Answer:
The binding energy is -882.5 MeV.
Explanation:
The nuclear binding energy of Boron-9 can be found as follows:
[tex] B = [Zm_{p} + Nm_{n} - M]c^{2} [/tex]
Where:
Z: is the number of protons = atomic number = 5
N: is the number of neutrons = A - Z = 9 - 5 = 4
[tex]m_{p}[/tex]: is the mass of the proton = 1.0078251 u
[tex]m_{n}[/tex]: is the mass of the neutron = 1.0086649 u
M: is the atomic mass of B-9 = 9.0133288 u
c² = 931.5 MeV/u
Hence, the nuclear binding energy is:
[tex] B = [Zm_{p} + Nm_{n} - M]c^{2} = [4*1.0078251 u + 4*1.0086649 u - 9.0133288 u]*931.5 MeV/u = -882.5 MeV [/tex]
Therefore, the binding energy is -882.5 MeV.
I hope it helps you!
