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Answer 1 · 2016-05-01T12:31:54

It is four.

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)