InterviewSolution
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InterviewSolution
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The nuclide ratio, `._(1)^(3) H` to `._(1)^(1) H` in a sample of water si `8.0xx10^(-18) : 1` Tritum undergoes decay with a half-line period of `12.3yr` How much tritum atoms would `10.0g` of such a sample conatins `40` year after the original sample is collected? |
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Answer» We know, `18g H_(2)O` has `2N H` atoms in it and `H^(3) : H^(1) : : 8xx10^(-18) : 1` `:. 18g H_(2)O` has `._(1)H^(3)` atoms `= 8xx10^(-18) x6.023xx10^(23) xx2` `:. 10g H_(2)O` has `._(1)H^(3)` atoms `= (8xx10^(-18)xx6.023xx10^(23)xx2xx10)/(18)` i.e., `N_(o) pf ._(1)H^(3) = 5.354 xx 10^(6)` atoms Now `t = (2.303)/(lamda) log_(10) (N_(o))/(N)` `40 = (2.303xx12.3)/(0.693) log_(10) (5.354xx10^(6))/(N)` `N = 5.624xx10^(5)` atoms |
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