Numerical

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

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#### Solution

**Data:** M = 4.00151 u, m_{p} = 1.00728 u, m_{n} = 1.00866 u, 1 u = 931.5 MeV/c^{2}

The binding energy of an alpha particle =

(Zm_{p} + Nm_{n} - M)c^{2}

= (2m_{p} + 2m_{n} - M)c^{2}

= [(2)(1.00728 u) + 2(1.00866 u) - 4.00151 u]c^{2}

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

= 28.289655 MeV

= 28.289655 × 10^{6} eV × 1.602 × 10^{-10} J

= 4.532002731 × 10^{-12} J

Concept: Mass-energy and Nuclear Binding Energy - Nuclear Binding Energy

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