1

Nucleons

size: ~1 fm

Nuclei

a bunch of nucleons bound together create a potential for an additional :

nucleons attract each other via the strong force ( range ~ 1 fm)

neutron

proton�(or any other charged particle)

V

r

R

V

r

R

Coulomb Barrier V_c

Potential

Potential

R ~ 1.3 x A^1/3 fm

🡪 Nucleons are bound by attractive force. Therefore, mass of nucleus� is smaller than the total mass of the nucleons by the binding energy dm=B/c²

2

Nuclear Masses and Binding Energy

Energy that is released when a nucleus is assembled from neutrons and protons

m_p = proton mass, m_n = neutron mass, m(Z,N) = mass of nucleus with Z,N

• B>0
• With B the mass of the nucleus is determined.
• B is very roughly ~A�

Most tables give atomic mass excess Δ in MeV:

Masses are usually tabulated as atomic masses

(so for ¹²C: Δ=0) (see nuclear wallet cards for a table)

Nuclear Mass

~ 1 GeV/A

Electron Mass

511 keV/Z

Electron Binding Energy�13.6 eV (H)�to 116 keV (K-shell U) / Z

m = m_nuc + Z m_e - B_e

3

Q-value

Energy released in a nuclear reaction (>0 if energy is released, <0 if energy is used)

Example: The sun is powered by the fusion of hydrogen into helium:

4p 🡪 ⁴He + 2 e⁺ + 2ν_e

Mass difference dM�released as energy�dE = dM c²

(using nuclear masses !)

4

In practice one often uses mass excess Δ and atomic masses.

As A is always conserved in nuclear reactions the mass excess Δ can always�be used instead of the masses (the Am_u term cancels)

Q-value with atomic masses:

If Z is conserved (no weak interaction) atomic masses can be used instead�of nuclear masses (Zme and most of the electron binding energy cancels)

Otherwise: For each positron emitted subtract 2m_e /c²= 1.022 MeV from the Q-value

Example:

4p 🡪 ⁴He + 2 e⁺ + 2ν_e

Z changes an 2 positrons are emitted

With atomic masses

Q-value with mass excess Δ�

(as nucleon masses cancel on both sides, its really the binding energies that�entirely determine the Q-values !)

With atomic mass excess

5

The liquid drop mass model for the binding energy: (Weizaecker Formula)

(assumes incompressible fluid (volume ~ A) and sharp surface)

x 1 ee

x 0 oe/eo

x (-1) oo

Volume Term

Surface Term ~ surface area (Surface nucleons less bound)

Coulomb term. Coulomb repulsion leads to reduction� uniformly charged sphere has E=3/5 Q²/R

Asymmetry term: Pauli principle to protons: symmetric filling� of p,n potential boxes has lowest energy (ignore Coulomb)

protons

neutrons

neutrons

protons

lower total�energy =�more bound

Pairing term: even number of like nucleons favoured

(e=even, o=odd referring to Z, N respectively)

(each nucleon gets bound by about same energy)

and in addition: p-n more bound than p-p or n-n (S=1,T=0 more�bound than S=0,T=1)