Q: The mean life of a radioactive substance are 1620 and 405 year for α – emission and β-emission respectively . Find out the time during which three fourth of a sample will decay if it is decaying both by α – emission and β-emission simultaneously .

Solution : When a substance decays by α – emission and β-emission simultaneously , the equivalent rate of disintegration is given by λ_{eq} = λ_{α} + λ_{β}

Mean life is given by $\large T_{eq} = \frac{1}{\lambda_{eq}} $

$\large T_{eq} = \frac{1}{\lambda_{eq}} $

λ_{eq} = λ_{α} + λ_{β}

$\large \frac{1}{T_{eq}} = \frac{1}{T_{\alpha}} + \frac{1}{T_{\beta}} $

$\large \frac{1}{T_{eq}} = \frac{1}{1620} + \frac{1}{405} $

$\large \lambda_{eq} = \frac{1}{1620} + \frac{1}{405} $

$\large \lambda_{eq} = \frac{5}{1620} year^{-1} $

As remaining sample is N_{0}/4

$\large N = N_0 (\frac{1}{2})^n$

$\large \frac{N_0}{4} = N_0 (\frac{1}{2})^n$

n = 2 = Number of half lives

$\large n = \frac{t}{T_{1/2}}$

$\large t = n T_{1/2}$

$\large t = n \frac{0.693}{\lambda_{eq}}$

$\large t = 2 \times \frac{0.693}{\frac{5}{1620}}$

t = 449 year