The radioactive sources A and B of half lives | vedclass

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Question

For a first order reaction, $\frac{0.693}{t_{1 / 2}}=\frac{2.303}{t} \log \frac{A_{0}}{A-t}$

For radioactive source $A$, the equation becomes $\fra

Vedclass · Accepted Answer

$\sqrt{2} : 1$

Vedclass · Answer

For a first order reaction, $\frac{0.693}{t_{1 / 2}}=\frac{2.303}{t} \log \frac{A_{0}}{A-t}$

For radioactive source $A$, the equation becomes $\frac{0.693}{t}=\frac{2.303}{t} \log \frac{A_{0}}{A_{t}}&hellip;..(1)$

For radioactive source $B$, the equation becomes $\frac{0.693}{t}=\frac{2.303}{2 t} \log \frac{A_{0}}{B_{t}}&hellip;.(2)$

Divide equation $(1)$ with equation $(2).$

$2 \log \frac{A_{0}}{A-t}=\log \frac{A_{0}}{B_{t}}$

At the end of t hours, their rates of disintegration are in the ratio $\sqrt{2}: 1$

The radioactive sources $A$ and $B$ of half lives of $2\, hr$ and $4\, hr$ respectively, initially contain the same number of radioactive atoms. At the end of $2\, hours,$ their rates of disintegration are in the ratio :

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