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To investigate the beta-decay of Mg^(23) radionuclide, a counter was activated at moment t=0. It registered N_(1) beta-particles by a moment t_(1)= 2.0s, and by a moment t_(2)=3t_(1) the number of registered beta-particles was 2.66 times greater. Find the mean lifetime of the given nuclei. |
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Answer» Solution :If `N_(0)` is the number of radionuclei present initially,then `N_(1)=N_(0)(1-e^(-t_(1)//tau))` `etaN_(1)=N_(0)(1-e^(-t_(2)//tau))` where `eta=2.66` and `t_(2)= 3t_(1)`. Then `eta=(1-e^(-t_(2)//tau))/(1-e^(-t_(1)//tau))` or `eta-eta e^(-t_(1)//tau)= 1-e^(-t_(2)//tau)` Substituting the VALUES `1.66= 2.66e^(-2//tau)-e^(-6//tau)` PUT `e^(-2//tau)=x`. Then `x^(3)-2.66x+1.66=0` `(x^(2)-1)x-1.66(x-1)=0` or `(x-1)(x^(2)+x-1.66)=0` Now `x~~1 so x^(2)+x- 1.66=0` `x=(-1+-sqrt(1+4xx1.66))/(2)` NEGATIVE sign has to be rejected as `xgt0`. Thus `x= 0.882` This gives `tau=(-2)/(In 0.882)= 15.9 sec`. |
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