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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.
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Solution
Given:
Half-life period of 40K, `T_"1/2"` = 1.30 × 109 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 × 1023 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 40K in natural potassium = (2 × 0.00063 × 100)% = 0.12%
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