**Chapter 43**

__Nuclear Physics__

In this chapter the following topics will be discussed:

- The Cause of Radioactivity
- Isotopes of an Element
- Types of Radiations
- Radioactive Decay
- Carbon Dating
- Nuclear Energy
- How Do Reactors Work?
- The Decay of a Radionuclide

1. The Cause of Radioactivity:

Radioactivity comes from the nucleus of atoms where repelling protons are
held together. Neutrons help protons to be held together by creating enough
distance between them. **A proton and a neutron** hold each other by **
strong attraction forces** that act over very short distances, **while two
protons repel** each other by the **Coulomb’s repulsive forces** regardless
of the distance between them. If by any means the distance between two protons
(separated by one or more neutrons) is disturbed or increased, then the
repulsive forces overcome the short-range attraction forces and the protons
eject like when compressed springs are released. The ejection of a single or a group of particles
from the nucleus of atoms is called radioactivity. Radioactivity can be ejected
particles or very energetic photons (strong **electromagnetic** radiation
called **"Gamma rays"**)**.**

**For very light elements**, the number of
protons is almost the same as the number of neutrons. As elements get heavier,
more and more neutrons are needed to hold protons together and therefore keep
the nucleus intact. **For very heavy elements**, even excess neutrons
cannot hold the protons together and radioactivity occurs more often. The
first 8 elements of the Periodic Table ( H, He, Li, Be, B, C, N, and Oxygen) are
shown below:

For each element, the **top number (A)** shows **the # of nucleons**
(protons plus neutrons), and the **bottom number (Z)** shows **the # of
protons**. In nuclear physics, any element (X) is shown as

where A = Z + N is the # of nucleons and Z = the # of protons. For example, the most abundant type of carbon has 6 protons and 6 neutrons that make 12 nucleons at its nucleus. After oxygen, fluorine has 9 protons but 10 neutrons written as

As the elements get heavier more neutrons become necessary to make their nuclei stable. For example the most stable iron nucleus has 26 protons, but 30 neutrons.

For much heavier elements the ratio approaches 1**.**5. For example, the
U-235 nucleus contains 92 protons and 143 neutrons, written as

2. Isotopes of an Element:

Every element is identified by the number of protons (Z) at its nucleus. For example, any element that contains 8 protons is known as oxygen. Oxygen can have 8, 9, 10 11, or 12 neutrons. As long as there are 8 protons at its nucleus, it is oxygen and chemically behaves as oxygen. The isotopes of oxygen are therefore,

As another example uranium, the 92^{nd} element in
the Periodic Table has the following isotopes:

Definition: Isotopes of an element have the same # of protons but different #s of neutrons.

3. Types of Radiation

Three important types of radiation are **Alpha**-rays,
**Beta**-rays, and **Gamma**-rays.

α-rays are **helium nuclei** that eject from some radioactive isotopes
that go under spontaneous decay as well as during fission processes in power
reactors. Therefore an α-ray is a
moving
helium nucleus.

Example 1: An α-ray has an energy of 25 KeV. Find its speed.

**Solution: **The kinetic energy in Jules is K.E. =
(25) (10^{+3}) (1**.**6x10^{-19 }) J = 4**.**0 x 10^{-15}
J

The mass of (He) nucleus is the mass of 2 protons + 2 neutrons.

Using the K.E. formula**: ** K.E. = ½ Mv^{2}, results
in v = 1**.**1x10^{6} m/s
(Verify).

β
–rays are fast moving electrons. A β-ray is often shown as β^{-}.
In the following example, you are asked to find the energy of the fast
moving electron in (eV).

Example 2: Calculate
the energy in eV of an electron that moves at 1**.**1x10^{6} m/s.

Answer : 3.44 eV.

