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$A$ and $B$ are the two radioactive elements. The mixtue of these elements show a total activity of 1200 disintergrations/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
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The correct answer is:
$150 \mathrm{dis} / \mathrm{min}$
We have, $A=\lambda(N)=\frac{0.693}{T_{1 / 2}}(N)$
Initial number of atoms is $A$ and $B$ are same
$\therefore \quad A_{0} \propto \frac{1}{T_{1 / 2}}$
$\Rightarrow \quad \frac{A_{0}(A)}{A_{0}(B)}=\frac{48 \mathrm{hr}}{24 \mathrm{hr}}=2$
Also, $A_{0}(A)+A_{0}(B)=1200$
$\Rightarrow \quad 3 A_{0}(B)=1200$
$\Rightarrow \quad A_{0}(B)=400$
and $\quad A_{0}(A)=800$
So, $\quad A(A)=\frac{A_{0}(A)}{2^{4}}=\frac{800}{16}=50$
and $\quad A(B)=\frac{A_{b}(B)}{(2)^{2}}=\frac{400}{4}=100$
Hence, total activity after 4 days is
$=A_{0}(A)+A_{0}(B)=50+100=150 \mathrm{dis} / \mathrm{min}$
Initial number of atoms is $A$ and $B$ are same
$\therefore \quad A_{0} \propto \frac{1}{T_{1 / 2}}$
$\Rightarrow \quad \frac{A_{0}(A)}{A_{0}(B)}=\frac{48 \mathrm{hr}}{24 \mathrm{hr}}=2$
Also, $A_{0}(A)+A_{0}(B)=1200$
$\Rightarrow \quad 3 A_{0}(B)=1200$
$\Rightarrow \quad A_{0}(B)=400$
and $\quad A_{0}(A)=800$
So, $\quad A(A)=\frac{A_{0}(A)}{2^{4}}=\frac{800}{16}=50$
and $\quad A(B)=\frac{A_{b}(B)}{(2)^{2}}=\frac{400}{4}=100$
Hence, total activity after 4 days is
$=A_{0}(A)+A_{0}(B)=50+100=150 \mathrm{dis} / \mathrm{min}$
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