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Nuclear Physics

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Published in: Physics
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General Notes On Nuclear Physics.

Akhilesh K / Lucknow

4 years of teaching experience

Qualification: M.Sc (NIT Rourkela - 2019)

Teaches: All Subjects, English, Mathematics, Science, Chemistry, Physics, Algebra, IIT JEE Mains, AIPMT, NEET

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  1. The Nucleus Nuclear Composition Nuclear Properties Stable Nuclei Binding Energy notes by akhil
  2. Chapter The Nucleus Introduction Most of the physical and chemical properties of matter which we are familiar with are a result of the number and configuration of atomic electrons. That's why we have spent most of the course to concentrating topics related to atomic electrons. notes by akhil
  3. Nevertheless, atomic nuclei are vitally important for a number of reasons, including: The number of electrons an atom can have depends on how many protons the nuclei has. Thus, the nucleus plays a large, if indirect, role in determining atomic structure. Most of the energies liberated in everyday processes involve nuclear reactions. The first couple of sections of this chapter describe several problems the nucleus presents us. notes by akhil
  4. For example, consider this problem. Take two protons at their approximate separation in the nucleus. Calculate the repulsive Coulomb energy between them. (1.6 x 10-19)2 = 1.4 MeV . 4TtE0 (10-15 That's a large repulsive potential! Far more than electronic bonding energies. Now imagine the repulsive Coulomb energy for sever I dozen protons packed tightly into a nucleus. We have a major problem here. How can a nucle s stick together? notes by akhil
  5. A logical guess is that the nucleus contains electrons, which reduce the Coulomb repulsion. 11.1 Nuclear Composition A good guess might be that half of an atom's electrons are contained within the nucleus, and reduce the electrostatic repulsive forces between protons. Other facts which suggest the nucleus might contain electrons are nuclide masses, which are nearly multiples of the hydrogen mass (which contains an electron). In addition, some nuclei undergo beta decay, in which electron is spontaneously emitted from the nucleus. But other experiments demonstrate that the nucleus cannot ncontain electrons...
  6. Reason 1 -- nuclear size. The Heisenberg uncertainty principle places a lower bound on the energies of particles confined to a nucleus. Take a typical nucleus of radius 5x10-15 m. Suppose an electron exists inside the nucleus. In the example on page 114, we estimated the minimum momentum such an electron must have. The estimated momentum corresponds to a kinetic energy of at least 20 MeV. Electrons emitted during nuclear decay are found to have only 2 or 3 MeV of energy—not nearly enough to correspond to electron escaping from a nucleus. Protons, with their much larger masses, would only need to have a few tenths of an MeV of energy to be confined to a nutléuS. this is possible.
  7. Reason 2 -- nuclear spin. Electrons and protons both have spins of 1/2. A deuteron (an isotope of hydrogen) has a mass roughly equal to two protons. If the deuterium nucleus contains two protons and one electron (whose mass is small enough to not worry about here), then deuterium should have a nuclear spin of or ±3/2 (from ± 1/2 ± 1/2 ± 1/2). The deuterium nuclear spin is measured to be 1. Its nucleus cannot contain an electron. (If it did, angular momentum would not be conserved.) notes by akhil
  8. Reason 3 -- nuclear magnetic moments. Electrons have magnetic moments about 6 times larger than protons. If nuclei contain electrons, their magnetic moments should be comparable to electron magnetic moments. Observed nuclear magnetic moments are comparable to proton magnetic moments. Nuclei cannot contain electrons. notes by akhil
  9. Reason 4 -- electron-nuclear interactions. The energies binding nuclear particles together are observed to be very large, on the order of 8 MeV per particle. Remember that atomic electronic binding energies are of the order ev to a few keV. Why, then, can some atomic electrons "escape" from being bound inside the nucleus? In other words, if you allow any electrons to be bound in the nucleus, you really must require all of them to. notes by akhil
