Question

$4$grams of radioactive substance A left $1/2$ gm after some time. $1$ gram of another radioactive substance B left $1/4$ gm in the same period. If half life of B is $2$ hours, the half life of $A$ is

• A. $3/4hours$
• B. $4/3hours$
• C. $1/4hours$
• D. $1/2hours$

Answer 1

The correct option is B $4/3hours$

From integrated rate law, half life of B $=2hours$

So, $\lambda =\frac{ln2}{2}$

For B, $\frac{1}{4}=e^{\frac{-ln2t}{2}}$

$ln\left(\frac{1}{4}\right)=\frac{-tln2}{2}\Rightarrow -2ln=\frac{-t}{2}ln2$

$\Rightarrow t=4seconds$

For A, after same period,

From rate law:

$\frac{1}{2}=4e^{-\lambda 4}\Rightarrow ln\left(\frac{1}{8}\right)=-4\lambda \Rightarrow \frac{3ln2}{4}=\lambda$

Half life $=\frac{4ln2}{3ln2}=\frac{4}{3}hours=\frac{4}{3}hrs$