
A
-decay: Z
X N  Y + 42
Conservation of Energy
2
2
p
pY
Q

2mY 2m
Q
p 2
2 m
A-4
Z-2 N-2
2
(mX  mY  m )c2  TY  T
Q
For a parent AX
nucleus at rest:
pY   p
1  m / mY   T 1  m / mY 

Q  T 1  m / mY

T 
Note: for heavy nuclei
Q
1
m
mY
Q

1  4A
T  Q
to within ~98% accuracy, anyway
We’ll see from a few examples that typically
T  4-5 MeV
Repeating an OLDIE but GOODIE from Lecture 13 on “Radiation”:
Is 236
94 Pu unstable to -decay?
Pu 
236
94
U+  +Q
232
92
4
2
Q = (MPu – MU  M)c2
= (236.046071u – 232.037168u – 4.002603u)931.5MeV/u
= 5.87 MeV > 0
Some (especially the heaviest) nuclei are unstable
with respect to the emission of heavy particles
•essentially the break up of a nucleus.
In one extreme: the emission of a single nucleon
but it includes the far more common alpha emission
and fission of the original nucleus
into smaller, approximately equal sized nuclei.
Table 8.1 Energy Release (Q value) for various modes of decay of 232U
Emitted particle
n
1H
2H
3H
3He
4He
5He
6He
6Li
7Li
Energy Released (MeV)
-7.26 MeV
-6.12
-10.70
-10.24
-9.92
+5.51 MeV
-2.59
-6.19
-3.79
-1.94
repulsive
Coulomb potential
attractive
nuclear
potential
Let’s follow:
The calculation
of the kinetic
energy of an
alpha particle
emitted by the
nucleus 238U.
The model for
this calculation
is illustrated
on the potential
energy diagram
at the right.
stepping through the details:
This potential energy curve
combines a nuclear well
of radius 7.75 fm
(from R = 1.25 x A1/3 fm)
and the Coulomb
potential energy
of an alpha in the electric field
of the daughter 234Th nucleus.
The mean binding energy per nucleon B/A
for 238U
(from the Semi-empirical mass formula)
is 7.5 MeV.
Thus to remove 4 average nucleons would require 30 MeV.
Compare to using the
semi-empirical mass formula
to calculate the energy
required to remove
2 protons and 2 neutrons
from the highest 238U
energy levels.
24.4 MeV
assumes they are the last two particles of each type
added to the 234Th nucleus.
For the alpha particle Dm= 0.03035 u which gives
28.3 MeV binding energy!
 
protons 2  1.00728 u
N
neutrons 2  1.00866 u
N
Mass of parts
4.03188 u
N

N
Alpha
particle
Mass of alpha 4.00153 u
1 u = 1.66054  10-27 kg = 931.494 MeV/c2
The model for alpha emission proposes that the
alpha particle is preformed inside the nucleus.
The binding energy released (28.3 MeV)
appears in part as kinetic energy of the alpha.
An alpha particle with positive energy
is created inside the nucleus
where it is trapped by the potential barrier.
According to quantum mechanics it has a finite probability of escape.
Let’s see how well quantum mechanics
and our model of the potential
can calculate that probability (decay rate)
Tunneling
finite (but small) probability
of being found outside
the nucleus at any time
Nuclear
potential
Coulomb potential
always some probability of a
piece of the nucleus escaping
the nuclear potential
with a STATIC POTENTIAL this probability is CONSTANT!
Let’s examine this through a simple model: an  forms inside the nucleus
and then escapes through quantum mechanical barrier penetration.
The potential seen by the  is spherically symmetric, so we can start by
first separating the variables -
(r )  Rnl (r )Ylm (, )
the functions Ylm are the same spherical harmonics
you saw for the wavefunctions for the hydrogen atom.
Then the equation for the radial function Rnl(r) can be written as
2

l
(
l

1
)

2
2
2
d (rRnl ) / dr  2m /   E  V (r ) 
rRnl  0
2 
2m r 

2

l
(
l

1
)

d 2 (rRnl ) / dr 2  2m /  2  E  V (r ) 
rRnl  0
2 
2m r 

For states without orbital angular momentum ( l = 0) this
reduces to an equation like that for a 1-dimensional barrier.
The transmitted part of the wave function X is of the form
X e

where
2m

V
(
r
)

