Name: Beta Emission
Uploaded: 2019-07-19
Duration: 8 min 33 s
Description: How do we measure the thickness of extremely thin sheets of paper or other material? Radioactivity plays a role! Learn how beta emissions are used in manufac...

Question

The isotope

K^{40}

can decay to either

{Ca}^{40}

or

{Ar}^{40}

. Assume both decays have a half-life of

1.26 \times 10^{9} y

. The ratio of

Ca

produced to

Ar

produced is

\frac{8.54}{1} = 8.54

. A sample originally had only

K^{40}

. It now has equal amounts of

K^{40}

and

{Ar}^{40}

, that is, the ratio of

K

and

Ar

is

\frac{1}{1} = 1

. How old is the sample? (Hint: Work this like other radioactive-dating problems, except that this decay has two products).

Accepted Answer

The law of exponential decay is given by,
$N = N_{0} e^{- λ t}$
where, $λ$ is the disintegration constant and it is given by,
$λ = \frac{\ln (2)}{T_{\frac{1}{2}}}$
Thus, for $K$ , we have,
$N_{K} = N_{K 0} e^{\frac{- \ln (2) f}{T_{\frac{1}{2}}}} . . . (1)$
The number of nuclei is conserved, so we can write,
$N_{KO} = N_{K} + N_{Ar} + N_{Ca} . . . (2)$

The ratio of $Cn$ produced to $Ar$ produced is _{$8.54 : 1$ ,} that is,
$N_{Ca} = 8.54 N_{Ar}$
Substituting in equation $(2)$ , we get,
$N_{K 0} = N_{K} + N_{Ar} + 8.54 N_{Ar}$

$N_{K 0} = N_{K} + 9.54 N_{Ar}$
Substituting in equation $(1)$ , we get,
$N_{K} = (N_{K} + 9.54 N_{Ar}) e^{- \frac{\ln (2) t}{T_{\frac{1}{2}}}}$

We need to find the time when the sample contains equal amounts of $K$ and $R$ , i.e., $N_{Ar} = N_{K}$ . So,
$N_{K} = (N_{K} + 9.54 N_{K}) e^{- \frac{\ln (2) t}{T_{\frac{1}{2}}}}$

$1 = (10.54) e^{- \frac{\ln (2) t}{T_{\frac{1}{2}}}}$
Solving for $t$ , we get,
$\frac{\ln (2) t}{T_{\frac{1}{2}}} = \ln (10.54)$
$l = \frac{T_{\frac{1}{2}} \ln (10.54)}{\ln (2)}$
Both the decays have a half-life time of $1.26 \times 10^{9} y$ . Substituting this,
$t = \frac{(1.26 \times 10^{9} y) \ln (10.54)}{\ln (2)}$

$= 4.28 \times 10^{9} y$

$t = 4.28 \times 10^{9} y$

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