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We are given the following atomic masses: `._(92)^(238)U=238.05079u ._(2)^(4)He=4.00260u` `._(90)^(234)Th=234.04363u ._(1)^(1)H=1.00783u` `._(91)^(237)Pa=237.05121u` Here the symbol `Pa` is for the element protactinium `(Z=91)`A. The energy released during the `alpha-` decay of `._(92)^(238)U` is `4.75MeV` approximatelyB. The energy released during the `alpha-` decay of `._(92)^(238)U` is `4.25MeV` approximatelyC. The emission of proton from `._(92)^(238)U` can be spontaneousD. The emission of proton from `._(92)^(238)U` can not be spontaneous. |
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Answer» Correct Answer - B::D The energy released in `alpha`- decay is given by `Q=(M_(U)-M_(Th)-M_(He))c^(2)` Substituting the atomic masses as given in the data, we find `Q=(238.05079-234.04363-4.00260)uxxc^(2)=(0.00456u)c^(2)=(0.00456u)c^(2)` `=(0.00456u)(931.5MeV//u)=4.25MeV` If `._(92)^(238)U` spontaneously emits a proton, the decay process would be `._(92)^(238)Uto_(92)^(237)Pa+._(1)^(1)H` The `Q` for this process to happen is `=(M_(U)-M_(Pa)-M_(H))c^(2)=(238.05079-237.05121-1.00783)uxxc^(2)=(-0.00825u)c^(2)` `=-(0.00825u)(931.5MeV//u)=-7.68MeV` Thus, the `Q` of the process is negative and therefore it cannot proceed spiontaneously. We will have to supply an energy of `7.68MeV` to `._(92)^(238)U` nucleus to make it emit a proton. |
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