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Calculate the energy in the reaction 2...

Calculate the energy in the reaction
`2 ._(1)^(1)H + 2 ._(0)^(1)n to ._(2)^(4)He`
Given, H = 1.00813 amu, n = 1.00897 amu and He = 4.00388

Text Solution

Verified by Experts

Loss of mass in the given nuclear reaction
= 2 (1.00813 + 1.00897) - 4.00388
= 0.03032 amu
energy released `= 0.03032 xx 931 = 28.3 MeV`
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