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Sunday, March 22, 2009

INTRODUCTION TO NUCLEAR PHYSICS

Q.1 In the previous lecture you learned how it was discovered that the atom is mostly empty
space with a cloud of electrons. At the centre is a small but very heavy nucleus that has
protons and neutrons. The word "nucleon" refers to both of these. So you can think of
the neutron or proton as being two different varieties of the nucleon. The masses of the
two are very similar, and they are roughly 2000 times heavier than the electron.
-27
-27
proton mass = 1.672 10
neutron mass = 1.675 10
electron
p
n
M kg
M kg
= ×
= ×
-31
-19
-19
mass = 9.109 10
The neutron is neutral, of course, but the charge on the proton is 1.6 10 while the
charge on the electron is the negative of this, 1.6 10 .
2. Using k
eM kg
C
C
= ×
×
− ×
2 2
ilograms is very awkward if you are dealing with such small particles. Instead we
we use to write the mass of a particle in terms of its rest energy, / . So
mass is measured in uni
E=mc m=E c
2
2
2
ts of / .
proton mass = 938 /
neutron mass = 940 /
p
n
MeV c
M MeVc
M MeVc
=
=
electron mass = 0.5 / 2
3. a) A hydrogen nucleus is just one proton.
b) A deuteron has a proton plus one neutron.
c) A triton (or tritium nucleus) has a proton
plu
eM= MeVc
s two neutrons.
All three atoms have one electron only, and
thus completely identical chemical properties.
4. Hydrogen, deuterium, and tritium are called isotopes. If a nucleus with Z protons has N
neutrons then its isotopes will have fewer, or more, neutrons. Since the number of electrons
is also Z, the chemical properties of all isotopes are exactly the same. But for any given
element, at most there is only one stable isotope.
161
1 3
6
4. A commonly used notation is where is the
element, and . Of the elements that you
see on the right, the most stable ones are H, He,
and Li. Now consider oxygen. The most stab
A
ZX X
A=Z+N
16
16 17
18
le
isotope is O. When you breathe in oxygen from
the atmosphere, 99.8% is O, 0.037% is O, and
0.163% is O. We shall see later what unstable
means and how nuclei decay.
1/ 3
0
0
5. The diameter of the nucleus is about 10 million times
smaller that the overall diameter of the atom. Nuclei
follow an approximate rule for the radius,
where 1.2 (remember,
r rA
r fm

= -13
1/ 3
1/ 3 1/3
208
1 fermi 10 cm) and
. Now, increases very slowly with . You
can check that 16 2.52 while 208 5.93. This
means that a very heavy lead nucleus is only about
2.4 t
A Z N A A
Pb
=
= +
= =
imes the size of the much lighter 16O nucleus.
6. To learn about how protons are distributed inside a
nucleus, we send a beam of electrons at a nucleus
and observe how they scatter in different directions.
The negatively charged electrons interact with the
positively charged protons, but they obviously will
not see the neutrons. The scattered electrons are
captured in a detector which can be moved around
to different angles. In this way one can reconstruct
the charge distribution which caused the electrons
to be scattered in that particular way.
7. What energy should electrons have in order to see a nucleus? We know that electrons are
waves with / (the De Broglie relation). To see something as small as 1 requires
a wave with wave
λ =h p fm
2 2
2
length at least 1 . A wave with longer wavelength would simply
pass over the nucleus without being disturbed. So the minimum electron energy is,
. Evaluation gives this to be a
2 2
fm
E p h
m m
λ
λ

= = few MeV, requiring an electron accelerator
of more than this minimum energy.
ρ(r)
R
8. From electron scattering we see that the proton
density is almost constant throughout the nucleus
and falls sharply at the surface. So nuclei should
be thought of as rather fuzzy balls. A
1/ 3
0
typical plot
of density versus distance from the centre looks
like this. Here the distance can be
called the nuclear radius. The distribution of
neutrons is very similar to this
R≈rA
plot, but requires
other techniques.
1/ 3 3 3
0 0 0
3
0
9. Let us consider the implication of the approximate formula for the nuclear radius,
where 1.2 . The volume of the nucleus is: 4 4 . From
3 3
this 3 0.14 nucle
4
r rA r fm V r r A
A
V r
π π
π
≈ = = =
= ≈ ons/fm3. This is the number of nucleons per cubic fermi, and
is independent of the nucleus considered. This is the density you would find at the centre
of any nucleus. Of course, this is approximately true only but it is quite remarkable.
10. Protons repel protons through the electrostatic force. So why does the nucleus not blow
apart. Obviously there must be some attractive force that is stronger than this repulsion.
It is, in fact, called the strong force. From what we have learned so far, we can guess
some of its important features:
a) Since neutron and proton distributions are almost the same, the N-P force cannot be
very different from the N-N or P-P force.
b) Since the density of nucleons in large nuclei is the same as in lighter nuclei, this means
that a given nucleon feels only the force due to its immediate neighbours, and does not
interact much with nucleons on the other side of the nucleus. In other words, the range
of the nucleon-nucleon force is very short and of the order of 1-2 fm only.
11. The force between two charges is always of one sign - repulsive if the signs are the same,
and attractive if they are opposite. In the early years of quantum theory, people realized
that this force comes about because of the exchange of photons between charges. The
nucleon-nucleon force is different. It has to be attractive to keep the nucleus together,
and has to short range (as discussed above). But, to prevent nucleons from sticking to each
other, it must be repulsive at short distances. Now here, "short" and long means distances
on the scale of fermis. Typically the distances between nucleons is on this scale as well.
r
V(r)
Nucleon-Nucleon potential
1.5 fm
r repulsive
attractive
F dV
dr
= −
p n π +
Here is roughly what the N-N potential looks like. Since
the force is the negative of the slope, you can see that for
large value of , the force is attractive (negative means
direct
r
ed towards smaller values of ). At roughly 1.4
the potential reaches its most attractive point. For values
of smaller than this, the force is repulsive. In fact there
is a very stro
r fm
r
ng core that almost completely forbids the
nucleons from getting closer than about 0.5 fm.
12. In 1935, the Japanese physicist Hideki Yukawa made an astonishing breakthrough in
understanding the basis for the attractive N-N force. He assumed that, just as between
charges, there must be a particle that is emitted by one nucleon and then captured by the
other. He called this a "pi-meson" or "pion", and made a good guess for what its mass
should be. His argument uses the time-energy uncertainty principle discussed earlier:
Let be the time that the pion takes between emission and capture. In this time it
will have travelled a distance approximat
• Δt
ely equal to because light particles can
travel no faster than light.
Creating a pion from "nothing" means that energy
conservation has been violated. The amount of
cΔt

