Question

The binding energy per nucleon of $_{20}^{40}Ca$ is…
($_{20}^{40}Ca = 39.962589$, $m_{p} = 1.007825 amu$, $m_{n} = 1.008665 amu$
 and $1\ amu$ is equivalent to $931.5 MeV$)

Answer 1

Hint: Nuclear binding energy is the energy needed to leave an atomic nucleus into its component protons and neutrons, or the energy released by joining individual protons and neutrons into a particular nucleus. The total mass of the forced particles is less than the total of the masses of the separate particles by a quantity equivalent to the binding energy.

Complete step-by-step solution:
$_{20}^{40}Ca$ has $20$ protons and $20$ neutrons.
Total Binding Energy of $_{20}^{40}Ca$ is $20 m_{p} + 20 m_{n} – M(_{20}^{40}Ca)$
$m_{p}$ is the mass of proton and it is equal to:
$m_{p} = 1.007825 amu$
$m_{n}$ is the mass of neutron and it is equal to:
$m_{n} = 1.008665 amu$
Now, we evaluate total binding energy = $20 m_{p} + 20 m_{n} – M(_{20}^{40}Ca)$
$=20 \times 1.007825 + 20 \times 1.008665 – 39.962589$
This gives,
Total binding energy = $0.367211$ amu
Now, we will convert it into MeV.
Total binding energy = $0.367211 \times 931.5 = 342.05$ MeV
We have to find the binding energy per nucleons. There are nucleons.
So, binding energy per nucleons = $\dfrac{342.05}{40} = 8.55$ MeV.
The binding energy per nucleon of $_{20}^{40}Ca$ is $8.55$ MeV.

 Note: Binding energy, the energy needed to separate a particle from an arrangement of particles or separate all the system particles. Binding energy is particularly applicable to subatomic particles in tiny nuclei, electrons forced to nuclei in atoms, and atoms and ions wrapped together in crystals.