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Nuclear Physics
Chapter 25
3
Li
lithium
6.941
1
atomic number (# of p+)
symbol
name
average atomic mass
electrons in outer energy level
in the nucleus…
Protons
positive charge
made of two up quarks and a down quark
Electric charge +2/3 e
Electric charge -1/3 e
Neutrons
made of two down quarks and an up quark
= proton
= neutron
Li 7
number of nucleons
(mass number)
Li 6
an isotope of Li
mass number talks about 1 atom
atomic mass is an average of all atoms
Mass is usually expressed in amu or u
1 amu = 1/12th the mass of a carbon 12 atom
1 amu = 1.66 x 10-27 kg = mass of p+ or n0
an n0 is actually about 5 x 10-4 u more massive
an e- has a mass of about 5 x 10-4 u…hmmm
rest energy (E0) = MeV
Mega (1 x 106) electron volt (1.60 x 10-19 J)
E0  mc
2
E0
m 2
c
Mass can also be expressed in MeV/c2
1 proton = 938.3 MeV/c2
mass number
atomic number
(charge)
7
3
Li
The electric force between protons is repulsive
q1q2
Felectric  kc ( 2 )
r
There must be a force stronger than the
electric force holding the nucleus together
The Strong Force is the force that holds
nucleons together.
It is independent of charge, it binds neutrons
and protons.
It only works over very small distances
(atomic).
10-15 m
rnucleus= (1.2x10-15 m )A1/3 where A is the atomic number
Stable “big” atoms have more neutrons
than protons, because the electric force
acts over greater distances.
Nucleons bound together in a nucleus
have less mass (energy) than unbound
nucleons
binding energy is that energy that holds
the nucleus together
Ebind = Dmc2
mass defect
the difference in mass
Find the total binding energy of Al 27 if it has a
mass of 26.981534 u. H 1 has a mass of 1.007825 u
and the mass of a n0 is 1.008665 u. There are
931.50 MeV/u.
Al is atomic # 13, so it has 13(1.007825 u) of p+
and e-, = 13.101725 u.
Al has 14 n0 = 14(1.008665 u) = 14.12131 u
Dm = (13.101725 + 14.12131) u – 26.981534 u =
.241501 u x 931.50 MeV/u = 225 MeV
Nuclear Decay
Alpha Decay: an a particle (He nucleus) escapes
the strong force by quantum tunneling. The
nucleus loses 2 p+ and 2 n0.
212
82
Pb
4
2
He 
208
Hg
80
Beta decay: the weak force causes an up
quark to change into a down quark, or vice
versa. This causes a proton to change into a
neutron or vice versa.
1
0
n p
1
1
neutron proton
0
1
e 
electron
b-
antineutrino
1
1
p  n  e 
proton
1
0
0
1
neutron
positron neutrino
b+
A thallium 208 nucleus emits a b- particle (an
electron). Write the equation.
Tl 
208
81
0
1
e  
208
82
Pb
Gamma Decay: a nucleon can be in an
excited state, needing to get rid of energy to
reach a stable energy level. The nucleon
emits a photon, but the structure of the
nucleus stays the same.
C  C 
14
6
14
6