Question

The nuclei of two radioactive isotopes of same substance $A^{236}$ and $A^{234}$ are present in the ratio 4 : 1 in an ore obtained from Mars. Their half lives are $30\text{min}$ and $60\text{min}$ respectively. Both isotopes are alpha emitters and the activity of the isotope with half life $30\text{min}$ is one Rutherford. Calculate after how much time (in min) their activities will become identical.

Answer 1

Given ratio of isotopes $A^{236}:A^{234},$ nuclei $4N_0:N_0$
Given ratio of half lives $30:60$
Activity $=\lambda N=\lambda N_0{e}^{-\lambda t}$
For activity to be same for both isotopes,
${\lambda }_{1}4N_0{e}^{-\frac{0.693}{30}t}={\lambda }_2{N}_0{e}^{-\frac{0.693}{60}t}\frac{0.693}{30}\times 4N_0{e}^{-\frac{0.693}{30}t}=\frac{0.693}{60}\times N_0{e}^{-\frac{0.693}{60}t}$
$8=e^{0.693t(\frac{1}{30}-\frac{1}{60})}$
$8=e^{+\frac{0.693}{60}t}3\times 0.693=\frac{0.693t}{60}$
$t=180\text{min}$