  10. Of course, it is obvious to us that nuclei don't contain elect 10 IIS. But that's mainly because we've been taught that way for so long. If we were starting from scratch 70 years ago, we would probably try to "put" electrons inside nuclei. Some definitions and facts. The most abundant type of carbon atom is defined to have mass of exactly 12 u, where u is one atomic mass unit: lu 27 kg - 931.48 MeV . notes by akhil
  11. Atomic masses always refer to neutral atoms. In other words, atomic masses include the masses of all of the electrons in the neutral atom. Isotopes are atoms of the same element having different masses. A nuclide is simply any particular nuclear species. Hydr e and deuterium are isotopes. They are also nuclides. C 12 is a nuclide, but it is not an isotope of hydrogen. notes by akhil
  12. neutrons Rutherford predicted the existence of the neutron in 1920. The neutron, being a neutral particle, proved difficult to detect. 'Many" tried and failed. In 1930, German Physicists Bothe and Becker were experimenting with alpha particles and beryllium. When they bombarded beryllium with alpha particles, the beryllium emitted a mysterious radiation. The radiation was neutral (they could test for th With magnetic fields) and passed through a whopping 200 millimeters of lead (one millimeter stops a proton). notes by akhil
  13. The only uncharged particle known at the time was the photon. Irene Curie and her husband Frederic tried putting a block of paraffin wax in front of the mysterious beam coming out of the beryllium. Huh? Paraffin? Paraffin is made of light hydrocarbons. It contains lots of protons. It is a good "test block" for studying collisions. notes by akhil w w.g'
  14. The "mystery rays" knocked protons out of the paraffin. alpha protons Be The protons come out with energies up to 5.7 MeV (big!). The gamma ray energy needed to produce such energetic protons is about 55 MeV. Gamma rays of this much energy were not observed. About 20% of the observed 5.7 MeV energy is the ost that can be produced by gamma rays. ---note y akhil
  15. Chadwick in 1932 proposed that the unknown radiation could be neutral particles having about the mass of protons. Charge neutrality is necessary radiation to easily penetrate matter. for Because a collision between particles the of equal mass can transfer all of the kinetic energy from the projectile to th target, the neutrons needed to have only 5.7 MeV of e which was a much more reasonable value. notes by akhil
  16. Chadwick devised an experiment to test his hypothesis. http://hyperphysics.phy- astr.gsu.edu/hbase/particles/neu trondis.html The discovery of the neutron won Chadwick the 1935 Nobel prize. The neutron: Polonium source Analyze Boron maximum target forward scattered N "in N? detector Target mass=1.00867u, just a little more than the proton charge=0 spin=1/2 is unstable outside of nuclei (lifetime is abou minutes), and decays into a proton, an electron, n antineutrino. neutrons produce attractive forces which help hold nuclei notes by akhil together.
  17. The Nucleus. The number of protons in a nucleus (and electrons in the atom, if the atom is not ionized) is represented by Z. N is the number of neutrons in the nucleus. The atomic mass number A is given by A=Z+N. Neutrons and protons are called nucleons, so A is the number of nucleons in a nucleus. We identify nuclides by writing For example, the mos abundant isotope of iron has 26 pro ons and electrons, mass number of 56, so we write notes by akhil
  18. Isotopes. Isotopes of an element have the same number of protons and electrons, but different numbers of neutrons. Because most physical and chemical properties are determined by the number and arrangement of atomic electrons, isotopes of an element are very similar in behavior. As Beiser states, all isotopes of chlorine make good bleach and are poisonous. Some properties, such as density and freezing points, ar different for different isotopes of the same element, bu differences are usually so slight that isotopes are diffic '[t to sepa rate. notes by akhil
