Question

1. Calculate the total binding energy of 12

Answer 1

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:

$B = (Zm_{p} + Nm_{n} - M)*931 MeV/C^{2}$

Where:

Z: is the number of protons

$m_{p}$: is the proton's mass = 1.00730 u

N: is the number of neutrons

$m_{n}$: is the neutron's mass = 1.00869 u

M: is the mass of the nucleus

1. The total binding energy of $^{12}_{6}C$ is:

$B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u$

$B = (6*1.00730 + 6*1.00869 - 12.011)*931.49 MeV/u = 79.12 MeV$

 

2. The average binding energy per nucleon of $^{24}_{12}Mg$ is:

$B = \frac{(Zm_{p} + Nm_{n} - M)}{A}*931.49 MeV/u$

Where: A = Z + N

$B = \frac{(12*1.00730 + 12*1.00869 - 24.305)}{(12 + 12)}*931.49 MeV/u = -4.39 MeV/nucleon$                                  

   

3. The average binding energy per nucleon of $^{85}_{37}Rb$ is:

$B = \frac{(Zm_{p} + Nm_{n} - M)}{A}*931.49 MeV/u$

$B = \frac{(37*1.00730 + 48*1.00869 - 85.468)}{85}*931.49 MeV/u = 2.40 MeV/nucleon$

     

4. The binding energy per nucleon of $^{238}_{92}U$ is:

$B = \frac{(92*1.00730 + 146*1.00869 - 238.03)}{238}*931.49 MeV/u = 7.48 MeV/nucleon$

 

5. The total binding energy of $^{20}_{10}Ne$ is:

$B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u$

$B = (10*1.00730 + 10*1.00869 - 20.180)*931.49 MeV/u = -18.72 MeV$

6. The total binding energy of $^{40}_{20}Ca$ is:

$B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u$

$B = (20*1.00730 + 20*1.00869 - 40.078)*931.49 MeV/u = 225.23 MeV$

I hope it helps you!