Question

PLEASE HELP ME WITH THIS ONE QUESTION

Given the atomic mass of Boron-9 is 9.0133288 u, what is the nuclear binding energy of Boron-9? (Mproton = 1.0078251, Mneutron = 1.0086649, c^2 = 931.5 eV/u)

Answer 1

The binding energy is -882.5 MeV.

Step-by-step explanation:

The nuclear binding energy of Boron-9 can be found as follows:

`B = [Zm_(p) + Nm_(n) - M]c^(2)`

Where:

Z: is the number of protons = atomic number = 5

N: is the number of neutrons = A - Z = 9 - 5 = 4

`m_(p)`: is the mass of the proton = 1.0078251 u

`m_(n)`: 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:

`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`

Therefore, the binding energy is -882.5 MeV.

I hope it helps you!