Question

A fresh radioactive mixture containing short live species $A$ and $B$. Both emitting $\alpha$ particles initially of $8000\alpha -$particles per minute. $20$ minutes later, they emit at the rate of $3500\alpha -$particles per minute. If the half-lives of the species $A$ and $B$ are $10$ minutes and $500$ hours respectively, then the ratio of $A:B$ in the initial mixture was:

• A. $4:6$
• B. $6:4$
• C. $3:4$
• D. $3:1$

Answer 1

The correct option is D $3:1$

The relationship between the half life period and the rate constant for first order reaction is $k=\frac{0.693}{t_{1/2}}$
For A and B, the expression becomes $k=\frac{0.693}{10}$ and $k=\frac{0.693}{500\times 60}$The integrated form of the first order rate law expression is $[A{]}_t=[A{]}_0{e}^{-kt}$
For A , the expression becomes $[A{]}_t=[A{]}_0{e}^{-\frac{0.693\times 20}{10}}$
For B , the expression becomes $[B{]}_t=[B{]}_0{e}^{-\frac{0.693\times 20}{500\times 60}}$
Also $[A{]}_0+[B{]}_0=8000$ and $[A{]}_t+[B{]}_t=3500$.
On solving these equations, we get the ratio A:B in the inital mixture as 3:1.