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A $\;^{118}cd\;$ radionuclide goes through transformation chain $\;^{118}Cd\; \to_{30 min}^{118}ln \to_{45 min} ^{118}Sn\;$(stable) The half lives are written below respective arrows ,A time t=0 only celuear present. Then the fraction of nuclei transformed into stable over 60 minutes:


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Answer : (c) 0.31
Explanation :
At time t=t
$N_{1}=N_{0}e^{-\lambda t }$
$N_{2}=\large\frac{N_{0} \lambda_{1}}{\lambda_{2} - \lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})$
$=N_{0}\;[1-e^{-\lambda_{1}t} - \large\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})]$
and t=60 minutes
answered Feb 25, 2014 by yamini.v

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