γ –rays are highly energetic
electromagnetic radiation. Electronic transitions back and forth between the
electronic shells can go as high as 10^{19} or 10^{20} Hz in
frequency. Nuclear (protons) transitions are more energetic and can give
electromagnetic bursts as high in frequency as 10^{25} Hz. Gamma rays
are very penetrative and therefore dangerous due to their extremely small
wavelengths. Alpha and Beta rays can be stopped by putting on suitable
clothing and even by human skin if they are not very energetic. The radiation
concern of nuclear reactors is mainly for Gamma-ray radiation. Gamma-rays require very thick shielding to stop them.

Example 3: Find the speed and frequency of 100MeV, 300MeV, and 600MeV Gamma-rays.

4. __Radioactive Decay__

Radioactive elements, as was mentioned, try to get rid of
their excess protons in order to become more stable. During the decay, either of
Alpha, Beta, or Gamma radiation can occur**.** As a result of the
decay, a new isotope of an element or an entirely different element can be
generated. The new isotope or element is called the “daughter” nucleus as
opposed to the original nucleus or the “parent nucleus**.**”

There is a special **decay rate (
λ )** for every decay process. This
gives us an idea of **how fast** the nuclei of an isotope keep converting to
a different element or isotope.

Another way to describe the speed at
which an element decays is via defining a **half-life** for it. The
half-life ( T _{½} ) of a
radioactive element is the time it takes for half of its nuclei (or amount) to
convert to a different type of nucleus. It is easy to show that

If at ( t = 0 ),
N_{o} nuclei of a parent
element are present, the number of parent nuclei present, N, after the
elapse of time ( t
)
can be found from the equation:

Example 4:** **A hypothetical radioactive element has a
half-life of 3**.**00 days. At time t = 0, there are 256x10^{15} nuclei of
this element present. Find (a) the number present or left after 9**.**00 days, and
(b) the number converted to a different element after 9**.**00 days.

Solution: N_{o} = 256x10^{15}
nuclei. (This number is roughly one million times smaller than the Avogadro
number. This means that, mass-wise, we are talking about micrograms
of such element).

(a)

N =
(256 x10^{15}) e ^{-(}^{0.231)(9.00)} =
32x10^{15} (nuclei present)

(b) N_{daughter} =
(256 - 32)x10^{15} = 224x10^{15} (nuclei
converted)

5. Carbon Dating

** Activity:** Activity is defined as the
product ( λN

Another unit for activity is Curie (Ci).
1 Ci = 3**.**70x10^{10}
decays /sec.

__C-14 Radioactive Dating:__

Out of every 7**.**2x10^{11} C-12 atoms,
one is C-14. C-14 is produced in the atmosphere at a constant rate.
Cosmic rays in the atmosphere cause neutron production and neutron collision
with N-14 converts it to C-14. The reaction is shown
below:

C-14 then decays back to N-14 by emitting a Beta-ray. For every C-14 decay, there is a single emission of a Bata-ray that can be counted.

Living organisms take in air (N_{2} and O_{2})
molecules some of which have been converted to (C-14). The N_{2}
molecules and C-14 atoms come in and go out of living organisms due to
breathing, eating, and elimination. Therefore, the rate of production of C-14
and conversion of it back to N-14 is the same in living organisms as it is in
the atmosphere. Once a living organism dies such as a tree, animal, or a human,
there will not be any more intake and output of C-14. The concentration of C-14
immediately after death starts decreasing. However, since the half-life of C-14
is 5730 years, after every 5730 years, the concentration of C-14 becomes half.
Since the concentration of C-14 can be measured by the rate of Beta-decay, a
Beta-radiation measurement of an ancient object can give us the age of it.

It is easy to show that the activity (λN) of C-14 in living organisms is 16 decays /(gr-min.).

As an example, If the activity measured in an ancient object is 8 decays /(gr-min.), the age of the object is then 5730 years. If the activity is 4 decays /(gr-min.), the age is 2 x 5730yrs = 11460 yrs and so on....

