Y=S+B
K0
–
K+
00
K–
+
I3 = Q + ½Y
K0
The Nucleus
Atoms consist of a positively charged nucleus plus electrons
•Nuclear charge is Ze, where Z is an integer called the atomic number
•This determines what chemical element it is
-e
-e
•The mass/potential energy (E0=mc2) of a neutral atom has three
+Ze
components:
-e
•The mass of the nucleus
-e
•The mass of the electrons – there are Z of these
•The binding energy of the electrons
•Binding energy is tiny, so
mAtom  mNucleus  Zme  mbinding
mH  938.27203 MeV/c 2  .51100 MeV/c 2  .00001 MeV/c 2
mNucleus  mAtom  Zme
Tables of isotopes give the
mass of the neutral atom in u
The Mass of an atom
•Not all neutral atoms of the same element have the same mass
•Atoms come in different isotopes with different masses
•All isotopes have masses that are approximately integer multiples of the same
common unit
•The atomic mass unit (u) is defined as 1/12 of 12C atom
•The integer closest to M/u is called A, the mass number
u
1
12
m  C   931.494 MeV/c
12
2
M Au
Avogadro’s number
•The ratio of u to g is called Avogadro’s number
•Useful for lots of problems
1g
NA 
 6.022  1023
1u
11Li:
11.0438 u
118Sn: 117.9016 u
Naming isotopes
•Isotopes are described by telling their charge Z, their atomic mass number A, and the
name of the chemical symbol
A
A
X
or
X
•The chemical symbol X tells you Z, so normally skipped Z
•Sometimes an isotope has a bit of extra energy – we call it an isomer
•Denoted by putting a * on it
A
X*
•Almost always very unstable
The size of the nucleus
•Can be measured in various ways
•My favorite: replace an electron by the 200 times heavier muon
•Wave function is 200 times smaller
•The wave function responds to the finite nuclear size
•Radius goes crudely as A1/3
1/3
R

R
A
0
•Volume roughly proportional to
number of nucleons
R0  1.2 fm
+Ze
The composition of the nucleus
All normal nuclei have only two types of particles in them:
•The proton has charge +e
0
# Particle
•There are Z of these
+e +e
Z Proton
0
•The neutron has charge 0
N Neutron
•There are N of these
Electron
•Electrons are not found in the nucleus
Mass
1.007276 u
1.008665 u
0.000549 u
•The mass of an atom is protons + neutrons + electrons + binding
•To a crude approximation, this is just the number of protons + neutrons
•This is why the mass is almost an integer
A Z N
What is Z, N, A, and the approximate mass of 235U?
A  235
M  235 u
Z  92
N  235  92  143
Q
+e
0
-e
m  Z  N  u
Radioactivity
•Many nuclei decay over time
•This is a quantum mechanical process – you can’t predict when it will happen
•If you have a lot of atoms, the rate at which they decay will be proportional to the
number of atoms
dN
dN
•The radioactivity destroys the atoms R   N  
  dt
dt
•Integrate to see how number changes
N
 t
with time
N

N
e
0
ln N  t  k
•N is number of atoms
•N0 is initial number of atoms
R  R0 e t
• is the decay rate
•Also, multiply by 
•R is the rate at which atoms are decaying
•R0 is the initial rate
•Half-life, t1/2 is the time it takes for half the atoms to decay
•Let’s find a formula for it
1
2
N 0  N 0e
  t 1/2
e
 t1/2
2
t1/2  ln 2
Sample problem
134Cs
has a half-life of 2.065 y.

•What is the decay rate ?
•If we start with 1.000 g,
what is the initial decay rate? N 
0
•How long must we wait until
the decay rate is less than
1.000 Ci = 3.700  104 s-1?
ln 2
0.692

 0.3357 y1
t1/2
2.065 y
6
23
6
10
6.022

10


M 1.000 10 g


m
133.9067 u
133.9067
 4.497 1015
1
R0   N0   0.3357 y 1  4.496 1015   1.509 10 7 y  4.784 107 s 1
3.166 10 s/y
7 1
R0 4.784 10 s
t
 t

