A primer on radiation and it interaction with matter

Bismarck State College
A primer on radiation and it's interactions with matter
Science of Radiation protection class (Instructor J Custer)
By Maurice (Jeff) Rhoades (Student)
The following primer is an attempt by this student to take applicable radiation information,
and data found in many sources, coalesce the information in a logical, and repetitive nature,
for each type of radiation. By doing this, I hope that learning how radiation interacts with
matter, (as our body is made up of matter), will help me to have a better understanding of what
Radiation protection is all about.




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Each of the following radiation types will be examined, (1) Alpha, ? (2) Beta, ?- (3) Positron,
?+ (4) Gamma ? and X rays, (5) Neutron, ? (6) Proton, ? . Other particles and rays will not be
covered, as they are outside the scope of this course. For each type of radiation, the following
items will be discussed.
a) A brief history of how, and by whom, it was discovered.
b) Some examples of the sources for the radiation.
c) How the radiation causes ionization and excitation.
d) The penetrating ability of the particle or ray.
e) Any formulas that describe the radiation interactions with matter.
f) Any other facts that might be useful to the student.
The Alpha particle. Credit for the discovery of the alpha (?) particle was given to Sir Ernest
Rutherford. One of Rutherford's now famous experiments, involved a radioactive alpha emitter
(radium) in which the particles were focused on a leaf of gold foil. His assistants then watched
the phosphorescent screens, which were placed at various positions around the foil. As the
particles passed through the foil, scatter angles were determined as well as those that simply
passed through. Rutherford was surprised when some of the ? particles even bounced back 180
degrees towards the alpha source. Using the information obtained from this experiment,
Rutherford announced his model for the atom. In 1908 Rutherford was awarded the Nobel Prize
for his work with radioactivity and radiation.
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Alpha particle radiation, ( 4
2 He++) results from the decay of heavy radioactive nuclides. The
? particle is ejected from the nucleus in order to obtain a more stable nuclei. Some examples of
alpha emitting nuclides are as follows:
U238 Th234 + 4
2 He++ (4.197 MeV)
Pu239 U235 + 4
2 He++ (5.156 MeV)
Po210 Pb206 + 4
2 He++ (5.3044 MeV)
Ra223 Rn219 + 4
2 He++ (5.7164 MeV)
Note: The energy of the emitted alpha is characteristic of the nuclide that emits it. For
example, U238 emits a 4.197 MeV alpha. This is true for most nuclides that are alpha emitters.
Also, these values listed are the actual ? particle Ke, (kinetic energy) and does not include
the recoil energy, of the nucleus that emitted it.

The alpha particle is a direct ionizing radiation that consists of 2 protons and 2 neutrons. It is
essentially a helium atom without its two electrons. (Notice that's how those two ++ signs got
there). It is emitted by the nucleus of an unstable and usually large atom as it decays to a more
stable state. The alpha particle rest mass is 4.00153 amu. In kg's that would simply be:
Kg = (4.00153amu) (1.66053 x 10-27 kg/amu) = 6.64466 x 10-27 kg.
Now that we know the mass, the Ke, and the charge (+2) of the ? particle, we can now go to
work on figuring out how this particle interacts with matter. Let's start by figuring out the
velocity of the alpha particles in Rutherford's experiment.
We know that: Ke = 1/2 mv2
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Where: Ke = kinetic energy, in joules.
m = mass in kg.
v = velocity, in meters per sec.
we also know that Radium's alpha has 5.7164 MeV's of energy when emitted. So let's convert
the MeVs to joules.
(5.7164 X 106 eV)(1.60219 X 10-19 joules/eV) = 9.1587 X 10-13 j
Ke = 1/2 mv2
9.1587 x 10-13j = 1/2 (6.64466 x 10-27kg) (V2)
(2) 9.1587 ?? 10?13 ?? =
6.64466 ?? 10?27????
v = 1.66033 x 107 m/sec
1.66033 ?? 107 ??/??????
Since the speed of light is = 2.99793 x 108 m/sec, then, ?speed = 2.99792 ?? 108 ??/?????? = .05538
about 5.5 % of the speed of light for this MeV alpha. This is slow compared to the other
radiations that we will discuss later. However, because of the slow speed and +2 charge of the
alpha particle, it is indeed, a very ionizing radiation. As the alpha particle passes through a
material, it either excites the atoms or removes electrons from the orbits of atoms it passes near.
In this way the alpha is causing direct ionization in its surroundings by creating ion-pairs. (The
ejected electron e-, and the now positively charged atom). In the case of water, which makes up
much of our bodies, the ion pairs produced are H20+ and e-. See figure 1 for a visualization for
what happens as the alpha approaches the molecule of water.



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Figure 1
If the alpha particle does not get close enou
gh to strip an electron
from the atom, it still causes the atom to go to an excited state
through coulomb interactions. These types of reactions are called

ionization and excitation. Energy is lost by the alpha and gained
by the atom. These same reactions were going on during
Rutherford's scattering experiment, but with gold taking the place

of water.