6. Nuclear Energy

Nuclear energy is the result of conversion of small amounts
of mass in nuclear reactions to energy. According to
Einstein’s formula
( E = Mc^{2} ), when M kilogram of mass is annihilated and converted to energy, (
Mc^{2 }) Joules of energy will
be generated. This energy appears in the form of K.E. in the fast ejection of
the fission fragments in nuclear processes. The collision of fission fragments
to the walls of fuel rods in a nuclear reactor, converts the motion energy to
heat causing the fuel rods to become very hot. The flow of water around the
fuel rods removes this heat and converts a secondary water flow to steam at high
temperature and pressure. The pressurized steam is then blown into turbine
blades which rotation turns an electric generator converting the mechanical
energy to electric energy for use.

Example 5: Calculate (a) the energy resulting from
the conversion (annihilation) of 1**.**00gram of mass into energy by using the
Einstein’s mass-energy relation. (b) How many tons of water can be pumped to
the top of a mountain 1**.**00 mile high if only 33**.**0% of this energy can be useful?

Solution:

E = Mc^{2} =
(0**.**00100kg)x(3**.**00x10^{8} m/s )^{2} = 9**.**00x10^{13} Joules

The weight of 1**.**00ton or 1000**.**kg of water is 9800N.

To pump each ton to a height of 1609m (1 mile), an energy of W = Fx = (9800N)(1609m) is needed.

Energy
/
ton = (15**.**8E6) Joules

Imagine pumping 1**.**88 million tons of water to a height of
1 mile by using the energy of 1 gram of mass annihilation! That is
unbelievable! Although we may still have to sacrifice another half of this
energy for safety and other losses, but still, in spite of radiation problems
and nuclear waste disposal issues, nuclear energy is one solution to a possible
energy crisis in the future.

7. __How Do
Reactors Work?__

** **As was discussed in the chapter
on the atomic physics, atoms are very empty. The way nuclear energy ( in a
Uranium Fuel Reactor ) is generated is that each atom of U-235 must be hit by a
neutron so that its energy balance is disturbed causing it to split into two or
three fission fragments of high speed giving (kinetic) nuclear energy. The
types of fission fragments and the way a fissile isotope (U-235 in this case) go
through fission, depends on the target nucleus energy as well as the energy of
the incident neutron. Uranium-235 is sensitive to slow
neutrons**.** What
we call slow neutrons are those at speeds around 2200 m/s like the average speed
of air molecules at the Room Temperature (25 degrees Celsius). Faster neutrons
can have speeds of 1000,000m/s or more. Neutrons slow down to the right speed
after enough collisions with other elements in the core. One good and
fortunate thing is that when a U-235 nucleus splits into fragments in a fission
process, it releases 1,2, or 3 new neutrons that can be used to split other
U-235 nuclei with and so on. This causes a chain reaction that if controlled
carefully, it can sustain fission occurrence and therefore nuclear energy
generation. The fission of each U-235 nucleus generates 200MeV of energy while
each atom of carbon that burns produces about one millionth of this energy.

U-235
+ 1 neutron =======**>**
Fragment A + Fragment B + 200 MeV of energy.

**The key to sustaining a chain
reaction is to keep neutrons from escaping the reactor core before they are
absorbed again by other U-235 nuclei**. Fuel rods contain fuel pellets and
fuel pellets contain Uranium in the form of Oxide. **Why do neutrons escape?**

As it was pointed out, atoms are very empty and a neutron that is not bothered by the attraction or the repulsion of protons and electrons can freely go through the atomic space of a U-235 atom without hitting its nucleus. See the following figure:

The solution to this problem is to increase the
number of layers of atoms by making a huge bulk of uranium fuel so that each
released neutron be absorbed before escaping the reactor core. The minimum bulk
of fuel that must be loaded into the core to sustain a chain reaction is called
the “critical mass”. In practice, more than the critical mass is placed in
the core in the form of **fuel rods**. With all the fuel rods in place just
by themselves, the reactor will explode in no time. This is because some
neutrons are constantly being ejected from some U-235 nuclei (spontaneous decay)
and can start a chain reaction to an avalanche in may be less than one
thousandth of a second. For this reason, not all rods are fuel rods. In a
symmetric way some **control rods** are placed in between the fuel rods.
Control rods are made of neutron absorbing material such as boron

