NEET-XII-Physics
46: The Nucleus
- #5Find the binding energy per nucleon of
79197Au if its atomic mass is 196.96 u.Ans : Given:
Atomic mass of Au, A = 196.96
Atomic number of Au, Z = 79
Number of neutrons, N = 118
Binding energy, `` B=(\,\mathrm{\,Z\,}{m}_{p}+N{m}_{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=\left[\right(79\times 1.007276+118\times 1.008665)\,\mathrm{\,u\,}-196.96\,\mathrm{\,u\,}]{c}^{2}``
`` =\left(198.597274-196.96\right)\times 931\,\mathrm{\,MeV\,}``
`` =1524.302094\,\mathrm{\,MeV\,}``
Binding energy per nucleon `` =\frac{1524.3}{197}=7.737\,\mathrm{\,MeV\,}``
Page No 442: