Question

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

Given:

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

Mass of proton $=1.007825u$

Mass of neutron $=1.008665u$

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

• A. $18.32MeV$
• B. $8.55MeV$
• 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 $=20\times 1.007825$
$=20.1565u$
Mass of 20 neutrons $=20\times 1.008665$
$=20.1733u$
Total mass $=40.3298u$
___________
Mass ${}_{20}^{40}\text{Ca}$ nucleus $=39.962589u$
Mass defect, $\Delta m$ $=0.367211u$
$Bindingenergy=0.367211\times 931=341.87MeV$
$Bindingenergy\text{per nucleon}=\frac{341.87}{40}=8.547MeV$