Control rods absorb enough of the
spontaneously released neutrons and avoid the occurrence of a chain reaction.
In order to start a reactor, control rods must be gradually and very carefully
withdrawn to a certain extent so that more of the released neutrons are absorbed
by the fuel nuclei to cause fission. Again, in each fission, some 2 or 3
new neutrons will be generated some of which will be absorbed by U-235
atoms, some by the control rods, and some do escape from the core. The
reactor can be **
sub-critical, critical, or super-critical.** The ratio of neutrons generated
to the ones absorbed by the fuel is very important and determines the
criticality. If the ratio is slightly greater than 1, the reactor becomes
super-critical and produces power. If the ratio is equal to 1, the reactor is
critical and can be at zero power. If the ratio is less than 1, the reactor is
sub-critical. Therefore, **it is the withdrawal of the control rods that can
start a reactor **by allowing more of the neutrons to be absorbed by the fuel
nuclei.

8. __The Decay of a
Radionuclide:__

__Half-life:__
The half life ( T_{1/2}
) of a radioactive element is the time it takes for half of the nuclei to
decay (either convert to another element or become an isotope of the same
element). For example, if the half-life of an element is 3 days and 400 atoms
of it are present at this moment (t=0), after 3 days, only 200 atoms of the same
element will be present and 200 converted. After another 3 days, 100 will be
present and 300 converted. After another 3 days, 50 will be present and 350
converted, and so on ...**.**

__Decay Rate (λ):__

The decay rate λ is
the number of decays occurring per second. By definition,
1 decay per second is called
1 Becquerel.
3.7x10^{10} decays per second is
called 1 Curie. Decay rate __
λ__ and Half-life
T_{1/2} are related by the
following formula that will be verified shortly:

The study of a radionuclide can be started at any given instant. The instant a study starts is considered t = 0.

If N_{o} is
the number of nuclei present at t = 0,
it is then possible to calculate the number of nuclei present
N(t) at any instant
t afterwards. If the decay rate is λ
(decays per second, for example), the rate of change of
N with respect to
t or simply
dN/dt, is proportional to the number
present N as well as
λ, the decay rate. This
proportionality may be written as:

Example 6:
Using the result above, verify the relation between
λ and T_{1/2},
or simply derive the formula**:**

Solution: Note that
after a time elapse equal to T_{1/2}
, the number present will be 1/2 of N_{o}.
This means that for a substitution of t = T_{1/2}
in the above formula, we may substitute (1/2)N_{o}
for N as follows:

Example 7:
The half-life of carbon 14 is 5730 years. If 3**.**60 grams of C-14 existed
in a live organism that died 25000 years ago, how many grams of C-14 is left in
it today?

Solution: The decay
rate λ may be found from the given
half-life. Since T_{1/2 } =
0.693/λ
;therefore,

**1)
Radioactivity comes from**

(a) electronic cloud. (b) nucleus. (c) just excess protons. (d) just excess neutrons. click here.

**2)
Cause of radioactivity is mainly due to**

(a) repulsion of electrons. (b) attraction between neutrons. (c) repulsion between protons.

(d) attraction between protons and neutrons. click here.

**3)
Radioactivity is the ejection from the nucleus of**

(a) electrons. (b) protons and Alpha-particles. (c) neutrons and Gamma-rays. (d) all of the above.

**4)
Gamma-rays are **

**
**(a)
high-energy protons (b)
fast-moving electrons (c)
fast moving neutron

(d) high-energy electromagnetic radiation click here.

