Question #1dc8f

May 1, 2016

It is four.

Explanation:

The half life of any substance is given by:
${t}_{\frac{1}{2}} = \frac{I n 2}{l} a m \mathrm{da}$

Hence

${t}_{\frac{1}{2} \left(1\right)} = \frac{I n 2}{l} a m {\mathrm{da}}_{1}$ .... and.... ${t}_{\frac{1}{2} \left(2\right)} = \frac{I n 2}{l} a m {\mathrm{da}}_{2}$

We want to find:

$R a t i o = {t}_{\frac{1}{2} \left(1\right)} / \left({t}_{\frac{1}{2} \left(2\right)}\right)$

Substituting the previous equations into the above gives
$R a t i o = {t}_{\frac{1}{2} \left(1\right)} / \left({t}_{\frac{1}{2} \left(2\right)}\right) = l a m {\mathrm{da}}_{2} / l a m {\mathrm{da}}_{1}$ [equation A]

For a sample of N atoms, the activity, A, is given by:

$A = l a m \mathrm{da} \cdot N$

Where $l a m \mathrm{da}$ is the decay constant.

So for sample 1: ${A}_{1} = l a m {\mathrm{da}}_{1} \cdot {N}_{1}$ [equation 1]

and for sample 2: ${A}_{2} = l a m {\mathrm{da}}_{2} \cdot {N}_{2}$ [equation 2]

We are told that that ${N}_{1} = 2 {N}_{2}$ or${N}_{1} / 2 = {N}_{2}$
and also ${A}_{2} = 2 {A}_{1}$

Substituting both of these into equation 2 we get:
$2 {A}_{1} = l a m {\mathrm{da}}_{2} \cdot {N}_{1} / 2$
So
${A}_{1} = l a m {\mathrm{da}}_{2} / 4 \cdot {N}_{1}$ [equation 3]

Comparing equation 1 and 3 we can see that:

$l a m {\mathrm{da}}_{1} = l a m {\mathrm{da}}_{2} / 4$

and so $l a m {\mathrm{da}}_{2} / l a m {\mathrm{da}}_{1} = 4$

Hence the ratio of halve lives is 4 (by comparing this result to equation A)