Question

Calculate the binding energy per nucleon for the formation of \(^4_2He\) nucleus. Mass of \(^4_2He\) atom = 4.0026 u.

Answer 1

Given: m = 4.0026 u

Z = 2, A = 4 

To find: Binding energy per nucleon (\(\bar{B}\) ) 

Formulae: i. Δm = Zm_H + (A – Z)m_n – m 

ii. B.E. = Δm × 931.4 MeV

iii. \(\bar{B}\) = \(\frac{B.E.}A\)

The mass defect, Δm = [Zm_H + (A – Z)m_n ] – m Δm = [(2 × 1.0078) + (2 × 1.0087)] – 4.0026 = 0.0304 u

Total binding energy, B.E. (MeV) = Δm (amu) × 931.4 

= 0.0304 × 931.4 

= 28.315 MeV

iv. B.E. per nucleon, \(\bar{B}\) = \(\frac{B.E.}A\)

 \(\bar{B}\) = \(\frac{28.315}4\) = 7.079 Mey/nucleon

Binding energy per nucleon for formation of \(^4_2He\) nucleus = 7.079 MeV/nucleon