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`.

Explanation

We have, `((N_(1))/(N_(2)))_(0)=(4)/(1)`
Half-life of two isotopes are,
` " "(T_(1//2))_(1)= 30" min "and (T_(1//2))_(2)=60 " min"`
Activity of first isotope is, `(R_(1))_(0)=10^(6)` dps
We know, Activity of radioactive substance is, `R_(0)=lamdaN_(0)` and decay constant is, `lamda=(0.693)/(t_(1//2))` , where `t_(1//2)` = half-life.
`therefore" "((R_(1))_(0))/((R_(2))_(0))=(lamda_(1)(N_(1))_(0))/(lamda_(2)(N_(2))_(0))=((T_(1//2))_(2))/((T_(1//2))_(1))((N_(1))/(N_(2)))_(0)`
`" "(10^(6))/((R_(2))_(0))=(60)/(30)xx(4)/(1)rArr(R_(2))_(0)=(1)/(8)xx10^(6)` dps
Let their activities become equal after time `t`,
`rArr" "(R_(1)=R_(2))`
We know, after `n` half-life, `R=R_(0)((1)/(2))^(n)`.
`rArr" "(R_(1))_(0)((1)/(2))^(n_(1))=(R_(2))_(0)((1)/(2))^(n_(2))andt=nT_(1//2)`
We have, `n_(1)=(t)/((T_(1//2))_(1))=(t)/(30)and n_(2)=(t)/((T_(1//2))_(2))=(t)/(60)`
`therefore" "10^(6)((1)/(2))^((t)/(30)-(t)/(60))=(1)/(8)=((1)/(2))^(3)rArr((1)/(2))^((t)/(60))=((1)/(2))^(3)rArr(t)/(60)=3`
`rArr" "Time, `t` = 180 min`
According to question, when `N_(1)=N_(2)`
After `n` half-life, number of nuclei left undecayed is,
`" "N =N_(0)((1)/(2))^(n)`,
`" "(N_(1))_(0)((1)/(2))^(t//30)=(N_(2))_(0)((1)/(2))^(t//60)`
` or" "((1)/(2))^(t//60)=((N_(2))/(N_(1)))_(0)=(1)/(4)=((1)/(2))^(2)`
`" "t//60=(1)/(2)rArr` time, `t` = 120 min.