2
2
violation is minimum energy of pion
and this must obey / 2. Hence,
2
and the distance travelled by the pion c .
2
Put 1.2 (range of
2
E mc
E t t
mc
t
mc
fm
mc
Δ = ≈
Δ Δ ≈ Δ ≈
≈ Δ ≈
• ≈
􀀽 􀀽
􀀽
􀀽
2
nuclear force). This gives
the mass of the pion as close to 124 . This
was an amazing prediction - the first time a particle had
been predicted to exist on the b
mc ≈ MeV
2
asis of a theoretical
argument. When experimentalists searched for it in
1947, they indeed found a particle of mass rather close
to it, with mc ≈ 138 MeV. It was a very d
+
ramatic
confirmation for which Yukawa got the Nobel prize.
Pions can rightfully be called the carriers of the strong nuclear force. They have 3
possible charge states: π ,π

0 , . They belong to a larger family of particles called
mesons. Other family members are rho-mesons, omega-mesons, K-mesons,...
Today we can produce mesons in huge amounts by sma
π −
shing nucleons against each
other in an accelerator.
13. A few nuclei are stable, most decay. The decay law is simply derived: if the number of
nuclei decreases by in time , then must be proportional to both the the number
of nuclei
dn dt dn
n that are decaying and , so (minus sign for decrease). with the
proportionality constant , we have , or . We have encountered
the solution of this type of equatio
dt dn ndt
dn ndt dn n
dt
λ λ λ
∝ −
= − = −
-
0 0
1
0 2
0
n before, . You can see that at 0, .
Taking the log, we have ln . We define the half-life as the time it takes for
half the original sample to decay. If / 2 then
n ne t t n n
n t T
n
n n
λ
λ
= = =
= −
= 0
0
1
2
log , from which the half
2
life is related to by, log 2 0.693. The larger , the more radioactively unstable
a nucleus is. Some typical half-lives are:
n t
n
T
λ
λ λ
λ λ
= −
= =
214 -4
84
89
36
90
38
Polonium 1.64 10
Krypton K 3.16 minutes
Strontium Sr 28.5 years
P × s
226
88
14
6
Radium Ra 1600 years
Carbon C 5730 years
Uranium 238
92 U 4.5 109 years
You can see how hugely different the lifetimes of different nuclei are!
14. Here is a plot of the number of unstable nuclei
left as a function of time. After each ha
×
lf-life,
the number of nuclei decreases in number by
half of the previous. Eventually there is only
one nucleus left, and that too will eventually
decay. So how can the derivation for the decay
law be correct? Strictly speaking, we are not
allowed to write down, or solve, a differential
equation like because this assumes
that ( ) is a continuous f
dn n
dt
n t
= −λ
unction. But this is
almost true because in real life we deal with very
large numbers of nuclei and so it makes a lot
of sense to think of n(t) as continuous
222
86
218 4
84 2
15. Just to get an idea, consider the decay of Rn (Radon, a very dangerous gas that is
found underground) into Po (Polonium, another terrible poison) and He (harmless,
fortunate 222
86
222 218
86 84
ly!). The half life is 3.8 days. So, if we started with 20,000 atoms of Rn,
then in 3.8 days we would have 10,000 atoms of Rn and 10,000 atoms of Po
In 7.6 days we would have 222 222
86 86 5000 atoms of Rn, in 11.4 days, 2500 , Rn etc.
16. The decay law can be used to see how old things are. This is called radioactive dating.
Carbon dating is widely used for living things
14
6
12
6
that died a few hundred or few thousand
years ago. How does it work?This uses the decay of the unstable isotope, . Of course,
the stable isotope of carbon is .
When a living
C
C
• 2
-
organism dies, CO is no longer absorbed. Thus the ratio of carbon 14:12
decreases by half every 5730 years. We can measure the rate of decrease through
t or the "activity"
o N=Neλ A with 0.23 / . (The becquerel Bq
is the unit of radioactivity, defined as the activity of a quantity of radioactive material
in which one nucleus decays per second
t
o o =A e−λ A = Bq g
14
6
7
. )
The amount of isotopes in the atmosphere is approximately constant, despite a half-life
of 5730 y because there is a constant replenishment of through the reaction,
C

