KCET · Physics · Nuclear Physics
\(A\) 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
- A \(200 \mathrm{dis} / \mathrm{min}\)
- B \(250 \mathrm{dis} / \mathrm{min}\)
- C \(500 \mathrm{dis} / \mathrm{min}\)
- D \(150 \mathrm{dis} / \mathrm{min}\)
Answer & Solution
Correct Answer
(D) \(150 \mathrm{dis} / \mathrm{min}\)
Step-by-step Solution
Detailed explanation
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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