# Question #59bcf

Dec 3, 2014

The answer is $t = 0.05 \sec o n \mathrm{ds}$

The exponential decay of an element can be written as

$N \left(t\right) = {N}_{0} \left(t\right) \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)}$ , where

$N \left(t\right)$ - the quantity that remains and has not yet decayed after a time t;
${N}_{0} \left(t\right)$ - the initial quantity of the substance that will decay;
${t}_{\frac{1}{2}}$ -the half-life of the decaying substance;

Given that ${N}_{0} \left(t\right)$ = $560$ grams and that we need the element to decay to $\frac{1}{4}$ of its original mass, $N \left(t\right)$ is equal to

${N}_{0} \left(t\right) \cdot \frac{1}{4} = 140$ grams

Therefore, we get $140 = 560 \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{0.025}}$, which yields

$\frac{140}{560} = {\left(\frac{1}{2}\right)}^{\frac{t}{0.025}}$ , and $\frac{t}{0.025} = {\log}_{\frac{1}{2}} \left(0.25\right)$

$t = 2 \cdot 0.025 = 0.05$ seconds

Dec 3, 2014

It will take 0.05s.

A quick way is to count the number of Half-Lives which have elapsed. To go 1/2 then 1/4 of the original amount = 2 Half-Lives. So time elapsed = 2 x 0.025 = 0.05 s.

If the numbers don't work out so nicely you will need to use the method described in Stefan's answer.