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

Answer : (c) 0.31

Explanation :

At time t=t

$N_{1}=N_{0}e^{-\lambda t }$

$N_{2}=\large\frac{N_{0} \lambda_{1}}{\lambda_{2} - \lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})$

$N_{3}=N_{0}-N_{1}-N_{2}$

$=N_{0}\;[1-e^{-\lambda_{1}t} - \large\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})]$

$\large\frac{N_{3}}{N_{0}}=1-e^{-\lambda_{1}t}-\large\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}\;(e^{-\lambda_{1}t}-e^{-\lambda_{2}t})$

$\lambda_{1}=\large\frac{0.693}{30}=0.231min^{-1}$

$\lambda_{1}=\large\frac{0.693}{45}=0.0154min^{-1}$

and t=60 minutes

$\large\frac{N_{3}}{N_{0}}=1-e^{-0.0231\times60}-\large\frac{0.0231}{0.0154-0.0231}\;(e^{-0.0231\times60}-e^{-0.0154\times60})$

$=1-0.25+3(0.25-0.4)$

$=0.31\;.$

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