Chapter 10 - Nuclear Physics

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Transcription:

The release of atomic energy has not created a new problem. It has merely made more urgent the necessity of solving an existing one. -Albert Einstein David J. Starling Penn State Hazleton PHYS 214

Ernest Rutherford proposed and discovered the nucleus in 1911.

Ernest Rutherford proposed and discovered the nucleus in 1911. Alpha particles are positively charged helium nuclei 4 He.

Their scattering angle theory matched predictions!

Their scattering angle theory matched predictions! Geiger and Marsden conducted the experiments with Radon gas.

Protons (Z) and Neutrons (N) combine to give the mass number (A) of the atom.

Protons (Z) and Neutrons (N) combine to give the mass number (A) of the atom. Note how the masses are not whole number multiples of Hydrogen.

Atoms tend to have more neutrons than protons, and can have multiple stable isotopes.

Atoms tend to have more neutrons than protons, and can have multiple stable isotopes. Bismuth (Z = 83) is the largest stable nucleus.

Stable nuclides have different abundances; radionuclides have different half-lives.

The radius of nuclides can be roughly approximated. r = r 0 A 1/3 with r 0 = 1.2 fm and A the mass number.

The radius of nuclides can be roughly approximated. r = r 0 A 1/3 with r 0 = 1.2 fm and A the mass number. This only applies to spherical nuclides and not to unstable halo nuclides (e.g., 11 3 Li).

Atomic masses are reported in atomic mass units with Carbon-12 defined to be 12 u. 1 u = 1.660538 10 27 kg

Atomic masses are reported in atomic mass units with Carbon-12 defined to be 12 u. 1 u = 1.660538 10 27 kg m p = 1.672623 10 27 kg m n = 1.674929 10 27 kg

Atomic masses are reported in atomic mass units with Carbon-12 defined to be 12 u. 1 u = 1.660538 10 27 kg m p = 1.672623 10 27 kg m n = 1.674929 10 27 kg Therefore, the mass number of an atom and its atomic mass are very similar (e.g., 197 Au has a mass of 196.966522 u).

It takes energy to pull protons and neutrons apart; therefore, they lose energy when they come together.

It takes energy to pull protons and neutrons apart; therefore, they lose energy when they come together.

It takes energy to pull protons and neutrons apart; therefore, they lose energy when they come together. This change in energy is equivalent to a change in mass (via E = mc 2 ).

The binding energy is just E b = i m i c 2 Mc 2 > 0.

The binding energy is just E b = i m i c 2 Mc 2 > 0.

The binding energy is just E b = i m i c 2 Mc 2 > 0. The mass excess is M A.

Once formed, nuclides have energy levels similar to the electrons.

Once formed, nuclides have energy levels similar to the electrons.

Once formed, nuclides have energy levels similar to the electrons. However, much more energy is required to excite the nucleus.

Lecture Question 10.1 Which statement is true about the forces inside an atom? (a) Gravity holds electrons, while the strong nuclear force holds nuclei together. (b) Gravity holds electrons in their orbits and nuclei together. (c) Gravity holds electrons, while the electromagnetic force holds nuclei together. (d) The strong nuclear force holds electrons, while the electromagnetic force holds nuclei together. (e) The electromagnetic force holds electrons, while the strong nuclear force holds nuclei together.

Unstable nuclei decay into other nuclei. For a sample of N particles, the rate is proportional to N with decay constant λ. dn dt = λn

Unstable nuclei decay into other nuclei. For a sample of N particles, the rate is proportional to N with decay constant λ. Solving for N: dn dt = λn dn N = λdt

Unstable nuclei decay into other nuclei. For a sample of N particles, the rate is proportional to N with decay constant λ. Solving for N: dn dt = λn N N 0 dn N = λdt dn N = λ t 0 dt

Unstable nuclei decay into other nuclei. For a sample of N particles, the rate is proportional to N with decay constant λ. Solving for N: dn dt = λn N N 0 dn N = λdt dn N = λ ln N N 0 = λt t 0 dt

Unstable nuclei decay into other nuclei. For a sample of N particles, the rate is proportional to N with decay constant λ. Solving for N: dn dt = λn N N 0 dn N = λdt dn N = λ ln N N 0 = λt t 0 N = N 0 e λt dt

The number of nuclei decays exponentially.

The number of nuclei decays exponentially. 1 becquerel = 1 Bq = 1 decay per second

The number of nuclei decays exponentially. 1 becquerel = 1 Bq = 1 decay per second 1 curie = 1 Ci = 3.7 10 10 Bq

The decay rate is minus the derivative of N. R = dn dt = λn 0 e λt = R 0 e λt

The decay rate is minus the derivative of N. R = dn dt = λn 0 e λt = R 0 e λt The half-life is the time to reduce the nuclei number (or decay rate) by 1/2. R = R 0 e λt 1/2 = R 0 /2 T 1/2 = ln 2/λ = τ ln 2

The decay rate is minus the derivative of N. R = dn dt = λn 0 e λt = R 0 e λt The half-life is the time to reduce the nuclei number (or decay rate) by 1/2. R = R 0 e λt 1/2 = R 0 /2 T 1/2 = ln 2/λ = τ ln 2 The mean life time τ = 1/λ is when N has been reduced to N 0 /e.

