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Calculate the energy released in the fol...

Calculate the energy released in the following:
`._(1)H^(2) + ._(1)H^(3) rarr ._(2)He^(4) + ._(0)n^(1)`
(Given masses : `H^(2) = 2.014, H^(3) = 3.016, He = 4.003, n = 1.009 m_(u))`

Text Solution

Verified by Experts

Mass on the reactant side
`= 2.014 + 3.016 = 5.030 m_(u)`
Mass on the product side
`= 4.003 + 1.009 = 5.012`
Mass lose `= 5.030 - 5.012 = 0`
Energy released per atom of helium
`(0.018 xx 931) MeV = 16.76 MeV`
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