Answer:
2.16 MeV
Explanation:
To determine the amount of work done that is needed to assemble the atomic mass; we need to apply the equation;
U = [tex]\frac{3}{4 \pi E_o} (\frac{e^2}{r})[/tex]
where:
[tex]\frac{1}{4 \pi E_o}[/tex] = proportionality constant = [tex](9*10^9N.m^2/C^2)[/tex]
e = magnitude of the charge of each electron = [tex](1.6*10^{-19} C)^2[/tex]
r = length of each side of the vertex = [tex](2.00*10^{-15}m)[/tex]
So; replacing our values into above equation; we have:
U = [tex]3\frac{(9*10^9N.m^2/C^2)(1.6*10^{-19} C)^2}{(2.00*10^{-15}m)}[/tex]
U = 3.456 × 10 ⁻¹³ J
If we have to convert our unit from J to Mev; then we are going to have:
U = 3.456 × 10 ⁻¹³ J [tex](\frac{1 MeV}{1.602*10^{-13}J})[/tex]
U = 2.16 MeV
Therefore, the amount of work done needed to assemble an atomic nucleus = 2.16 MeV