As the alpha approaches the H2O
nuclei, it starts feeling the

coulomb repulsive force from the
+
molecule.
-


+
-
The negative electron is
+
stripped away from its
?
-
-
quantum level.
H2O
+





The coulomb force felt by the ? is given by the formula:
K q1q2

F

e =
??
where: Fe = electrostatic force in newtons
q1 = charge of the first particle ( coulombs )

q2 = charge of the second particle ( coulombs )
k = 9.0 x 109 N-m2/C2 ( Boltzmann constant )

r = distance between the particles ( meters )
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Let's go back to Rutherford's experiment and take a look at the equation given in figure 1.
As I stated earlier, some of the alpha particles that hit the gold foil bounced back at almost 180o
toward the source. This raises the question, did the alpha particle actually collide head on with a
nucleus of the gold, or did the Coulomb force repel the alpha? We can use the equation in figure
1, but to find this out, we will need some more information. For an actual collision to occur, the
two nuclei would have to come in contact. So, we need to first find the radii of the two nuclei
involved. The formula for this is:
Rnucleus = Rnucleon x A1/3 where: Rnucleus = radius of the nucleus
Rnucleon = radius of one nucleon (1.3 x 10-15 m)
A = the atomic number (number of nucleons)
So for alpha R? = (1.3 x 10-15m) (4)1/3 = 2.06 x 10-15 m
And for gold RAu = (1.3 x 10-15m) (197)1/3 = 7.56 x 10-15 m
Now we have enough information to answer the question, but first let's sketch this out to make
sure we have a clear picture of what's happening.

Au
?
nucleus
particle


radius of Au
radius of the alpha particle

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Now, the distance between the centers would become the "r" in the formula, and that would be,
R? + RAu, which is, 2.06 x 10-15m + 7.56 x 10-15 = 9.62 x 10-15 m.

K q1 q2
So: Fe =
??

8.99 ?? 109 ????2???2 79 ?? 2?? ( 1.6 ?? 10?19 )2
=
*
9.62 ?? 10?15 ??
= 3.778 x 10-12 j
= 23.61 MeV
* (The 1.6 x 10-19 is the c in the 79 and 2 it converts the unit charge to coulombs.)
Our alpha particle only has 5.7164 MeV, so there is no way that an actual collision could have
occurred. (That is barring some quantum tunneling possibility)
Ionizing radiation, like the alpha, is rated by the intensity to which it creates the ion-pairs.
This is called the ionizing power, and is proportional to the number of ion-pairs formed per
centimeter of travel. The following formula gives a good approximation of the ionizing
capability of the different direct ionizing radiations and is given as:
?? ??2
I =

????
where: I = is the ionizing power (factor)
m = mass of the particle
z = the number of unit charge
Ke = is the kinetic energy
(Note: This equation is for comparison only. So as long as you use the same units for each direct
ionizing radiation type, you will have a good comparison between the different types by its
ionizing capabilities. )

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So for our alpha the following result is obtained. We will use this value to compare with the
other direct ionizing particles later on in the study.
6.64466 ?? 10?27 ???? ( 2 )2
I =
=
9.1587 ?? 10?13 ??
2.902 x 10-14
Let's now look at the penetrating power for the alpha. Because the relatively slow speed of
the alpha particle, and its plus two charge, it is not a very good penetrator. The range of an alpha
is finite; it is dependent on the number of atoms the particle encounters when it travels through a
medium. We can analyze this for penetration of human skin by a simple thumb rule equation as
follows:
????
R? =

1000 gcm-2 Where: R? = the range of the material in mater.
E? = the energy of the alpha in MeV.
1000 = constant for ?
5.7164
So, for our alpha; R? =

1000 = 5.7164 x 10-3 gcm-2 then ,you divide by the density of the
material of interest. Skin density is about the same as that for water which is 1.0g/cm3
5.7164 ??10?3?????? ?2
R? =
1??/???? 3
= 5.7164 x 10-3 cm. Outer skin layer is ? 7 x 10-3 cm thick. So
you can see that even though the ionizing power is high its penetration is very low. After the
alpha slows to the same energies as its surrounding, it becomes a simple atom of helium. (After it
steals two electrons) See table one, next page, for densities of some common materials.


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table 1

Density
( g/cm3 )
Material
Air at STP
.001293
Lead
11.3
Iron
7.8
Gold
19.3
H20 @ 4oC
1.0

You may ask then, why are we even concerned about alpha particles? Alpha's are not much
of a threat to us externally and are easily shielded (paper or clothing is enough). That is true as
long as the particles stay outside of the body. It's when it gets inside are bodies that it becomes
the real threat. For example, Radon is a naturally occurring radioactive gas that can be taken into
the lungs in our homes and even outside to some extent. Because Radon is an inert gas, it should
not be a problem either. This is because, by being inert, as you breathe it in, you also exhale it
right back out. However, if it should decay while in your lungs, its decay products are not gasses
and are still radioactive. I am sure, that internal hazards with alpha producing nuclides will be
examined in much more detail during the course.


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The Beta particle. Credit for the discovery of the beta (?-) particle has been given to Henri
Becquerel. He was working with the particle in 1896, and the particle was even called a
"Becquerel ray" for some time before the particles were renamed as beta particles. ?- Particles
are electrons that are emitted from the nucleus of unstable radioactive atoms. It has a negative 1
charge and is much lighter than the alpha. The rest mass for the beta is only 9.109 x 10-31 kg.
Some examples of radioactive atoms that emit beta particles are as follows:
27 Co 60 28 Ni 60 + ?- (1.6 MeV)
38 Sr 90 39 Y 90 + ?- (.564 Mev)
53 I 131 54 Xe 131 + ?- (.606 Mev)
As you can see in each of the above reactions the Z number, (number of protons) goes up by
one, and the A number, (number of nucleons) stays the same. This is because a neutron has
changed into a proton and an electron is ejected from the nucleus becoming a beta particle. Also,
the above Mev listings are an average ?- energy for the nuclide. Unlike alpha, the beta is emitted
across a spectrum of energies from low to Emax for a particular nuclide with the majority being
emitted about at the 1/3 point. If the spectrum is graphed it looks like graph 1.
Energy range where most ?-
are emitted for each nuclide.
Graph 1.
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