Question

A radioactive material has a half life of $600$ years and $300$years for $\alpha$ and $\beta$ emissions respectively. The material decays by simultaneous $\alpha$ and $\beta$ emissions. What is time in which $\frac{3}{4}$th of the material will decay?

• A. $600$years
• B. $900$years
• C. $220$years
• D. $400$years

Answer 1

The correct option is C $220$years

A radioactive material has a half life$=600$ years for $\alpha$ emission.

For $\beta$ emission $=300years$ respectively

The material decays by simultaneous $\alpha$ and $\beta$ emissions

The material decays time t when, the decay will be

$\frac{3}{4}th$ of the material

Solution

The decay constant $=\lambda$ is reciprocal of mean life

${\lambda }_{\alpha }=\frac{1}{600}peryear$

and ${\lambda }_{\beta }=\frac{1}{300}$ per year

Total decay constant

$\lambda ={\lambda }_{\alpha }+{\lambda }_{\beta }=\frac{1}{600}+\frac{1}{300}=\frac{1+2}{600}=\frac{3}{600}=\frac{1}{200}$

per year

When $\frac{3}{4}th$ per of sample has disintegrated.

$N=\frac{N_0}{4}$

Let total sample $=N_0$

$1-\frac{3}{4}=\frac{N_0}{4}thremaining\frac{N_0}{4}=N_0{e}^{-\lambda t}{e}^{-\lambda t}=4\lambda t=ln4t=\frac{1}{\lambda }ln{2}^{2}t=\frac{2}{\lambda }ln2=2\times 200ln2=277.25years$