Find energy released in the alpha decay, Given `_92^238Urarr_90^234Th+_2^4He` `M(`_92^238U)=238.050784u` `M(`_90^234Th)=234.0435

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`

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)=4.002

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