# 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:

$(a)\;0.49\qquad(b)\;0.38\qquad(c)\;0.31\qquad(d)\;0.26$

Explanation :
At time t=t
$N_{1}=N_{0}e^{-\lambda t }$
$N_{2}=\large\frac{N_{0} \lambda_{1}}{\lambda_{2} - \lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})$
$N_{3}=N_{0}-N_{1}-N_{2}$
$=N_{0}\;[1-e^{-\lambda_{1}t} - \large\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})]$
$\large\frac{N_{3}}{N_{0}}=1-e^{-\lambda_{1}t}-\large\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})$
$\lambda_{1}=\large\frac{0.693}{30}=0.231min^{-1}$
$\lambda_{1}=\large\frac{0.693}{45}=0.0154min^{-1}$
and t=60 minutes
$\large\frac{N_{3}}{N_{0}}=1-e^{-0.0231\times60}-\large\frac{0.0231}{0.0154-0.0231}\;(e^{-0.0231\times60}-e^{-0.0154\times60})$
$=1-0.25+3(0.25-0.4)$
$=0.31\;.$