Question

A certain radioactive material is known to decay at a rate proportional to the amount present. If after one hour it is observed that $10$ percent of the material has decayed, find the half-life (period of time it takes for the amount of material to decrease by half) of the material (in hrs.).

• A. $6.58$
• B. $8.58$
• C. $10.58$
• D. $12.58$

Answer 1

The correct option is A $6.58$

let the amount of radioactive material be $\alpha$

Given that,$⟹-\frac{d\alpha }{dt}=k\alpha$

Where $k$ is the proportionality constant

$⟹\frac{1}{\alpha }d\alpha =\frac{-1}{k}dt$

let ${\alpha }_0$ be the initial amount taken,

solving the differential equation gives,

$\alpha ={\alpha }_0{e}^{-kt}$

Given that after an hour $10percent$ of the given material has been decayed,

substituting this in the obtained equation gives,

$\frac{9{\alpha }_0}{10}={\alpha }_0{e}^{-k}$

Thus, we can obtain the value of $k$ from this equation.

$k=0.105379$

Now for finding the half life,

$\frac{{\alpha }_0}{2}={\alpha }_0{e}^{-tk}$

$\therefore t_{\frac{1}{2}}=\frac{0.693}{k}$

$⟹t_{\frac{1}{2}}=\frac{0.693}{0.105379}$

$\therefore t_{\frac{1}{2}}=6.58hrs$