c
Given : \(\lambda_{A}=5 \lambda\), \(\lambda_{B}=\lambda\)

At \(t=0,\left(N_{0}\right)_{A}=\left(N_{0}\right)_{B}\)

\(\frac{N_{A}}{N_{B}}=\left(\frac{1}{e}\right)^{2}\)

According to radioactive decay, \(\frac{N}{N_{0}}=e^{-\lambda t}\)

\(\therefore \,\frac{N_{A}}{\left(N_{0}\right)_{A}} =e^{-\lambda_{A} t} \)  ..... \((i)\)

\(\frac{N_{B}}{\left(N_{0}\right)_{B}} =e^{-\lambda_{B} t}\)  ..... \((ii)\)

Divide \((i)\) by \((ii)\), we get

\(\frac{N_{A}}{N_{B}}=e^{-\left(\lambda_{A}-\lambda_{B}\right) t}\)  or, \(\frac{N_{A}}{N_{B}}=e^{-(5 \lambda-\lambda) t}\)

or, \(\left(\frac{1}{e}\right)^{2}=e^{-4 \lambda t}\) or,  \(\left(\frac{1}{e}\right)^{2}=\left(\frac{1}{e}\right)^{4 \lambda t}\)

or, \(4 \lambda t=2\) \( \Rightarrow \quad t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}\)