Question

Calculate the binding energy per nucleon of \[\ce{^84_36Kr}\] whose atomic mass is 83.913 u. (Mass of neutron is 1.0087 u and that of H atom is 1.0078 u).

Answer 1

Given: A = 84, Z = 36, m = 83.913 u, m_n = 1.0087 u, m_H = 1.0078 u

To find: Binding energy per nucleon `(bar"B")`

Formulae: 

1. ∆m = Zm_H + (A - Z)m_n - m
2. B.E. = ∆m × 931.4 MeV
3. `bar"B" = "B.E."/"A"`

Calculation: 

1. ∆m = Zm_H + (A - Z)m_n - m
   = (36 × 1.0078) + (48 × 1.0087) - 83.913
   = 36.2808 + 48.4176 - 83.913
   = 0.7854 u
2. B.E. = ∆m × 931.4 MeV
   = 0.7854 × 931.4
   = 731.4 MeV (by using log table)
3. `bar"B" = "B.E."/"A" = 731.4/84`
   = 8.706 MeV (by using log table)

Binding energy per nucleon of \[\ce{^84_36Kr}\] = 8.706 MeV