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Question
How much energy is released in the following reaction : 7Li + p → α + α.
Atomic mass of 7Li = 7.0160 u and that of 4He = 4.0026 u.
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
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Solution
Given:-
Mass of 7Li = 7.0160 u
Mass of 4He = 4.0026 u.
Reaction :-
`L_i^7 + p → alpha + alpha + E` ,
Energy release (E) is given by
`E = [m(""^7L_i) + (m_p) - 2 xx m ("^4H_e)] c^2`
`= [(7.0160 "u" + 1.007276 "u") - 2(4.0026 "u")]c^2`
`= (8.023273 "u" - 8.0052 "u") c^2`
`= 0.018076 xx 931 "MeV"`
`= 16.83 "MeV"`
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