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The half-life of ^{40}K is 1.30 × 10^{9} y. A sample of 1.00 g of pure KCI gives 160 counts s^{−1}. Calculate the relative abundance of ^{40}K (fraction of ^{40}K present) in natural potassium.

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#### Solution

Given:

Half-life period of ^{40}K, `T_"1/2"` = 1.30 × 10^{9} years

Count given by 1 g of pure KCI, A = 160 counts/s

Disintegration constant, `lambda = 0.693/T_"1/2"`

Now, activity, A = λN

⇒ `160 = 0.693/t_"1/2" xx N`

⇒ `160 = (0.693/(1.30 xx 10^9 xx 365 xx 86400)) xx N`

⇒ `N = (160 xx 1.30 xx 365 xx 86400 xx 10^9)/0.693`

⇒ `N = 9.5 xx 10^18`

6.023 × 10^{23} atoms are present in 40 gm.

Thus , `9.5 xx 1018` atoms will be present in `(40 xx 9.5 xx 10^18)/(6.023 xx 10^23)` gm.

= `(4 xx 9.5 xx 10^-4)/6.023` gm

= `6.309 xx 10^-4`

= 0.00063 gm

Relative abundance of ^{40}K in natural potassium = (2 × 0.00063 × 100)% = 0.12%

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