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Question
Find the binding energy per nucleon of `""_79^197"Au"` if its atomic mass is 196.96 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.)
Solution
Given:-
Atomic mass of Au, A = 196.96
Atomic number of Au, Z = 79
Number of neutrons, N = 118
Binding energy,
`B = (Zm_p + Nm_n - M)c^2`
Here, mp = Mass of proton
M = Mass of nucleus
mn = Mass of neutron
c = Speed of light
On substituting the respective values, we get
`B = [(79 xx 1.007276 + 118 xx 1.008665) "u" - 196.96 "u" ] c^2`
`= (198.597274 - 196.96) xx 931 "MeV"`
`= 1524.302094 "MeV"`
Binding energy per nucleon = `1524.3/197 = 7.737 "MeV"`
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