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Find energy released in the alpha decay, Given `_92^238Urarr_90^234Th+_2^4He` `M(`_92^238U)=238.050784u` `M(`_90^234Th)=234.043593u` `M(`_2^4He)=4.002602u` |
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Answer» Correct Answer - B::D Mass defect `summ_1-summ_f=Deltam` `=(238.050784)-(234.043593+4.002602)` `=4.589xx10^-3u` Energy released `=Deltamxx931.48MeV` `=4.27 MeV` |
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