E
dr
2 

The integral is carried out over the range of the potential barrier.
A solution can be found by approximating the shape of
the potential as a succession of thin rectangular barriers.
In this case the inner limit of the integral is effectively the nuclear
radius R, and the outer limit is taken as the point at which the
’s kinetic energy is equal in magnitude to its potential energy.
In simple 1-dimensional case
V
E
I
II ( x)  Ce
where
III
II
k2 x
 k2 x
 De
2m
k2 
(
V

E
)
2
In simple 1-dimensional case
V
E
x = r1
I
x = r2
III
II
probability of tunneling to here
2 2k2r2
| II ( x  r2 ) |  D e
2
the point at
which the
’s kinetic
energy is
equal to its
potential
energy.
Where
1 2(Z  2)e
T 
40
r2
2
E
r2
R
So let’s just write
1 2(Z  2)e
V (r ) 
40
r
2
r2
as V ( r )  T
r
2m
hence
  2 2  V ( r )  E dr

2mT
r2
with E=T becomes  2

1
dr

2
r
then with the substitutions:
r / r2  cos 
2
dr  2r2 sin  cosd
r2
tan 
1
r
2m T
2
(tan)(2r2 sin  cosd) 
2 

2m T
2
 2
2r sin d
2  2

Performing the integral yields:
2 ax dx  x  sin 2ax
sin

2
4a

2mT (Z  2)e2
1
2
2
cos
R / r2 
2
40T

r / r2  cos2 
1 2(Z  2)e2
T 
40
r2
R / r2 1  R / r2 
and for R << r2 the term in the square brackets reduces to

 2  2

R / r2 

2mT (Z  2)e2
2
2
2
40T



 2  2 R / r2 


into which we can again substitute for r2 from
1 2(Z  2)e2
T 
40
r2
1 2( Z  2)e 2
r2 
40
T
and get
( Z  2)e

2 0 
2
2m 8e m R( Z  2)

T

4 0
When the result is substituted into the exponential the expression
for the transmission becomes
X | (r ) |  e
2



 A( Z  2)

 exp 
 B R( Z  2) 
T




The decay probability is  = f X where f is the frequency with
which the alpha particle hits the inside of the barrier. Thus
A(Z  2)
ln   ln f  B R(Z  2) 
T
f can be estimated from crude time between striking nuclear barrier
2( A1 / 3r0 ) 2(6 fm)

v
v
of 4-8 MeV “pre-formed” alpha
f  10 times/ second
22
Easily giving estimates for  = 106/sec – 10-21/sec
Some Alpha Decay Energies and Half-lives
Isotope
232Th
238U
230Th
238Pu
230U
220Rn
222Ac
216Rn
212Po
216Rn
T(MeV)
t1/2
(sec-1)
4.01
4.19
4.69
5.50
5.89
6.29
7.01
8.05
8.78
8.78
1.41010 y
4.5109 y
8.0104 y
88 years
20.8 days
56 seconds
5 seconds
45.0 msec
0.30 msec
0.10 msec
1.61018
4.91018
2.81013
2.51010
3.9107
1.2102
0.14
1.5104
2.3106
6.9106
this quantum mechanically-motivated relation
A(Z  2)
ln   ln f  B R(Z  2) 
T
should be compared with the emperical Geiger-Nuttall law
ln   C ln T  D
232Th
Q=4.08 MeV
t=1.4×1010 yr
218Th
Q=9.85 MeV
t=1.0×10-7 sec
The dependence of alpha-decay half-life on the kinetic energy of the alpha particle.
Values are marked for some isotopes of thorium.
For each series of isotopes the
experimental data agree (1911)
The potential seen by an electron in the hydrogen atom
1 e
U (r) 
4 0 r
2
is spherically symmetric
(depends only on r, not its direction)!
Recognizing that we write Schrödinger’s equation in spherical polar coordinates

1
2 m r 2 sin 
2
2

  2    
 
1  
 U ( r )( r, , )  E( r, , )

 sin 

sin   r
2 

r

r




sin

 





To solve we apply a separation of variables: (r, , )  R(r)P( )F ( )

