Question

The value of binding energy per nucleon of $_{20}^{40} Ca$ nucleus is

Given:

Mass of $_{20}^{40} Ca$ nucleus $=39.962589 \, u$

Mass of proton $= 1.007825 \, u$

Mass of neutron $= 1.008665 \, u$

and $1 \, u=931 \, MeV \, C^{- 2}$

• A. $18.32MeV$
• B. $8.55 \, MeV$
• C. $9.94MeV$
• D. $14.72MeV$

Answer 1

Answer: (B)

The nucleus $_{20}^{40} \text{Ca}$ contains 20 protons and 20 neutrons.
Mass of 20 protons $= 20 \times 1.007825$
$= 20.1565 \, u$
Mass of 20 neutrons $= 20 \times 1.008665$
$= 20.1733 \, u$
Total mass $= 40.3298 \, u$
___________
Mass $_{20}^{40} \text{Ca}$ nucleus $= 39.962589 \, u$
Mass defect, $\Delta m$ $= 0.367211 \, u$
$Binding \, energy=0.367211\times 931=341.87 \, MeV$
$Binding \, energy \, \text{per nucleon}=\frac{341.87}{40}=8.547 \, MeV$