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How many disintegrations per second will occur in one gram of `._92U^(238)`, if its half life against alpha decay is `1.42xx10^(17)s` ? |
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Answer» Given Half-life period `(T)=(0.693)/(lambda)=1.42xx10^(17)s` `lambda=(0.693)/(1.42xx10^(17))=4.88xx10^(-18)` Avagadro number `(N)=6.023xx10^(23)` atoms `n=` Number of atoms present in `1 g` of `._(92)^(238)U = (N)/(A)` `=(6.023xx10^(23))/(238)=25.30xx10^(20)` Number of disintegrations`=(dN)/(dt)=lambdan` `=4.88xx10^(-18)xx25.30xx10^(20)` `=1.2346xx10^(4)` disintergrates/sec |
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