Calculate the mass defect and binding energy of \(^{59}_{27}Co\) wh

1.	Calculate the mass defect and binding energy of \(^{59}_{27}Co\) which has a nucleus of mass 58.933 u. [Take mp = 1.0078 u, mn = 1.0087 u]
Answer» Data: m_p = 1.0078 u, m_n = 1.0087 u, m_Co = 58.933 u, 1 u = 931.5 MeV/c² For \(^{59}_{27}Co\), A = 59, Z = 27 ∴ N = A – Z = 59 – 27 = 32 The mass defect, ∆m=(Zm_p + N_n) – m_Co = (27 × 1.0078 + 32 × 1.0087) – 58.933 = (27.2106 + 32.2784) – 58.933 = 59.4890 – 58.933 = 0.556 u ∴ The binding energy = ∆mc² = 0.556 uc² × 931.5 MeV/uc² = 517.8 MeV

Calculate the mass defect and binding energy of \(^{59}_{27}Co\) which has a nucleus of mass 58.933 u. [Take mp = 1.0078 u, mn = 1.0087 u]

Answer»

Data: m_p = 1.0078 u, m_n = 1.0087 u,

m_Co = 58.933 u,

1 u = 931.5 MeV/c²

For \(^{59}_{27}Co\), A = 59, Z = 27

∴ N = A – Z = 59 – 27 = 32

The mass defect,

∆m=(Zm_p + N_n) – m_Co

= (27 × 1.0078 + 32 × 1.0087) – 58.933

= (27.2106 + 32.2784) – 58.933

= 59.4890 – 58.933 = 0.556 u

∴ The binding energy

= ∆mc²

= 0.556 uc² × 931.5 MeV/uc²

= 517.8 MeV