Question

The half-life of 40K is 1.30 × 109 y. A sample of 1.00 g of pure KCI gives 160 counts s−1. Calculate the relative abundance of 40K (fraction of 40K present) in natural potassium.

Answer 1

Explanation:

It is given:

$40 \mathrm{K}$ has the half-life period,

$\mathrm{T} 12=1.30 \times 10^{9} \mathrm{years}$

By 1 g of pure KCI, the given count is A = 160 counts/s

Constant of disintegration,

$\lambda=0.693 \mathrm{T} 12$

Now, activity, A = λN

$\Rightarrow 160=N \times 0.693 t 1 / 2$

$\Rightarrow N=1018 \times 9.5$

In 40 grams, $6.023 \times 10^{23}$ atoms are existing.

Thus, in $40 \times 10186.023 \times 9.5 \times 1023 \mathrm{gm}, 9.5 \times 1018$ atoms will be available = 0.00063 gm.

In natural potassium, 40K has the abundance relatively = $(2 \times 0.00063 \times 100) \%=0.12 \%$