  19. 11.2 Some Nuclear Properties Nuclear sizes are usually measured by "scattering." Experimentally, using neutrons of energy 20 MeV or more, or electrons of 1 GeV (109 eV) or greater, it is found that the volume of a nucleus is proportional to the number of nucleons (neutrons and protons) it contains. Since the mass number A is proportional to volume and volume is proportional to the R3, where R is the nuclear radius, it follows that R is proportional to Al/3. We usually write R = R Al/3 where Rp is a constant and Ro m notes by ak 1.2 fm.
  20. The nucleus does not have a sharp boundary, so the 'constant" Ro is only approximate; also, nuclear matter and nuclear charge do not seem to be identically distributed. The unit of length 10-15 m is called a femtometer, abbreviated fm, and also often called a fermi, so R=1.2A1/3 in units of fm. Example: the radius of the 5.7 fm. nucleus is R = If a nucleus is not spherically symmetric, it will produce an electric field that will perturb atomic electronic energy levels. Such an effect is, in fact, observed, but it is small -- "hyperfine." The departures from spherical symmetry are småll. y akhil
  21. 11.3 Stable Nuclei We can begin to understand why certain nuclei are stable and others unstable by realizing that nucleons have spins of 1/2 and obey the Pauli exclusion principle. Nucleons, like electrons and their electronic energy levels, occupy discrete nuclear energy levels. Minimum energy configurations (i.e., nucleons in the lo possible energy levels) give the most stable nuclei. notes by akhil
  22. A plot of N versus Z for the stable nuclides looks like this: (I see two "typos," not mine!) Neutrons produce attractive forces within nuclei, and help hold the protons together. For small numbers of protons, about an equal number of neutrons is enough to provide stability, hence NZZ for small Z. 140 120 S 100 80 E 0 60 40 20 Proton Nurrber (Z) As the number of protons gets larger, an excess of neu o needed to overcome the proton-proton repulsion. notes by akhil is
  23. The stability of nuclei follows a definite pattern. The majority of stable nuclei have both even Z and even N nuclides). Most of the rest have either even Z and odd N ("even-odd") or odd Z and even N ("odd-even"). Very few stable nuclei have both Z and N odd. The reasons for this pattern are the Pauli exclusion principle and the existence of nuclear energy levels. Each nuclear energy level can contain two nucleons of opposite spin. notes by akhil
  24. The neutrons and protons occupy separate sets of energy levels. When both Z and N are even, the energy levels can be filled. The nucleus doesn't "want" to gain or lose nucleons by participating in nuclear reactions. The nucleus is stable. When both Z and N are odd, the nucleus is much more likely to "want" to participate in nuclear reactions or nuclear decay, because it has unfilled nuclear energy levels. notes by akhil neutrons neutrons neutrons
  25. Example: lic All of its neutrons and protons in filled energy levels-......-very Energy level diagrams illustrative only; not quantitatively accurate! Example: 'Extra" neutron in higher energy level; therefore unstable. Decays via ß decay intoÜ neutrons neutrons
  26. Attractive nuclear forces are limited in range and primarily operate between nearest neighbors ("saturation"), so there is a nuclear size beyond which neutrons are unable to overcome the proton-proton repulsion. The heaviest stable nuclide is . Heavier ones deca into lighter nuclides through alpha ecay (the emission of a nucleus): X A} 42 y + . z X is called the parentnucleus and Y is called the daugh nucleus. notes by akhil
  27. It may be that a nucleus produced by alpha decay has too many neutrons to be stable. In this case, it may decay via beta decay: 0 If the nucleus has too few neutrons, it may decay via positron emission or by electron capture 0 Note that Z decreases by 1 as a result of positron emiss- electron capture, but increases by 1 as a result of beta decay. notes by akhil
  28. 11.4 Binding Energy The binding energy that holds nuclei together "shows up" as 'missing" mass. Deuterium is an isotope of hydrogen which contains a neutron, a proton, and an "orbiting" electron. mass of hydrogen mass of neutron sum mass of deuterium difference notes by akhil 1.0078 u 1.0087 U 2.0165 u 2.014 0.0024 u
