` yr
After time t, remaining the amount of element A,
`N_(A)=1/e` kg
and remaining amount of element B,
`N_(B)=1` kg
For element A, after time t, amount of element left undecayed is given as
`N_(A)=(N_(0))_(A)(1/2)^(t/((t_(1//2))_(A)))`
`1/e=(N_(0))_(A)(1/2)^(t/10)` ...(ii)
Similarly, for element B,
`N_(B)=(N_(0))_(B)(1/2)^(t/20)`
`1=(N_(0))_(A)(1/2)^(t/20)` ...(iii)
Dividing Eqs. (ii) and (iii), and using Eq. (i) we get
`1/e=((1/2)^(t/10))/((1/2)^(t//20))=(1/2)^(t/20)`
taking log on the both sides, we get
`"log"1/e=t/20"log" 1/2`
`t=20/(log 2)`
`=20/0.7=200/7` yr `[ :' log 2=0.7]`