14 14
6
17. Let us use the above idea to find the time when this
man died. His body was found a few years ago buried
under deep snow in a mountain pass, so it it did not
decay as usual
N+n→ C+p
14
6
14
6
. By looking at the radioactivity in his
body, it was found that that the activity of was
0.121 / of body tissue. This is less than the normal
activity 0.23 / . because has
C
Bq g
Bq g C
4
4 1
1 2
1.21 10
4
been decaying
away. First find , 0.693 0.693 1.21 10
5730
Then use, 0.121 0.23 which gives,
ln 0.121 1.21 10 and so 5300 years is
0.23
when this poor
t
y
T
e
t t
λ λ

− −
− × ×

= = = ×
=
= − × × =
2
man was killed (or died somehow)!
18. The most famous formula of physics, , is the basis for nuclear energy. In 1935,
it was discovered by two German physicists, Otto Robert Frisch and Lise
E=mc
Meitner, that
a heavy nucleus can fission (or break up) into two or more smaller nuclei. The total
energy is, of course, conserved but the mass is not. This is completely different from the
MA Mb + Mc
usual situation. In the picture below you see an example of fission.
2
The masses of the two nuclei add up to less than the mass of the parent nucleus, and the
energy released is ( ) . This goes into kinetic energy and sends the
two daughter nucl
A b c Q= M −M −M c
ei flying apart at a large velocity. There happen to be NO completely
stable nuclei above 82, and no naturally occurring nuclei above 92. Above these
limits the nuclei decay or fall apa
Z= Z=
rt in some fashion to get below these limits.
19. A very useful concept is "binding energy". Suppose you want to take a nucleon out of a
nucleus. The binding energy is the amount of energy that you would have to provide to
pull it on the average. Nuclei with the largest BE per nucleon are the most stable. As you
can see from the graph below, the most stable element is iron, 56 with a BE per nucleon
of about 8.6 MeV. This is why iron is the heavy element found in the largest quantity on
ear
Fe
4
235
th and inside stars. The curve is not smmoth and you see that a He nucleus (i.e. an
parrticle, has a relatively high binding energy and so is relatively stable. In contrast,
Uranium U
α
or deuterium 2H are much less bound and they decay.
20. Nuclei can be unstable in different ways. A nucleus can emit , , and radiations.
Usually a nucleus will emit one of these three, but
α β γ
it is possible to emit two, or even all
three of these. (In addition, as we have discussed above, a nucleus can break up into
two or more smaller fragments through the process
of fission. ) What is the nature of , , radiations?
In the experiment shown here, you see that a piece
of radium has been put
α β γ
in a lead block (as shielding).
A magnetic field bends the charged particles emitted
during the decay. We find that they are of three types:
a)Heavy positively charged particles are bent to one
side by the magnetic field. The amount of bending
shows that they are particles.
b)Other particles are bent much more, and in the other
direction. They are ele
α
ctrons ( particles).
c)Some particles are not bent at all, hence must be neutral. They are particles (or rays,
or photons, same thing!).
21. Let's first consider alpha decay, A
Z Z X
β
γ
γ
− → 4 4
2 2 . As you see A changes by 4 and
Z by 2. You can think of particles as a gang of 4 particles that always stays together
in a big nucleus. Unstable nuclei simply cannot overcome
A X He
α
− ′ +
the proton repulsion and an
particle ultimately succeeds in esaping the nucleus. indefinitely.
22. The simplest beta decay reaction is when a neutron decays, . Other than
a proton
n p e
α
→ + − +ν
and electron, an anti-neutrino is also emitted. As discussed in the lecture, the
(anti) neutrino is a neutral particle with a very tiny mass that interacts very weakly with
matter. A nucleus
1
1
-
can undergo beta decay with either an electron being produced, i.e
(called decay) or with an anti-electron (positron, or positive
electron), , (called
A A
Z Z
A A
Z Z
X X e
X X e
ν β
ν β

+
+

→ ′+ +
→ ′+ + + decay). Beta decay involves the weak
nuclear force. This is one of the four fundamental forces in the world, and its small
strength means that the decay happens much more slowly than most other reactions.
23. Just as for an atom, a nucleus can only exist in
certain definite energy states. When a nucleus
goes from one state to the other, it can emit a
photon (γ ray). Because the spacing between
nuclear levels is of the order of one MeV (i.e.
a million times more than in atoms), the photon
is much more energetic than an optical photon.
In going between levels, emission also happens
as can be seen in this diagram.
β

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