A radioactive nucleus decays by two different process. The half...

Decay law for both process can be written as,

$\dfrac{d{N}_{1}}{dt} = - {\lambda}_{1}N . . . Eq ( 1 )$ and

$\dfrac{d{N}_{2}}{dt} = - {\lambda}_{2}N . . . Eq ( 2 )$

Hence, net decay will become

$\dfrac{dN}{dt} = \dfrac{d{N}_{1}}{dt} + \dfrac{d{N}_{2}}{dt} = - ( {\lambda}_{1} + {\lambda}_{2} ) N$

$⇒ {\lambda}_{eq} = {\lambda}_{1} + {\lambda}_{2}$

$⇒ \dfrac{\ln 2}{{t}_{\dfrac{1}{2}}} = \dfrac{\ln 2}{300} + \dfrac{\ln 2}{30}$

$⇒ t\dfrac{1}{2} = \dfrac{300}{11}$

Hence, required value of $\alpha = 300$.