 1293
e 
R  R0 e
4 1
R 3.700 10 s
7.164
7.164
t  ln 1292  7.164 t 
t  21.34 y

1

0.3357 y
15
Particles and anti-particles
•Several particles are important for
understanding nuclear processes
•Protons, neutrons, and electrons have already
been discussed
•The photon is a particle of light
•The neutrino is a massless (or nearly
massless) neutral particle
Anti-Particles
p+
n0
e-


e+

Particle
Proton
Neutron
Electron
Photon
Neutrino
anti-Elec.
anti-Neut.
Mass (MeV)
1.007276 u
1.008665 u
0.000549 u
0.000000 u
0.000000 u
0.000549 u
0.000000 u
•For every particle, there is an anti-particle
•Same mass, opposite charge
•Some particles (the photon) are their own anti-particles
•For nuclear physics, the important ones are the anti-electron and anti-neutrino
Sym.
p+
n0
e

e+

Neutron decay and anti-particles
Particle processes are a lot like equations
•You can turn them around and they still work
•You can move particles to the other side by “subtracting them”
•This means replacing them with anti-particles
•(However, you have to make sure energy works)
•The neutron (in isolation) is an unstable particle
•Decays to proton + electron + anti-neutrino
•This occurs in – decay
n0
p+
•Turn the reaction around
•Put the neutrino on the other side
•This occurs in electron capture
•Put the electron on the other side
•This occurs in + decay
p+
p+
p+
+
+
e-
+
+
+

+

en0
e-
n0
n0
+


+
e+
Calculating Energetics in a decay
Nuclear decay is when an isolated nucleus spontaneously breaks apart
•Typically (not always), there is one Parent nucleus and one Daughter nucleus
•Also, typically, some other particles too
P
D + ?
We want to know how much energy is released
•The potential energy of each component is just mc2
•The difference between these values is Q – the energy available
•Unfortunately, we aren’t given the nuclear masses, just the atomic
QM
c M
nuc 2
P
c  m?c
nuc 2
D
2
M Nuc  M Atom  Zme
Q   M PAtom  Z P me  c 2   M DAtom  Z D me  c 2  m?c 2
Q   M P  M D  c2   Z D  Z P  mec 2  m?c 2
This formula is just a
bridge to the formulas
we really want
•This energy generally appears as kinetic
energy, mostly of the lighter products on the right (the ? particles)
Nuclear Decay Processes
•There are many types of decay processes, we will focus on only the most common
•Our goal is to figure out how to calculate, for those we consider:
•The daughter isotope (Z,A)
P
D + ?
•The energy Q produced
•Whether the process actually occurs
•Processes can occur if Q > 0
•We won’t worry about
•How slowly it goes (some virtually never occur) (higher Q helps)
•Which are more likely than others (higher Q helps)
•– decay
•Electron capture
•+ decay
•Spontaneous fission
• decay
• decay
 decay
–
n0
p+
+
•– is another name for the electron and + for the positron
•A neutron inside a nucleus can decay to a proton
•Example: 3H  3He
+
p
n0 +0 p
ne
•The daughter nucleus:
•Total number of nucleons unchanged
•Charge increases by 1
•(Z,A)  (Z+1,A)
•The change in energy (Q):
Q   M P  M D  c   Z D  Z P  mec  mec  m c
2
2
  M P  M D  c  1 mec  mec  0
2
2
Q  MP  MD c
2
2
2
2
e-
+

Electron capture
p+
+
e-
•A proton in the nucleus captures one of the electrons in the atom
•Example: 7Be  7Li
e•The daughter nucleus:
•Total number of nucleons unchanged
•Charge decreases by 1
•(Z,A)  (Z-1,A)
•The change in energy (Q):
Q   M P  M D  c2   Z D  Z P  mec 2  mec 2  m c 2
  M P  M D  c2   1 mec2  mec2  0
Q   M P  M D  c2
n0
+

p+ 0
n
+
0
p
n +

+
p 0p
n
n0
 decay
+
p+
•A proton in the nucleus decays to a neutron
•Example: 11C  11Be
•The daughter nucleus:
•Total number of nucleons unchanged
•Charge decreases by 1
•(Z,A)  (Z-1,A)
•The change in energy (Q):
Q   M P  M D  c2   Z D  Z P  mec 2  mec 2  m c 2
  M P  M D  c2   1 mec2  mec 2  0
Q   M P  M D  c2  2mec2
n0
+