Alpha decay is when a heavy nucleus emits an alpha particle (i.e., 4 He nucleus).

Alpha decay is when a heavy nucleus emits an alpha particle (i.e., 4 He nucleus). The alpha particle is bound within the 238 U nucleus but can tunnel out with some small probability, leaving behind a 234 Th nuclues.

Beta decay is when a proton or neutron decays into the other, emitting an electron or positron to preserve charge. 32 15P 32 16S + e + ν 64 29Cu 64 28Ni + e + + ν

Beta decay is when a proton or neutron decays into the other, emitting an electron or positron to preserve charge. 32 15P 32 16S + e + ν 64 29Cu 64 28Ni + e + + ν The half life for these reactions is 14.3 d and 12.7 h, respectively. The very low mass ν particle is a neutrino.

Since two particles share the energy of beta emission, the positron/electron can have a range of energies.

Since two particles share the energy of beta emission, the positron/electron can have a range of energies. For beta decay of copper to nickel, the most probable positron emission energy is 0.15 MeV.

Neutrinos interact very weakly with matter and so large detectors are required to see even the most massive supernova events.

Neutrinos interact very weakly with matter and so large detectors are required to see even the most massive supernova events. This data is from the Super-Kamiokande detector in Japan, 1000 m underground, holding 50,000 tons of ultra-pure water.

Unstable nuclides decay toward stable ones over time; the nuclidic chart (abbreviated) helps understand how this works.

Unstable nuclides decay toward stable ones over time; the nuclidic chart (abbreviated) helps understand how this works. Radioactive nuclides transform through alpha decay, beta decay, emission of protons or neutrons, or fission into daughter particles.

Using the half-life of radioactive decay and measurements of various nuclides, we can date the age of objects.

Radiocarbon dating places the age of the Dead Sea Scrolls from the West Bank at around 2,000 years old. Using the half-life of radioactive decay and measurements of various nuclides, we can date the age of objects.

There are two ways to measure radiation doses. Absorbed Dose Independent of type of radiation 1 gray = 1 Gy = 1 J/kg 1 Gy = 100 rad

There are two ways to measure radiation doses. Absorbed Dose Independent of type of radiation 1 gray = 1 Gy = 1 J/kg 1 Gy = 100 rad Dose Equivalent Accounts for type of radiation for biological purposes RBE factors (relative biological effectiveness) 1 sievert = 1 Sv 1 Sv = 100 rem

There are two ways to measure radiation doses. Absorbed Dose Independent of type of radiation 1 gray = 1 Gy = 1 J/kg 1 Gy = 100 rad Dose Equivalent Accounts for type of radiation for biological purposes RBE factors (relative biological effectiveness) 1 sievert = 1 Sv 1 Sv = 100 rem Personal monitoring devices measure Dose Equivalent.

Why does the nucleus behave like this? Collective Model nucleons bounce around at random short mean free path helps describe nuclear reactions

Why does the nucleus behave like this? Collective Model nucleons bounce around at random short mean free path helps describe nuclear reactions Independent Particle Model nucleons exist in stable quantum states explains why some nuclei show closed-shell effects collisions only occur for open quantum states

Why does the nucleus behave like this? Collective Model nucleons bounce around at random short mean free path helps describe nuclear reactions Independent Particle Model nucleons exist in stable quantum states explains why some nuclei show closed-shell effects collisions only occur for open quantum states Combined Model combines features of previous models shell of outside nucleons occupy quantized states these nucleons interact with core and create tidal waves

Lecture Question 10.2 When bismuth 211 Bi undergoes alpha decay, what nucleus is produced in addition to a helium nucleus? (a) 207 Bi (b) 207 Tl (c) 209 Au (d) 211 Au (e) 209 Tl

Energy can be generated in a variety of ways.

Energy can be generated in a variety of ways. Nuclear reactions tend to release much more energy.

Fission happens when a species becomes unstable.

Fission happens when a species becomes unstable. Energy is released by fast moving neutrons.

As the nucleus distorts (r), the nuclide s potential energy varies.

As the nucleus distorts (r), the nuclide s potential energy varies. Once the distortion grows beyond 5 fm, the nuclear reaction occurs.

When a neutron is captured, does the added energy exceed the potential barrier?

When a neutron is captured, does the added energy exceed the potential barrier? Only certain radionuclides are fissionable by neutron capture.

When a radionuclide undergoes fission, its fragments can vary.

When a radionuclide undergoes fission, its fragments can vary. The most probable mass numbers are around A 95 and A 140.

