Chapter 43
Nuclear Physics
In this chapter the following topics will be discussed:
1. The Cause of Radioactivity:
Protons and neutrons form the nucleus of atoms. Radioactivity comes from the nucleus of atoms where repelling protons are somehow held together. Neutrons help protons to be held together by creating a certain distance between them to reduce repulsion. In addition to the repulsive force between every two protons, an attraction force is also at work at the nucleus that is called the "short range attraction force." Every two protons repel each other by the Coulomb's repulsive force regardless of the distance between them while the short range attraction force is pulling them together. Within very small distances the short range attraction force overcomes the repulsive force and holds the protons together. If by any means the distance between the protons (usually separated by neutrons in between them) 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 is also the emergence of very energetic photons (strong electromagnetic radiation) called the "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 their protons together in order to 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 O) 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: Its K.E. in Jules is K.E. = 25x10^{+3}(1.6x10^{-19})J = 4.0x10^{-15}J.
The mass of He nucleus is the mass of 2 protons + 2 neutrons.
Using the K.E. formula: K.E. = (1/2)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 a β-ray 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.
Solution: Left to students.
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, alpha, beta, gamma, and some other types of radiation can occur. As a result of the decay of an element, a new isotope of that 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. The λ of an element or isotope is a measure of how fast that element keeps 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_{1/2} of a radioactive element is the time it takes for half of its nuclei (or amount) to convert to different type of nuclei. It is easy to show that
At t = 0 , if 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 at the end of the 9.00th day, and (b) the number converted to a different element in that 9.00-day period.
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. Since N, or N(t) is the number of radioactive nuclei present at time (t), this product or activity gives the number of decays or conversions per second at any given time (t). The simplest unit for activity λN is Becquerel (Bq). 1 Bq = 1 decay/sec.
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 the collision of neutrons with N-14 atoms convert them to C-14 atoms or isotopes. The nuclear reaction is shown below:
C-14 then decays back to N-14 by emitting a Beta-ray. For every C-14 decay, one β-ray is emitted.
Living organisms take in N_{2} & O_{2} molecules some of which convert to C-14. The N_{2} molecules and C-14 atoms enter and leave 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 starts decreasing immediately after death. However, since the half-life of C-14 is 5730 years, after every 5730 years, the concentration of C-14 halves. Since C-14 concentration can be measured by the rate of beta-decay, a beta-radiation measurement of an ancient object can give us the age of it.
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 2x5730yrs = 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 is 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 used due to safety reasons?
Solution: E = Mc^{2} = 0.00100kg (3.00x10^{8}m/s)^{2} = 9.00x10^{13} J.
The weight of 1.00ton or 1000kg of water is 9800N.
To pump 1 ton to a height of 1609m, an energy of W = Fx = 9800N(1609m) is needed.
Energy/ton = 15.8 x10^{6} J.
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 a few percentage more 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 extremely empty. The way nuclear energy (in a Uranium Fuel Reactor) is generated is that each atom of U-235 must be collided by a neutron so that its energy balance is disturbed causing it to split into two or three fragments. The splitting of a nucleus into fragments of high speed is called "fission." It is the kinetic energy of the fast moving fission fragments that is called the "nuclear energy."
The types of fission fragments and the way a fissile isotope (U-235 in this case) goes through fission, depend on the energy of the target nucleus as well as the energy of the colliding neutron. Uranium-235 that is the target nucleus in this case is sensitive to the "slow neutrons."
What we call "slow neutrons" are those at speeds around 2200m/s like the average speed of air molecules at Room Temperature (25 degrees Celsius). Fast 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 reactor 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 extra neutrons that can be used to split other U-235 nuclei with. 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 twenty millionth of this energy!
U-235 + 1 neutron → Fragment A + Fragment B + 200 MeV.
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 extremely 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 U-235 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 the 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 become super critical 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 uncontrolled occurrence of a chain reaction. Many safety measures are in place that make explosion impossible to occur. 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 very 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 produce zero to very low 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. The above few paragraphs are not meant to reveal a precise explanation of reactor dynamics. They just give you a general idea of how nuclear energy is produced and how a reactor is controlled.
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 measurement on a radionuclide can be started at any given instant. The instant at which a study or measurement starts is considered to be t = 0.
If N_{o} is the number of nuclei present at t = 0, it is possible to calculate the number of nuclei present at any instant t afterwards that we write it as N(t). If the decay rate is λ decays per second, for example, the rate of change of N or N(t) with respect to t or simply dN/dt, is proportional to the number present N as well as λ. This proportionality may be written as:
Example 6: Using the above result, 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: Decay rate λ may be found from half-life. Since T_{1/2 }= 0.693/λ ;therefore,
1) Radioactivity comes from
(a) electronic cloud (b) nucleus (c) excess protons (d) excess neutrons.
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 a, b, and c.
4) Gamma-rays are
(a) high-energy protons (b) fast-moving electrons (c) fast moving neutrons
(d) high-energy electromagnetic radiation. click here.
5) The speed of Gamma-rays are
(a) 331m/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 (in Hz) is in the range of
(a) 10^{14}- 0^{15} (b) 10^{15}- 10^{19} (c) 10^{20} -10^{25} (d) 10^{10} - 10^{14}.
7) For every element
(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.
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 N/Z ratio
(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 electrons
(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
(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_{1/2} = 0.693 /(decay rate)^{2 }_{ }(b) T_{1/2} = 0.693 * (decay rate)
(c) T_{1/2} = 0.693 + (decay rate) (d) T_{1/2} = 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 one out of
(a) 1x10^{3} (b)1x10^{6} (c)1x10^{9} (d) 7.2x10^{11} 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 yr (b) 57.3 yr (c) 573 yr (d) 5730 yr. click here.
32) The rate of conversion of N-14 to C-14 and back to N-14 again is
(a) equal in live 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) one Beta-ray emits (b) two Beta-rays emit
(c) 5 Beta-rays emit (d) 5730 Beta-rays emit. 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, the rate of Beta-rays emission
(a) decreases (b) increases (c) stabilizes (d) quickly drops to zero.
35) The activity λN of C-14 or Beta emission rate 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). click here.
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 walls 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 a, b, or c.