Question

Calculate the binding energy per nucleon for 12C, 78Se, 9Be, and 14N. (For the atomic masses, see this table. Enter your answers to at least two decimal places.)

Answer 1

(a) For ¹⁴N, we have

made up of 7H and 7n.

7 (1.007825) + 7(1.008665) = 14.11543 u

Mass defect is given by,   m = [(14.11543 u) - (14.003074 u)]

m = 0.112356 u

Binding energy is given by, B.E = m c² = (0.112356 u) (931.5 MeV/c²)

B.E = 104.659614 MeV

Binding energy per nucleon will be given as -

B.E / N = (104.659614 MeV) / (14)

B.E / N = 7.477 MeV/nucleon

(b) For ¹²C, we have

made up of 6H and 6n.

6 (1.007825) + 6 (1.008665) = 12.09894 u

Mass defect is given by,   m = [(12.09894 u) - (12 u)]

m = 0.09894 u

Binding energy is given by, B.E = m c² = (0.09894 u) (931.5 MeV/c²)

B.E = 92.1 MeV

Binding energy per nucleon will be given as -

B.E / N = (92.1 MeV) / (12)

B.E / N = 7.67 MeV/nucleon