Answer:
1. B = 79.12 MeV
2. B = -4.39 MeV/nucleon
3. B = 2.40 MeV/nucleon
4. B = 7.48 MeV/nucleon
5. B = -18.72 MeV
6. B = 225.23 MeV
Explanation:
The binding energy can be calculated using the followng equation:
[tex] B = (Zm_{p} + Nm_{n} - M)*931 MeV/C^{2} [/tex]
Where:
Z: is the number of protons
[tex]m_{p}[/tex]: is the proton's mass = 1.00730 u
N: is the number of neutrons
[tex]m_{n}[/tex]: is the neutron's mass = 1.00869 u
M: is the mass of the nucleus
1. The total binding energy of [tex]^{12}_{6}C[/tex] is:
[tex] B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u [/tex]
[tex] B = (6*1.00730 + 6*1.00869 - 12.011)*931.49 MeV/u = 79.12 MeV [/tex]
2. The average binding energy per nucleon of [tex]^{24}_{12}Mg[/tex] is:
[tex] B = \frac{(Zm_{p} + Nm_{n} - M)}{A}*931.49 MeV/u [/tex]
Where: A = Z + N
[tex] B = \frac{(12*1.00730 + 12*1.00869 - 24.305)}{(12 + 12)}*931.49 MeV/u = -4.39 MeV/nucleon [/tex]
3. The average binding energy per nucleon of [tex]^{85}_{37}Rb[/tex] is:
[tex] B = \frac{(Zm_{p} + Nm_{n} - M)}{A}*931.49 MeV/u [/tex]
[tex] B = \frac{(37*1.00730 + 48*1.00869 - 85.468)}{85}*931.49 MeV/u = 2.40 MeV/nucleon [/tex]
4. The binding energy per nucleon of [tex]^{238}_{92}U[/tex] is:
[tex] B = \frac{(92*1.00730 + 146*1.00869 - 238.03)}{238}*931.49 MeV/u = 7.48 MeV/nucleon [/tex]
5. The total binding energy of [tex]^{20}_{10}Ne[/tex] is:
[tex] B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u [/tex]
[tex] B = (10*1.00730 + 10*1.00869 - 20.180)*931.49 MeV/u = -18.72 MeV [/tex]
6. The total binding energy of [tex]^{40}_{20}Ca[/tex] is:
[tex] B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u [/tex]
[tex] B = (20*1.00730 + 20*1.00869 - 40.078)*931.49 MeV/u = 225.23 MeV [/tex]
I hope it helps you!