1
2 m r 2 sin 
2
2

  2    
 
1  
 U ( r ) ( r, , )  E( r, , )

 sin 

sin   r
2 
r  r   
  sin   

with (r, , )  R(r)P( )F ( )
 2 1
2 m r 2 sin 
2

  2 R 
 
P 
1 2F 
  R( r ) F ( )  sin 
  R( r ) P( )
 P( ) F ( ) sin   r
2

r

r




sin









 U ( r ) R( r ) P( ) F ( )  ER( r ) P( ) F ( )

1
2 m r 2 sin 
2
1 e
U (r) 
4 0 r
2
 sin    2 R 
1  
P 
1
1  F
 U (r)  E
r

 sin 


2
  F ( ) sin   
 R( r ) r  r  P( )  
2
 1   2 R 
1
 
P 
1
1
 F  2m
2m
r

sin



U
(
r
)


E




 2
2
2
2
2
2
2

r

r




F
(

)
r sin    

 r sin  P( )


 r R( r ) 
2
 1   2 R 
1
 
P 
1
1
 F  2m
2m
r

sin



U
(
r
)


E

 2


 2
2
2
2
2
2
  F ( ) r sin    

 r R( r ) r  r  r sin  P( )  
2
2

 F /   2 m
2m
X
(
r
,

)

Y
(
r
,

)

U
(
r
)


E


2
2
F ( )  


 2 F /  2
 Z ( r , ) =
F ( )
 1
  2 R 
1

r


 2
 2

r

r
r
R
(
r
)

 r sin  P( ) 

K (some constant)
 2m
P 
1
2m

sin


K

U
(
r
)


E

 2 2

2
2
  r sin   


 1
 2m
  2 R  1
2m
r

Q
(

)

U
(
r
)


E

 2
 2

2
2

 r R( r ) r  r  r
 
 1   2 R  2 mr
 R ( r ) r  r r    2 ( E  U ( r ))  Q ( ) =


2
K2 (also some constant)
Then the problem becomes finding solutions to
the separate “stand alone” equations
each of which uniquely constraints the wavefunction:
A solution to the radial equation can exist only when a constant arising in its
solution is restricted to integer values (giving the principal quantum number)
Similarly, a constant arises in the colatitude
equation giving an orbital quantum number
Finally, constraints on the azimuthal equation
give a magnetic quantum number
Let’s examine this through a simple model: an  forms inside the nucleus
and then escapes through quantum mechanical barrier penetration.
The potential seen by the  is spherically symmetric, so we can start by
first separating the variables -
(r )  Rnl (r )Ylm (, )
the functions Ylm are the same spherical harmonics
you saw for the wavefunctions for the hydrogen atom.
Then the equation for the radial function Rnl(r) can be written as
2

l
(
l

1
)

2
2
2
d (rRnl ) / dr  2m /   E  V (r ) 
rRnl  0
2 
2m r 

PARITY TRANSFORMATIONS ALL are equivalent
to a reflection
y
(axis inversion)
plus a rotation
x'
y
x
z
x
z
y
y
x
z
x
z
y'
y
y
z'
x'
z
y'
x
The PARITY OPERATOR on
3-dim space vectors
every point is carried through the origin
to the diametrically opposite location
x
z
Wave functions MAY or MAY NOT have a well-defined parity
(even or odd functions…or NEITHER)
  cos x
P   cos( x)  cos x  
P
= +1
  sin x
P  sin(  x)   sin x  
P
= 1
but the more general
  cos x  sin x
P   cos x  sin x  
However for any spherically symmetric potential, the Hamiltonian:
→
→
H(-r)
= H(r)
 H(r)
[ P, H ] = 0
So they bound states of such a system have DEFINITE PARITY!
That means, for example, all the wave functions of the hydrogen atom!
52
1 Z
 Zr / 2 a
 211 
sin  ei
  re
8  a 
32
 100 
32
 200 
1  z   zr / a
  e
 a
1  Z   Zr  Zr / 2a
  1  e
  2a   2a 
52
 210
1 Z 
 Zr / 2 a