**5)
The speed of Gamma-rays are**

(a)
331 m/s (b)
3**.**00x10^{8} m/s (c)
8**.**99x10^{9} m/s (d)
9**.**11x10^{-31}m/s

**6)
The frequency of Gamma-rays is in the range of**

(a)
(10^{14} – 10^{15} )Hz (b)
(10^{15} – 10^{19} )Hz (c)
(10^{20} – 10^{25} )Hz (d)
(10^{10} – 10^{14} )Hz

**7)
For every element**
click here.

(a) the # of protons equals the # of neutrons (b) the # of protons > the # of neutrons

(c) the # of protons less than or equal to the # of neutrons (d) the # of protons equals the # of nucleons

** 8)
Isotopes of an element contain**

(a) equal # of protons but different # of neutrons (b) equal # of neutrons but different # of protons

(c) equal # of neutrons but different # of nucleons (d) equal # of protons but different # of electrons

**9)
If A = #
of nucleon, Z = # of protons, and N = # of neutrons, then**

(a) A = Z + N (b) A = Z – N (c) A = 2Z (d) A = 2N click here.

**10)
The ratio of N / Z**

(a) decreases for heavier elements. (b) remains constant for all elements.

(c) increases for heavier elements. (d) increases for lighter elements. click here.

**11)
Alpha-rays are**

(a) high energy protons. (b) fast moving helium nuclei.

(c) fast moving electrons. (d) Infrared waves.

**12)
Beta-rays are**

(a) stationary electrons. (b) slow moving neutrons.

(c) the same thing as X-rays. (d) fast moving electrons click here.

**13)
Gamma-rays are**

(a) fast moving neutrons. (b) slow eletrons.

(c)
ultraviolet radiation. (d)
high energy electromagnetic waves of frequencies 10^{20} – 10^{25}
Hz.

**14)
Gamma-rays come from**

(a) outer shells electronic transitions. (b) inner shells electronic transitions.

(c) infrared radiation. (d) nuclear transitions. click here.

**15)
Gamma-rays are**

(a) not penetrative into human tissue. (b) penetrative but can be stopped by having a shirt on.

(c) very penetrative and cannot be stopped by just having a shirt on. (d) are milder than Beta-rays.

**16)
A radioactive element
becomes more stable by**

(a) releasing an Alpha-ray (b) releasing a Beta-ray

(c) releasing a Gamma-ray (d) by any or a combination of the above click here.

**17)
As a result of
radioactive decay of an element**

(a) always a new element is produced. (b) always an isotope of the decaying element is produced.

(c) an isotope of the element or a new element is produced. (d) normally a more massive element is produced.

**18)
The decay rate (λ)
is a measure of**
click here.

(a) how fast matter vanishes. (b) how fast an element converts to another element or isotope.

(c) how fast an element loses its outer shell electrons. (d) how fast a radioactive isotope gains helium nuclei.

**19)
The half-life of a
radioactive element is the time**

(a) it takes for half of the radioactive nuclei to decay. (b) that is related to the decay rate.

(c) it takes for half of the mass of the radioactive element to convert. (d) all of the above. click here.

**20)
The way half-life is related
to decay rate is**

(a)
T ½ = 0**.**693 / (decay rate)^{2 }_{
}(b)
T ½ = 0**.**693 * (decay rate)

(c)
T ½ = 0**.**693 + (decay rate)
(d)
T ½ = 0**.**693 / (decay rate)

**21)
If N _{o} is the #
atoms of a radioactive isotope at t = 0, then after a time elapse of 3
half-lives, the percentage present will be**

(a) 50 (b) 25 (c) 12.5 (d) 6.25 click here.

**22)
If N _{o} is the #
atoms of a radioactive isotope at t = 0, then after a time elapse of 4
half-lives, the percentage decayed will be**

(a) 93.75 (b) 87.5 (c) 75 (d) 50

**23)
If N _{o} is the #
atoms of a radioactive isotope at t = 0, then after a time elapse of 5
half-lives, the fraction present will be**

(a) 1/5 (b) 1/32 (c) 1/64 (d) 1/128 click here.

**24)
If N _{o} is the #
atoms of a radioactive isotope at t = 0, then after a time elapse of 6
half-lives, the fraction decayed will be**

(a) 63/64 (b) 15/16 (c) 31/32 (d) 5/6

**25)
Activity is defined as**

(a) the # of radioactive nuclei present at time (t). (b) the # of decays per unit of time at time (t).

