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/c²,1 u = 931 MeV/c².)

Answer 1

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"`