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Nuclei — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsNuclei3 Marks
Question
Calculate the binding energy per $^{40}_{20}\text{CA}$ nucleon nucleus.
[Given: m $\big(^{40}_{20}\text{Ca}\big)=39.962589\text{u}$
$m_n($mass od a neutron$) = 1.008665u$
$m_p($mass of a proton$) =1.007825u$
$1u = 931\ MeV/c^2]$
✓

Answer

Total Binding energy of $^{40}_{20}\text{Ca}$ nucleus $=20\text{m}_\text{p}+20\text{m}_\text{n}-\text{M}\big(^{40}_{20}\text{Ca}\big)$
$= 20 × 1.007825 + 20 × 1.008665 - 39.962589$
$= 0.367211 u = 0.367211 × 931\ MeV = 341.87\ MeV$
$\therefore$ Binding energy per nucleus $=\frac{341.87}{40}\text{MeV}/ \text{nucleon}$
$=8.55\text{MeV}/ \text{nucleon}$

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