(c) the # of radioactive nuclei decayed. (d) the # of protons coming out of nucleus. click here.

**26)
Becquerel is defined as**

(a) the number of decays per second. (b) one decay per second.

(c) one decay per minute. (d) 100 decays per minute click here.

**27)
1 Curie is defined as**

(a) 1 decay per second. (b) 60 decays per minute.

(c)
3**.**7x10^{10} decays per second. (d)
8.99X10^{9} decays per second.

**28)
Concentration of C-14 in the Earth’s atmosphere is**

(a) one C-14 out of 1000 C-12 atoms. (b) one C-14 out of 1000,000 C-12 atoms.

(c) one C-14 out of 1000.000,000 C-12 atoms. (d) one C-14 out of 7.2x100,000,000,000, C-12 atoms.

**29)
C-14 atoms are generated in
atmosphere via the reaction:**

(a)
(N^{14})_{7}+ (n^{1})_{1} == (C^{14})_{6}
+ (H^{1})_{2} . (b)
(N^{14})_{7}+ (n^{1})_{1} == (C^{14})_{7}
+ (H^{1})_{1}.

(c)
(N^{14})_{8}+ (n^{1})_{0} == (C^{14})_{7}
+ (H^{1})_{1}. (d)
(N^{14})_{7}+ (n^{1})_{0} == (C^{14})_{6}
+ (H^{1})_{1}.
click here.

**30)
The way C-14 converts back
to N-14 is via emitting**

(a) an Alpha-ray. (b) a Beta-ray. (c) a Gamma-ray. (d) a proton.

**31)
The half-life of C-14 in its conversion back to N-14 is**

(a) 5.73 years. (b) 57.3 years. (c) 573 years. (d) 5730 years. click here.

**32)
The rate of conversion of
N-14 to C-14 and back to N-14 again is**

(a) equal in living organisms and atmosphere. (b) more in atmosphere than organisms that were once alive.

(c) decreasing in organisms that were once alive. (d) all of the above.

**33)
For every conversion of
(C-14 to N-14) in a live organism or a once alive one**

(a) a Beta-ray is emitted. (b) two Beta-rays are emitted.

(c) 5 Beta-rays are emitted. (d) 5730 Beta-rays are emitted. click here.

**34)
Since the number of C-14
atoms decreases as a result of their conversion back to N-14 in a dead organism**

(a) the rate of Beta-rays emission decreases. (b) the rate of Beta-rays emission increases.

(c) the rate of Beta-rays becomes stable. (d) the rate of Beta-rays quickly drops to zero.

**35)
The activity (λN) of
C-14 (or rate of Beta emission) in live organisms and atmosphere is**

(a) 16 decays/(gr-min). (b) 160 decays/(gr-min).

(c)
1600 decays/(gr-min). (d)
16000 decays/(gr-min).

**36)
If the activity of C-14 (
Beta emission rate) in a once alive organism is 4 decays/(gr-min), the dead
organism is**

(a) 3x (5730) years old. (b) 4x (5730) years old.

(c) (5730) / 2 years old. (d) 2 x (5730) years old.

**37)
Nuclear energy is the result
of**

(a) conversion of mass to energy. (b) annihilation of mass.

(c) the collision of fission fragments and heat generation. (d) all of the above. click here.

**38)
When fission fragments
collide with the wall of the fuel rods,**

(a) electric energy is converted to heat. (b) chemical reaction generates heat.

(c) mechanical energy is converted to heat. (d) some cooling occurs.

**39)
The heat that appears in the
fuel rods of a reactor is**

(a) absorbed by a flow of water. (b) a result of the collision of fission fragments to the rods.

(c) because of nuclear fission. (d) all of the above. click here.

**40)
The formula that calculates mass to energy conversion is**

(a)
E = M^{2}c. (b)
E = Mc^{2} where c is the sound speed.

(c)
The Einstein formula E = Mc^{2}. (d) None of the above.