MHT CET · Physics · Nuclear Physics
Two different radioactive elements with half lives ' \(\mathrm{T}_1\) ' and ' \(\mathrm{T}_2\) ' have undecayed atoms ' \(\mathrm{N}_1\) ' and ' \(\mathrm{N}_2\) ' respectively present at a given instant. The ratio of their activities at that instant is
- A \(\frac{\mathrm{N}_1 \mathrm{~T}_1}{\mathrm{~N}_2 \mathrm{~T}_2}\)
- B \(\frac{\mathrm{N}_2 \mathrm{~T}_2}{\mathrm{~N}_1 \mathrm{~T}_1}\)
- C \(\frac{\mathrm{N}_1 \mathrm{~T}_2}{\mathrm{~N}_2 \mathrm{~T}_1}\)
- D \(\frac{\mathrm{N}_1 \mathrm{~N}_2}{\mathrm{~T}_1 \mathrm{~T}_2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{N}_1 \mathrm{~T}_2}{\mathrm{~N}_2 \mathrm{~T}_1}\)
Step-by-step Solution
Detailed explanation
\(\therefore\) Activity is given as:
\(
\begin{aligned}
\therefore \mathrm{A} & =\lambda \mathrm{N} \\
\lambda & =\frac{\ln 2}{\mathrm{~T}}
\end{aligned}
\)
\(\therefore\) Ratio of two different radioactive elements will be:
\(
\begin{aligned}
& \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\lambda_1 \mathrm{~N}_1}{\lambda_2 \mathrm{~N}_2} \\
& \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\frac{\ln 2}{\mathrm{~T}_1} \mathrm{~N}_1}{\frac{\ln 2}{\mathrm{~T}_2} \mathrm{~N}_2} \\
& \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\mathrm{N}_1 \mathrm{~T}_2}{\mathrm{~N}_2 \mathrm{~T}_1}
\end{aligned}
\)
\(
\begin{aligned}
\therefore \mathrm{A} & =\lambda \mathrm{N} \\
\lambda & =\frac{\ln 2}{\mathrm{~T}}
\end{aligned}
\)
\(\therefore\) Ratio of two different radioactive elements will be:
\(
\begin{aligned}
& \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\lambda_1 \mathrm{~N}_1}{\lambda_2 \mathrm{~N}_2} \\
& \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\frac{\ln 2}{\mathrm{~T}_1} \mathrm{~N}_1}{\frac{\ln 2}{\mathrm{~T}_2} \mathrm{~N}_2} \\
& \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\mathrm{N}_1 \mathrm{~T}_2}{\mathrm{~N}_2 \mathrm{~T}_1}
\end{aligned}
\)
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