A radioactive nucleus decays by two different processes. The...

Using the equation of radioactive decay law,

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

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

Where,

$- \dfrac{d N}{d t}$ is the rate of decay

${\lambda}_{1,}{\lambda}_{2}$ are decay constant for two different processes

$N$ is the number of atoms.

Now applying the decay law for the combine process, we get

$- \dfrac{dN}{dt} = {\lambda}_{eff}N . . . ( 3 )$

Equating equation (1) and (2) we get

${\lambda}_{eff} = \left({\lambda}_{1} + {\lambda}_{2}\right) . . . ( 4 )$

Now the half life of the radioactive nucleus is given by,

$\lambda = \dfrac{\ln \left(2\right)}{T} . . . ( 5 )$

Where, $T$ is the half life of the radioactive nucleus

Using equation (5) equation (4) becomes,

$\dfrac{\ln \left(2\right)}{T} = \dfrac{\ln \left(2\right)}{{T}_{1}} + \dfrac{\ln \left(2\right)}{{T}_{2}}$

$\dfrac{1}{T} = \dfrac{1}{{T}_{1}} + \dfrac{1}{{T}_{2}}$

$\dfrac{1}{T} = \dfrac{1}{10} + \dfrac{1}{100} = \dfrac{11}{100}$

$⇒ T = \dfrac{100}{11} \approx 9s$