Question

Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.

Answer 1

Data : M = 4.00151 u, = 1.00728 u, 

m_n = 1.00866 u, 1 u = 931.5 MeV/c² 

The binding energy of an alpha particle

(Zm_p + N_n - M)c² 

=(2m_p + 2m_n - M)c² 

= [(2)(1.00728u) + 2(1.00866 u) – 4.00151 u]c²

= (2.01456 + 2.01732 – 4.00151)(931.5) MeV 

= 28.289655 MeV 

= 28.289655 × 10⁶ eV × 1.602 × 10^-19 J 

= 4.532002731 × 10^-12 J