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mrs_skeptik [129]
3 years ago
10

1. Calculate the total binding energy of 12

Physics
1 answer:
ahrayia [7]3 years ago
8 0

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}

<u>Where</u>:

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

<u>Where</u>: 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!

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8 0
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Vadim26 [7]

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<u>Explanation:</u>

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7 0
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