JEE Mains · Physics · STD 12 - 13. Nuclei
A radioactive sample is undergoing \(\alpha\) decay. At any time \(t_{1}\), its activity is \(A\) and another time \(t _{2}\), the activity is \(\frac{ A }{5}\). What is the average life time for the sample?
- A \(\frac{\ell n 5}{ t _{2}- t _{1}}\)
- B \(\frac{ t _{1}- t _{2}}{\ell n 5}\)
- C \(\frac{ t _{2}- t _{1}}{\ell n 5}\)
- D \(\frac{\ell n \left( t _{2}+ t _{1}\right)}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{ t _{2}- t _{1}}{\ell n 5}\)
Step-by-step Solution
Detailed explanation
Let initial activity be \(A _{0}\) \(A = A _{0} e ^{-\lambda t_{2}}....(i)\) \(\frac{ A }{5}= A _{0} e ^{-\lambda t_{2}}....(ii)\) \(( i ) \div ( ii )\) \(5= e ^{\lambda\left(t_{2}-t_{1}\right)}\) \(\lambda=\frac{\ell n 5}{t_{2}-t_{1}}=\frac{1}{\tau}\)…
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