Question

The nuclei of two radioactive isotopes of same substance `A^(236)` and `A^(234)` are present in the ratio of `4:1` in an ore obtained from some other planet. Their half lives are `30min` and `60min` respectively. Both isotopes are alpha emitters and the activity of the isotope with half-life `30min` is `10^(6)dps`. Calculate after how much time their activites will become identical. Also calculate the time required to bring the ratio of their atoms to `1:1`.

Answer 1

Let `A^(236): 1, A^(234):2`
`((N_(1))/(N_(2)))_(0) = (4)/(1)`
`(T_(1//2))_(1) = 30min, (T_(1//2))_(2) = 60 min`
`(A_(1))_(0) = 10^(6)dps`
`((A_(1))_(0))/((A_(2))_(0)) = (lambda_(1)(N_(1))_(0))/(lambda_(2)(N_(2))_(0)) =((T_(1//2))_(2))/((T_(1//2))_(1))((N_(1))/(N_(2)))_(0)`
`(10^(6))/((A_(2))_(0)) =(60)/(30)xx(4)/(1) rArr (A_(2))_(0) = (1)/(8) xx 10^(6)dps`
Let their activities become equal after time `t`
`A_(1) = A_(2)`
`(A_(1))_(0) ((1)/(2))^(n_(1)) = (A_(2))_(0) ((1)/(2))^(n_(2))`
`n_(1) = (t)/((T_(1//2))_(1)) = (t)/(30), n_(2) = (t)/((T_(1//2))_(2)) = (t)/(60)`,
`10^(6) ((1)/(2))^((t)/(30)) = (1)/(8) xx 10^(6)((1)/(2))^((t)/(60))`
`((1)/(2))^((t)/(30)-(t)/(60)) = (1)/(8) = ((1)/(2))^(3)`
`((1)/(2))^((t)/(60)) = ((1)/(2))^(3) rArr (t)/(60) = 3`
`t = 180 min`
`N_(1) = N_(2)`
`(N_(1))_(0) ((1)/(2))^(t//30) = (N_(2))_(0)((1)/(2))^(t//60)`
`((1)/(2))^(t//60) = ((N_(2))/(N_(1)))_(0) = (1)/(4) = ((1)/(2))^(2)`
`t//60 = (1)/(2) rArr t = 120 min`