For 236 U, when a neutron is ejected into a neighbor, we may get: 235 U + n + 236 U 140 Xe + 94 Sr + 2n

For 236 U, when a neutron is ejected into a neighbor, we may get: 235 U + n + 236 U 140 Xe + 94 Sr + 2n But 140 Xe and 94 Sr are unstable and will decay further: 140 54 Xe 14 s 140 55 Cs 64 s 140 56 Ba 13 d 140 57 La 40 h 140 58 Ce 94 38 Sr 75 s 94 39 Y 19 m 94 40 Zr

To calculate the energy released, you compare the binding energies before and after the reaction has completed. Q = j E ben,jn j i E ben,i N i

To calculate the energy released, you compare the binding energies before and after the reaction has completed. Q = j E ben,jn j i E ben,i N i E ben,i is the binding energy per nucleon for species i N i is the number of nucleons of species i Primes indicate after reaction Sum i is over initial nuclides (e.g., 236 U) Sum j is over final products (e.g., 140 Xe and 94 Sr)

Consider previous example: 236 U 140 Xe + 94 Sr + 2n )

Consider previous example: 236 U 140 Xe + 94 Sr + 2n Q = j E ben,jn j i E ben,i N i ) Binding energies: Uranium-236 (7.59 MeV), Xenon-140 (8.29 MeV) and Strontium-94 (8.59 MeV).

Consider previous example: 236 U 140 Xe + 94 Sr + 2n Q = j E ben,jn j i E ben,i N i = ( E140 Xe 140 + E94 Sr 94) ( E236 U 236) Binding energies: Uranium-236 (7.59 MeV), Xenon-140 (8.29 MeV) and Strontium-94 (8.59 MeV).

Consider previous example: 236 U 140 Xe + 94 Sr + 2n Q = j E ben,jn j i E ben,i N i = ( E140 Xe 140 + E94 Sr 94) ( E236 U 236) = 8.29 140 + 8.59 94 7.59 236 Binding energies: Uranium-236 (7.59 MeV), Xenon-140 (8.29 MeV) and Strontium-94 (8.59 MeV).

Consider previous example: 236 U 140 Xe + 94 Sr + 2n Q = j E ben,jn j i E ben,i N i = ( E140 Xe 140 + E94 Sr 94) ( E236 U 236) = 8.29 140 + 8.59 94 7.59 236 = 177 MeV Binding energies: Uranium-236 (7.59 MeV), Xenon-140 (8.29 MeV) and Strontium-94 (8.59 MeV).

A nuclear reactor must operate in a stable configuration by managing neutron number and speed.

A nuclear reactor must operate in a stable configuration by managing neutron number and speed. For fission, fast neutrons are not as effective as slow, thermal neutrons.

The first man-made reactor was built on a squash court at the University of Chicago during World War II.

The first man-made reactor was built on a squash court at the University of Chicago during World War II. The fuel was lumps of Uranium embedded in blocks of graphite.

So you spit a bunch of atoms. Now what?

So you spit a bunch of atoms. Now what? You heat water!

Lecture Question 10.3 How many neutrons are released in the following reaction: 235 92 U + 1 0n 88 38Sr + 136 54 Xe + 1 0n (a) 1 (b) 3 (c) 6 (d) 8 (e) 12

Nuclear fusion combines lighter atoms (with low E ben ) to form heavier atoms (with higher E ben ).

Nuclear fusion combines lighter atoms (with low E ben ) to form heavier atoms (with higher E ben ). Combining hydrogen into helium is how our sun produces so much energy.

Thermonuclear fusion is when the temperature of bulk matter is high enough for atoms to overcome Coulomb repulsion during collisions.

Thermonuclear fusion is when the temperature of bulk matter is high enough for atoms to overcome Coulomb repulsion during collisions. In the sun, hydrogen above 3 kev has an appreciable chance to fuse.

In the sun, six 1 H atoms and an electron can produce one 4 He atom, two 1 H atoms, six photons and two neutrinos.

In the sun, six 1 H atoms and an electron can produce one 4 He atom, two 1 H atoms, six photons and two neutrinos. This proton-proton chain produces 26.7 MeV total.

When a star burns most of its hydrogen, it either begins to fuse Helium to create heavier elements, or begins to contract and die.

For very massive stars (e.g., Sanduleak), this death can result in a supernova followed by a neutron star, pulsar, black hole or total disintegration. When a star burns most of its hydrogen, it either begins to fuse Helium to create heavier elements, or begins to contract and die.

A controlled thermonuclear fusion reaction is possible using one of the following chains: 2 H + 2 H 2 H + 2 H 2 H + 3 H +3.27 MeV +4.03 MeV +17.59 MeV 3 He + n 3 H + 1 H 4 He + n

A controlled thermonuclear fusion reaction is possible using one of the following chains: 2 H + 2 H 2 H + 2 H 2 H + 3 H +3.27 MeV +4.03 MeV +17.59 MeV 3 He + n 3 H + 1 H 4 He + n Requirements: High density High temperature Long confinement time

Nuclear fusion has two main research paths.

Nuclear fusion has two main research paths. Magnetic confinement of hot plasma, and laser confinement of fuel pellets.

Deuterium ( 2 H) and tritium ( 3 H) occur naturally in abundance and can be harvested from sea water.

Deuterium ( 2 H) and tritium ( 3 H) occur naturally in abundance and can be harvested from sea water. Formed into pellets, they can be fused via laser confinement.

Lecture Question 10.4 Consider the following nuclear fusion reaction and identify particle X. 2 1H + 3 1H 4 2He + X (a) a photon (b) a proton (c) a neutron (d) a positron (e) an electron