Question

The nuclei of two radioactive isotopes of same substance $A^{236}\text{and}{A}^{234}$ are present in the ratio $4:1$ in an ore obtained from Mars. Their halflives are $30\text{min}and60\text{min}$ 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

For given isotopes $A^{236}:A^{234}$
Ratio of Nuclei is $4N_0:N_0$
Ratio of their Half life is $30:60$
Activity = $\lambda N=\lambda N_0{e}^{-\lambda t}$

Let after t time activtieis will become identical

${\lambda }_{1}4N_0{e}^{-\frac{0.693}{30}t}={\lambda }_2{N}_0{e}^{-\frac{0.693}{60}t}\{\because T_{1/2}=\frac{0.693}{\lambda }\}$

$\frac{0.693}{30}\times 4N_0{e}^{-\frac{0.693}{30}t}=\frac{0.693}{60}\times N_0{e}^{-\frac{0.693}{60}t}$

$8=e^{0.693(\frac{1}{30}-\frac{1}{60})t}$
$8=e^{+\frac{0.693}{60}t}$

$3\times 0.693=\frac{0.693}{60}t$

$t=180\text{min}$