Question

Calculate the minimum energy required to remove a neutron from the ⁴³Ca nucleus.
MeV

Answer 1

The minimum energy required to remove a neutron from calcium nucleus can be determined by calculating its binding energy.

Now, to calculate the binding energy is to look at the measured mass of the nucleus and the total mass of the constituent nucleons.

E = deltam * c^2

Calcium (Ca) has 20 protons so mp = 20 * 1.00728u = 20.1456u

And -  

23 neutrons so mn = 23 * 1.00866u = 23.19918u

The measured mass of calcium (Ca) = 42.958766u

deltam = 20.1456 + 23.19918 - 42.958766 = 0.386014u

Now -  

1 u = 931.494 MeV / c^2

E = 0.386014 * 931.494 = 359.57 MeV is the nuclear binding energy of a Ca-43 nucleus

The binding energy per nucleon is 359.57 / 43 = 8.36 MeV

Therefore, minimum energy required to remove a neutron from 43Ca nucleus = 8.36 MeV (Answer)