cos
  re
  2a 
52
1 Z
 Zr / 2 a
 211 
sin  ei
  re
8  a
 300
1

81 3
Z
 
a
32

Zr
Z 2 r 2   Zr / 3a
 27  18  2 2 e
a
a 

-dE/dx = (4Noz2e4/mev2)(Z/A)[ln{2mev2/I(1-b2)}-b2]
I = mean excitation (ionization) potential of atoms in target ~ Z10 GeV
-dE/dx [ MeV·g-1cm2 ]
10
8
6
4
Minimum
Ionizing:
3
2
1
0.01
1 – 1.5 MeV2
g/cm
0.1
1.0
10
100
Muon momentum [ GeV/c ]
1000
A typical gamma detector
has a light-sensitive
photomultiplier attached
to a small NaI crystal.
The scinitillator responds
to the dE/dx of each
MIP track
passing through
If an incoming particle initiates a shower,
each track segment (averaging an interaction length)
will leave behind an ionization trail with about
the same energy deposition.
The total signal strength  Number of track segments
Basically
Emeasured  Ntracks  EMIP
avg
Measuring energy in a calorimeter is a counting experiment governed
by the statistical fluctuations expected in counting random events.
Since E
 Ntracks
and DN
= N
we should expect DE
 E
and the relative error
DE
E

E
E
1
=
E
DE = AE
a constant that
characterizes the resolution
of a calorimeter
(r)=c(r)ml (, )
ml (,
)=
the angular part of the solutions
are the SPHERICAL HARMONICS
(2l + 1)( l m)! m
im
P
(cos)e
l
4( l + m)!
Pml (cos) = (1)msinm [(
d
d (cos)
1
d
Pl (cos) = l [( d (cos)
2 l!
)m Pl (cos)]
)l (-sin2)l ]
The Spherical Harmonics Yℓ,m(,)
ℓ=0
ℓ=1
1
Y00 
4
3
Y11  
8
3
Y10 
4
15
4
2
Y21  
15
Y20 
sin 
Y33  
i
e
Y32 
cos
1
Y22 
ℓ=2
ℓ=3
8
2
sin 
e
2i
sin  cos
15  3
i
e
1
2
 cos   2 
4  2

1
35
4
4
1
105
4
2
Y31  
3
sin 
3i
2
sin  cos
1
21
4
4
sin
e
2i


5

Y30 
cos3 

4  2
7
e
i
2
5 cos   1 e
3

 cos 
2

z
then note r  r means



 
y

so: eim  eimeim
x
(eim=(1)m
and: Pml (cos) Pml (cos()) = Pml (-cos)
(-sin2)l = (1cos2l
but d/d(cos)

 d/d(cos)
ml (,
)=
(2l + 1)( l m)! m
im
P
(cos)e
l
4( l + m)!
d
Pml (cos) = (1)m(1-cos2m [( d (cos)
1
d
Pl (cos) = l [( d (cos)
2 l!
)m Pl (cos)]
)l (-sin2)l ]
So under the parity transformation:
P:ml (, ) =ml (-, )=(-1)l(-1)m(-1)m ml (, )
= (-1)l(-1)2m ml (, ) )=(-1)l ml (, )
An atomic state’s parity is determined by its angular momentum
l=0 (s-state)
l=1 (p-state)
l=2 (d-state)
 constant
 cos
 (3cos2-1)
parity = +1
parity = 1
parity = +1
Spherical harmonics have (-1)l parity.
In its rest frame, the initial momentum of the parent nuclei is just its
spin: Iinitial = sX
and: Ifinal = sX' + s + ℓ
1p1/2
4He
1p3/2
1s1/2
S = 0
So |sX' – sX| < ℓ< sX' + sX
Since the emitted  is described by a wavefunction:
 ~ Y m
ℓ
the parity of the emitted  particle is (1) 
Which defines a selection rule:
restricting us to conservation of angular momentum and parity.
If P X' = P X
then
If P X' = P X
then
ℓ = even
ℓ = odd
|sX' – sX| < ℓ< sX' + sX
If the 2p, 2n not plucked from the outermost shells
(though highest probability is that they are)
then they will leave gaps (unfilled subshells) anywhere:
Excited nuclei left behind!
EXAMPLE:
If SX = 0
ℓ = sX'
0  3 nuclear transition would mean
(conservation of angular momentum)
ℓ = 3 so PX' = PX
i.e. 0+3 is possible, but
0+3 is NOT possible
02
do not change the parity of the nucleus
04
So 0+2 would both be impossible
0+4
so
PX' = PX