PX282 - D6 - nuclear binding energies

E = m c^{2} \neq (Z m_{p} + N m_{n}) c^{2}

where, $m$ is the rest mass of the atom, $Z$ and $m_{p}$ are the number and mass of protons, and $N$ and $m_{n}$ are the number and mass of neutrons

this inequality shows that there is an energy difference, called the 'binding energy'

E_{n u c l e u s} < E_{p} + E_{n}

the binding energy is the amount of energy required to separate the components of an atom:

E_{b i n d i n g} = (Z m_{p} + N m_{n} - m_{n u c l e u s}) c^{2}

image: david moriaf

4_{1}^{1} H \to_{2}^{4} He

the energy released: $Δ E = 0.0084 m_{H}$ per nucleon
going past He gives energy, whereas going past Fe costs energy
Fe is the most tightly bound per nucleon
eg: $_{6}^{12} C, 6 p, 6 n$
- using the atomic mass unit: $u = 931.494 M e V c^{- 2} = 1.6605 \times 10^{- 27} k g$
- $m_{p} = 1.007276 u$ and $m_{n} = 1.008665 u$
- the mass of individual nucleons: $6 m_{p} + 6 m_{n} = 12.095646 u$
- the mass of the atom: $m_{a t o m} = 11.9967 u$
- the mass difference: $Δ = 0.09895 u = 92.2 M e V c^{- 2}$
- the binding energy: $B E = 92.2 M e V$
eg: $3_{2}^{4} He \to_{6}^{12} C$
- the mass difference: $Δ = 0.0111 u$
- the binding energy: $B E = 10.3 M e V$