  29. Since 1 u of mass has an energy equivalent of 931 MeV, the missing mass is equal to MeV = 2.2 MeV. The fact that this mass deficit is the binding energy is demonstrated by experiments which show that it takes 2.2 MeV of energy to split a deuterium into a neutron and a proton. Nuclear binding energies range from 2.2 MeV for deuterium to 1640 MeV for bismuth-209. These binding energies are enormous; millions of times greater than even the energies given off in highly energetic chemical reactions. We usually talk in terms of binding energy per nucleon Ich is 2.2/2=1.1 MeV per nucleon for deuterium, or 1640/209=7.8 MeVepernucIeon for bismuth-209.
  30. The figure below shows a plot of binding energy per nucleon as a function of nucleon number. 10 8 6 4 2 0 0 50 100 150 Mass Num ber, A 200 250 Keep in mind that energies are reduced on binding. The binding energy is negative, but when we say the words "binding energy" we associate them with the magnitud Vt e binding energy. notes by akhil
  31. 56 Notice the local minimum at nucleus. 100 150 250 Mass number, A which is a very stable Notice the absolute minimum at nucleus of all. which is the st stable
  32. Heavier elements are less stable, but stable enough to exist. It takes enormous energies to make elements heavier than iron- 522Fe . The only place in the universe where those 56 energies are available are supernovae. Nuclear Fusion and fission Two remarkable conclusions can be drawn from the curve o Fig. 11.12* T e irst is that if we can somehow split a heavy nucleus into two medium-sized ones* each of the new„v nuclei will have more binding energy per nucleon than the original nucleus did. The extra energy will be given off* and it can be a lot* For instance, if the uranium nucleus 921-J is broken into two smaller nuclei* the binding energy difference per nucleon is about 0.8 MeVt The total energy given off is therefore MeV (235 nucleons) = 188 McV nucleon This is a truly enormous amount of energy to be produced in a single atomic event, As know, ordinary chemical reactions involve rearrangements of the electrons in atoms liberate only a few electronvolts per reacting atom. Splitting a heavy nucleust note by a Il whicll is called nuclear fission, thus involves 100 million times more energy per atom than, sa)4 the burning of coal or oil.
  33. Nuclear Fusion and fission The other notable conclusion is that joining two light nuclei together to give a single nucleus of medium size also means more binding energy per nucleon in the new nucleus, For instance, if two IH deuterium nuclei combine to fom a {He helium nucleus* over 23 Mel/ is' released, Such a proces •called nuclear fusion) is also a veil effective way to 0b tain ener In nuclear fusion is the main energy source of the sun and other stars notes by akhil
  34. 10 56 50 100 150 Mass Num ber, A 200 250 Consider a nucleus with a large A. If we could split it into two smaller nuclei, with A's closer to iron, the two nuclei would have more binding energy per nucleon. But remember, binding energies are negative. If the resu nuclei have more negative energy than the starting ele e some positive energy must have been released in Itting the Sta elernent.
  35. The positive energy is the energy released in the reaction. 10 26 50 100 150 Mass Num ber, A 200 250 If we begin with two nuclei significantly lighter than iroo- and somehow make them fuse, the resulting nUCIeUS I I h ve more binding energy per nucleon. notes by akhil
  36. sample binding energy problem Home work problem 11.16 Find the binding energy per nucleon in 1%Au. [M- zmH - 931.5 mass of atom mass of hydrogen mass of neutron converts to MeV Note that all masses must be in units of u, and all e ectrons are automatically counted in this calculation.
  37. [M- zrnH - 931.5 This gives the total nuclear binding energy for the atom. usually want the binding energy per nucleon, so: b_per_nucleon We Because is in units of MeV, the binding energy per nucleon is in units of MeV/nucleon. Now, back to our problem. The mass of gold-197 is 196.966560 u. notes by akhil
  38. The mass of hydrogen is 1.007825 u and the mass of a neutron is 1.008665 u. Yes, you do need to keep all the decimal places. Note the hydrogen mass includes the mass of one electron— enough to make a difference! For gold-197, A=197 and Z=79. [M- zmH - 931.5 = [(196.966560)- ((197)- 931.5 = -1559 MeV b_per_nucleon = -7.916 MeV . 197 Thetbjnding energy is negative, as it must be.

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