The neutral helium atom, ${ } _ { 2 } ^ { 4 }$ He, has a mass of $4.002603 \mathrm { u }$, a neutron has a mass of $1.008665 \mathrm { u }$, a proton has a mass of $1.007277 \mathrm { u }$, and a neutral hydrogen atom has a mass of $1.007825 \mathrm { u }$. What is the binding energy of the ${ } _ { 2 } ^ { 4 } \mathrm { He }$ nucleus? $\left( 1 \mathrm { u } = 931.5 \mathrm { MeV } / c ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The binding energy is calculated by finding the mass defect and converting it to energy. The mass defect is the difference between the mass of the separate nucleons and the mass of the nucleus. For ${ } _ { 2 } ^ { 4 } \mathrm { He }$, the mass of 2 protons and 2 neutrons is $2 \times 1.007277 + 2 \times 1.008665 = 4.031884 \mathrm { u }$. The mass defect is $4.031884 - 4.002603 = 0.029281 \mathrm { u }$. The binding energy is $0.029281 \mathrm { u} \times 931.5 \mathrm { MeV/u} = 27.3 \mathrm { MeV }$.

The Neutral Helium Atom 24{ } _ { 2 } ^ { 4 }24​

The Neutral Helium Atom ${ } _ { 2 } ^ { 4 }$