The half-life of `40 k` is `1.30xx10^9` y. A sample if 1.00g of pure KCI gives 160 counts `s^(-1)`.Calculate the relative abundance of ``^40K`(fraction of ``^40K`present ) in natural potassium .

Answer

A, B

Explanation

Given: Half life period
` t_(1/2) = 1.30 xx 10^(9) year , `
` A = 160 counts//s `
` = 1.30 xx 10^(9) xx 365 xx 86400`
` A = lambda N`
`rArr 160 = (0.693)/(t_(1/2)) N `
` rArr 160 = ((0.693)/(1.30 xx 10^(9) xx 365 xx 86400)) N `
` rArr N = (160 xx 1.30 xx 365 xx 86400 xx 10^(9))/(0.693) `
` = 9.5 xx 10^(18) `
`:. 6.023 xx 10^(23) ` No. of present in 40 gm
` :. 9.5 xx 10 ` present in `
`= ( 40 xx 9.5 xx 10^(18))/(6.023 xx10^(23)) `
` = ( 4 xx 9.5 xx 10^(-4))/(6.023) gm`
` 6.309 xx 10 ^(-4) = 0.00063 gm `
:. The relative at 40 K in natural potassium
` = (2 xx 0.00063 xx 100) % = 0.12 %` .