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The half- life of .^238U for alpha-decay is 4.5xx10^19 years. How many disintegrations per second occur in 1g of .^238U ? (Avogadro.s number = 6.023 xx 10^23 "mol"^(-1) ) |
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Answer» Solution :Half-life `T=4.5xx10^9xx365xx86400s=1.419xx10^17` s `lambda=0.693/T = 0.693/(1.429xx10^17)=4.882xx10^(-19) s^(-1)` Number of `.^238U` atoms in 1g , `N="Avogardo.s number"/"MASS number"=(6.023xx10^23)/238=2.530xx10^21` Number of disintegration per second, `(dN)/(dt)=lambdaN=4.882xx10^(-18)xx2.530xx10^21` `=1.235xx10^4 s^(-1)` |
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