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The mean life of a radio nuclide which decays by given parallel path is:

X \stackrel{\lambda_{1}}{\longrightarrow} Y \lambda_{1}=3.6 \times 10^{-3} \mathrm{~s}^{-1}
2 \times \stackrel{\lambda_{2}}{\longrightarrow} Y \lambda_{2}=10^{-3} \mathrm{~s}^{-1}

Option: 1

17.85 s

Option: 2

178.5 s

Option: 3

78.5 s

Option: 4

50 s

Answers (1)


\lambda_{\text {net }}=\lambda_{1}+2 \lambda_{2}=3.6 \times 10^{-3}+2 \times 10^{-3}=5.6 \times 10^{-3} \mathrm{~s}^{-1}

\text { Average life } \tau =\frac{1}{\lambda_{\text {net }}}
                            \tau =\frac{1}{5.6 \times 10^{-3}}
                             \tau =\frac{1000}{5.6}
                            \tau =178.5\, \mathrm{sec}

Posted by

Deependra Verma

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