Question #1dc8f

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Question #1dc8f

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Answer 1 · 2016-05-01T12:31:54

It is four.

Explanation:

The half life of any substance is given by:
$t_{\frac{1}{2}} = \frac{I n 2}{λ}$ $t_(1/2)=(In2)/lamda$

Hence

$t_{\frac{1}{2} (1)} = \frac{I n 2}{λ_{1}}$ $t_(1/2(1))=(In2)/lamda_1$ .... and.... $t_{\frac{1}{2} (2)} = \frac{I n 2}{λ_{2}}$ $t_(1/2(2))=(In2)/lamda_2$

We want to find:

$R a t i o = \frac{t_{\frac{1}{2} (1)}}{t_{\frac{1}{2} (2)}}$ $Ratio=t_(1/2(1))/(t_(1/2(2)))$

Substituting the previous equations into the above gives
$R a t i o = \frac{t_{\frac{1}{2} (1)}}{t_{\frac{1}{2} (2)}} = \frac{λ_{2}}{λ_{1}}$ $Ratio=t_(1/2(1))/(t_(1/2(2)))=lamda_2/lamda_1$ [equation A]

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

$A = λ \cdot N$ $A=lamda*N$

Where $λ$ $lamda$ is the decay constant.

So for sample 1: $A_{1} = λ_{1} \cdot N_{1}$ $A_1=lamda_1*N_1$ [equation 1]

and for sample 2: $A_{2} = λ_{2} \cdot N_{2}$ $A_2=lamda_2*N_2$ [equation 2]

We are told that that $N_{1} = 2 N_{2}$ $N_1=2N_2$ or $\frac{N_{1}}{2} = N_{2}$ $N_1/2=N_2$
and also $A_{2} = 2 A_{1}$ $A_2=2A_1$

Substituting both of these into equation 2 we get:
$2 A_{1} = λ_{2} \cdot \frac{N_{1}}{2}$ $2A_1=lamda_2*N_1/2$
So
$A_{1} = \frac{λ_{2}}{4} \cdot N_{1}$ $A_1=lamda_2/4*N_1$ [equation 3]

Comparing equation 1 and 3 we can see that:

$λ_{1} = \frac{λ_{2}}{4}$ $lamda_1=lamda_2/4$

and so $\frac{λ_{2}}{λ_{1}} = 4$ $lamda_2/lamda_1=4$

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

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Question #1dc8f

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