JEE Mains 12-Apr-2023 Shift 1 Solved Paper
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Question : 53 of 90
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A common example of alpha decay is
92 238 U → 90 234 Th + 2 He 4 + Q
Given :
92 238 U = 238.05060 u ,
234 Th = 234.04360 u ,
90
2 4 He = 4.00260 u and
lu = 931.5
The energy released (Q) during the alpha decay of238 U is _______ MeV
The energy released (Q) during the alpha decay of
[12-Apr-2023 shift 1]
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