+
p0+n0p+0
n p
ne+0 +n +
+
p 0 0p
n p+n
e+
Sample problem
What would be the resulting isotope and the
Q-value for each of the following decays of 40K?
(a) - decay (b) electron capture (c) + decay
-
decay: (Z,A)  (Z+1,A)
•Daughter is 40Ca
u  931.494 MeV/c 2
2me c 2  1.022 MeV
Q   M P  M D  c2   39.964000  39.962591 uc2
 .001409 931.5 MeV   1.312 MeV
Z el. A
18 Ar 36
37
38
39
40
42
19 K 39
40
41
42
43
20 Ca 40
41
42
43
44
46
48
mass (u)
35.967547
36.966776
37.965903
38.964314
39.962384
41.963049
38.963708
39.964000
40.961827
41.962404
42.960716
39.962591
40.962279
41.958618
42.858767
43.955481
45.953687
47.952534
Sample problem
What would be the resulting isotope and the
Q-value for each of the following decays of 40K?
(a) - decay (b) electron capture (c) + decay
Electron capture: (Z,A)  (Z-1,A)
•Daughter is 40Ar
Q  MP  MD c
2
u  931.494 MeV/c 2
2me c 2  1.022 MeV
  39.964000  39.962384 uc2
 .001616 931.5 MeV   1.505 MeV
+ decay: (Z,A)  (Z-1,A)
•Daughter is 40Ar
Q   M P  M D  c2  2mec2  1.505 1.022 MeV
 0.483 MeV
Z el. A
18 Ar 36
37
38
39
40
42
19 K 39
40
41
42
43
20 Ca 40
41
42
43
44
46
48
mass (u)
35.967547
36.966776
37.965903
38.964314
39.962384
41.963049
38.963708
39.964000
40.961827
41.962404
42.960716
39.962591
40.962279
41.958618
42.858767
43.955481
45.953687
47.952534
Spontaneous Fission
•A large nucleus has a lot of electrostatic repulsion
•It would like to separate, but strong forces hold it together
•More on this later
•It is possible, but rare for it to break apart into two (or more) pieces
•Commonly, neutrons are emitted as well.
D2
P
0
n
D
n0 1
•A quantum tunneling process
•Very rare when large chunks are involved
•No naturally occurring elements
•We need a small, very stable
chunk to make this work better
•The  particle is such a chunk
 Decay
•The  particle is the nucleus of Helium – it is very stable
•Two protons and two neutrons
•Because it is light, it has a good chance of tunneling out
•The daughter nucleus:
•Nucleons decrease by four
•Charge decreases by two
•(Z,A)  (Z–2,A–4 )
•The change in energy (Q):
0
n
p+ 0p+
D
P
n
Q   M P  M D  c2   Z D  Z P  mec2  m c2
  M P  M D  c2  2mec 2  m c 2
•m  + 2me is just the mass of a helium atom
Q   M P  M D  M 4 He  c2
0
n
p+ 0p+
n
 Decay
•Sometimes, nuclei have internal energy
•Like an atom in an excited state
•Like an atom, the energy comes out in the form of a photon
•The daughter nucleus:
•No change in nucleons
•(Z,A)*  (Z,A )
D
P
•The change in energy (Q):
Q   M P  M D  c2  m c2
Q   M P  M D  c2
How did we get an excited nucleus in the first place?
•Usually a byproduct of a previous nuclear decay
60
To us, this just looks like it
Co 60 Ni**  e 
came from the Cobalt
60
**
60
*
Ni  Ni   1.17 MeV 
60
Ni* 60 Ni   1.33 MeV 
Summary
Decay Z

+2
–
+1
e.c.
–1
+
–1

0
A
+4
0
0
0
0
Formula for Q
(MP – MD – M4He)c2
(MP – MD)c2
(MP – MD)c2
(MP – MD)c2 – 2 mec2
(MP – MD)c2
Radiation Hazards
All of these processes (except electron capture) produce high-energy ionizing
radiation that can be extremely damaging to you
• particles are easily stopped, by paper or dead skin, if they are outside your body
• radiation can penetrate more deeply, so they are more dangerous
• radiation is very penetrating, and hence is most dangerous