Question

$A$ and $B$ are the two radioactive elements. The mixture of these elements show a total activity of $1200$ disintegrations/minute. The half life of $A$ is $1$ day and that of $B$ is $2$ days.What will be the total activity after $4$ days ? Given : The initial number of atoms in $A$ and $B$ are equal.

• A. 200 dis/min
• B. 250 dis/min
• C. 500 dis/min
• D. 150 dis/min

Answer 1

The Correct Option is D

We have, $A=\lambda(N)=\frac{0.693}{T_{1 / 2}}(N)$

Initial number of atoms is $A$ and $B$ are same

$\therefore A_{0} \propto \frac{1}{T_{1 / 2}} $

$\Rightarrow \frac{A_{0}(A)}{A_{0}(B)} =\frac{48 hr }{24 hr }=2 $

Also, $A_{0}(A)+A_{0}(B) =1200 $

$\Rightarrow 3 A_{0}(B)=1200 $

$\Rightarrow A_{0}(B)=400$

and $A_{0}(A)=800$

So, $A(A)=\frac{A_{0}(A)}{2^{4}}=\frac{800}{16}=50$

and $A(B)=\frac{A_{0}(B)}{(2)^{2}}=\frac{400}{4}=100$

Hence, total activity after $4$ days is

$=A_{0}(A)+A_{0}(B)=50+100=150$ dis / min