#### Question

Find the binding energy per nucleon of `""_79^197"Au"` if its atomic mass is 196.96 u.

(Use Mass of proton m_{p} = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron m_{n} = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c^{2},1 u = 931 MeV/c^{2}.)

#### 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, m_{p} = Mass of proton

M = Mass of nucleus

m_{n} = 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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Solution Find the Binding Energy per Nucleon of 197 79 Au If Its Atomic Mass is 196.96 U. Concept: Mass-energy and Nuclear Binding Energy - Nuclear Binding Energy.