The neutral deuterium atom, $^{2}_{1}$H, has a mass of 2.014102 u; a neutral ordinary hydrogen atom has a mass of 1.007825 u; a neutron has a mass of 1.008665 u; and a proton has a mass of 1.007277 u. What is the binding energy of the deuterium nucleus? (1 u = 931.5 MeV/c²)

Question

Quizplus · Accepted Answer

The mass of the deuterium nucleus is:
$2(1.007277 	ext{ u})+1(1.008665 	ext{ u}) = 2.014719 	ext{ u}$
The binding energy is then given by the mass defect:
$\Delta m = 2.014719 	ext{ u} - 2.014102 	ext{ u} = 0.000617 	ext{ u}$
$E = \Delta m 	imes 931.5 	ext{ MeV/u} = 0.000617 	ext{ u} 	imes 931.5 	ext{ MeV/u} = 0.5751 	ext{ MeV} \approx 2.2 	ext{ MeV}$

The Neutral Deuterium Atom 12^{2}_{1}12​ H, Has a Mass of 2

The Neutral Deuterium Atom $^{2}_{1}$ H, Has a Mass of 2