NEET-XII-Physics
46: The Nucleus
- #36The 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.Ans : Given:
Half-life period of 40K, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}`` = 1.30 × 109 years
Count given by 1 g of pure KCI, A = 160 counts/s
Disintegration constant, `` \lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}``
Now, activity, A = λN
`` \Rightarrow 160=\frac{0.693}{{t}_{1/2}}\times N``
`` \Rightarrow 160=\left(\frac{0.693}{1.30\times {10}^{9}\times 365\times 86400}\right)\times N``
`` \Rightarrow N=\frac{160\times 1.30\times 365\times 86400\times {10}^{9}}{0.693}``
`` \Rightarrow N=9.5\times {10}^{18}``
6.023 × 1023 atoms are present in 40 gm.
`` \,\mathrm{\,Thus,\,}\,\mathrm{\,9.5\; \times \; 1018\,}\,\mathrm{\,atoms\,}\,\mathrm{\,will\,}\,\mathrm{\,be\,}\,\mathrm{\,present\,}in\frac{40\times 9.5\times {10}^{18}}{6.023\times {10}^{23}}\,\mathrm{\,gm\,}.``
`` =\frac{4\times 9.5\times {10}^{-4}}{6.023}\,\mathrm{\,gm\,}``
`` =6.309\times {10}^{-4}``
`` =0.00063\,\mathrm{\,gm\,}``
Relative abundance of 40K in natural potassium = (2 × 0.00063 × 100)% = 0.12%
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