Question

Calculate the binding energy per nucleon of ₂₀Ca⁴⁰ nucleus M(₂₀Ca⁴⁰) = 39.962589 u, m_n = 1.008665 u, m_p = 1.007825 u:

पर्याय

• 8.54 MeV/N

• 9.54 MeV/N

• 8.46 MeV/N

• 10.56 MeV/N

Answer 1

8.54 MeV/N

Explanation:

Number of protons = 20

Number of neutrons = 40 - 20 = 20

Total mass of 20 protons and 20 Neutrons

= 20 m_p + 20 m_n = 20(m_p + m_m)

= 20(1.007825 + 1.008665)

= 40.32984

Mass defect, Δ m = 40.3298 - 39.962589

= 0.367211 u

Total B.E.  0.367211 u

= 341.873441 mev

B.E/N = `(341.873441)/41`

= 8.54 MeV/N