Question

Question #1dc8f

Answer 1

It is four.

Explanation:

The half life of any substance is given by:
`t_(1/2)=(In2)/lamda`

Hence

`t_(1/2(1))=(In2)/lamda_1` .... and.... `t_(1/2(2))=(In2)/lamda_2`

We want to find:

`Ratio=t_(1/2(1))/(t_(1/2(2)))`

Substituting the previous equations into the above gives
`Ratio=t_(1/2(1))/(t_(1/2(2)))=lamda_2/lamda_1` [equation A]

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

`A=lamda*N`

Where `lamda` is the decay constant.

So for sample 1: `A_1=lamda_1*N_1` [equation 1]

and for sample 2: `A_2=lamda_2*N_2` [equation 2]

We are told that that `N_1=2N_2` or`N_1/2=N_2`
and also `A_2=2A_1`

Substituting both of these into equation 2 we get:
`2A_1=lamda_2*N_1/2`
So
`A_1=lamda_2/4*N_1` [equation 3]

Comparing equation 1 and 3 we can see that:

`lamda_1=lamda_2/4`

and so `lamda_2/lamda_1=4`

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