Nuclear Physics PHY232 Remco Zegers zegers@nscl.msu.edu Room W109 cyclotron building http://www.nscl.msu.edu/~zegers/phy232.html
Periodic table of elements We saw that the periodic table of elements can be used to visualize the atomic structure of atoms. It describes the structure of the electron shells in the atom which is important to understand chemical properties of the elements PHY232 - Remco Zegers - nuclear physics 2
quiz (extra credit) Photon A is emitted when an electron in a hydrogen atom drop from the n=3 level to the n=2 level. Photon B is emitted when an electron in a hydrogen atom drops from the n=4 to the n=2 level. In which case is the wavelength of the light larger? a) Photon A b) photon B c) the wavelengths are the same PHY232 - Remco Zegers - nuclear physics 3
the nucleus The core of the atom is the nucleus and it consists of neutron and protons. To understand the properties of the nucleus, a quantum-mechanical treatment is necessary (just like for the electrons orbiting the nucleus). The quantum mechanical description is quite similar to that for the electrons, resulting in a shell-model that describes the structure of the nucleus and its excited states. each constituent is described by a wave function that gives the probability of finding the nucleon at a certain place PHY232 - Remco Zegers - nuclear physics 4
Z identifying nuclei N In the periodic table of elements, there is only one entry for a given atom (e.g. Carbon). However, for a given number of protons (Z), which determines the name of the element, a variable amount of neutrons can be present. For example, in the case of Carbon, two stable isotopes exist, one with 6 neutrons ( 12 C, which is most abundant in nature) and one with ( 13 C) 7 neutrons, which is relatively rare. The other isotopes are unstable (e.g. 14 C) and only live for a short period of time PHY232 - Remco Zegers - nuclear physics 5
identification of nuclei Mass number A: total number of nucleons. A=Z(protons)+N(neutrons) Atomic number Z: number of protons Neutron number N: number of neutrons Note that a notation using just the character for the element and the mass number is sufficient to identify the nucleus. For example: 26 Mg: Mg tells you that Z=12 and A=26, so N=14 The radius of the nucleus can be calculated using: r=r 0 A 1/3 with r0 =1.25x10-15 m (1.25 fm) PHY232 - Remco Zegers - nuclear physics 6
Nuclear masses Particle kg u MeV/c 2 Proton 1.673x10-27 1.007276 938.28 Neutron 1.675x10-27 1.008665 939.57 Electron 9.11x10-31 5.486x10-4 0.511 1 unified mass unit: mass( 12 C)/12 Einstein: E=mc 2 so: m=e/c 2 1eV=1.60217733x10-19 J 1MeV=1.60217733x10-13 J 1u=931.494 MeV/c 2 PHY232 - Remco Zegers - nuclear physics 7
example: what is the radius of 12 C? what is the radius of 208 Pb? 12 C: r=r 0 A 1/3 =1.25x12 1/3 =2.86 fm 208 Pb: r=r 0 A 1/3 =1.25x208 1/3 =7.40 fm PHY232 - Remco Zegers - nuclear physics 8
what keeps the nucleons together? The nucleus consists of positive charges (protons) and neutral particles (neutrons). The positive charges should repel and the nucleus thus fall apart. However, they stick together!! The reason is the so-called strong force that acts between the particles. It is very strong, but its range very short! Nevertheless, not all nuclei can stay together PHY232 - Remco Zegers - nuclear physics 9
Binding energy The total energy (mass) of a bound system is less than the combined energy (mass) of the separated nucleons Example: deuteron 2 H (1 proton + 1 neutron+1 electron) m p =1.007276u (proton mass, not 1 H mass!) m n =1.008665 u m e =5.486x10-4 u (if 1 H is used for m p ignore m e!) m p+n+e =2.016490 u (sum) m 2H =2.014102 u m p+n+e -m 2H =0.002338u (2.224 MeV) The deuteron is 0.002338 u lighter than the sum of the separate proton and the neutron. This is the binding energy and is the energy needed to break that nucleus apart in its separate constituents PHY232 - Remco Zegers - nuclear physics 10
example What are the binding energies (MeV) of 15 O and 16 O? 15 O: atomic mass: 15.003065 u 8 p+7 n+8e - : 8x1.007276u+7x1.008665u+8x5.486x10-4 u=15.12325u binding energy: M 8p+7n+8e -M 15O =15.12325-15.003065=0.1202 u 1u=931.494 MeV/c 2 so: binding energy: 111.95 MeV usually given in binding energy per nucleon:111.95/15=7.463 MeV/c 2 16 O: atomic mass: 15.994914 u 8 p+8 n+8e - : 8x1.007276u+8x1.008665u+ 8x5.486x10-4 u =16.1319u binding energy: M 8p+8n+8e -M 16O =16.1319u-15.994914=0.136975 u 1u=931.494 MeV/c 2 so: binding energy: 127.591 MeV usually given in binding energy per nucleon:127.591/16=7.974 MeV/c 2 Go here to get atomic masses: http://ie.lbl.gov/toi2003/masssearch.asp PHY232 - Remco Zegers - nuclear physics 11
Binding energy M(A)=M(Z)+M(N)-B(N,Z) MeV per Nucleon Most stable fusion fission Atomic mass A source of energy! PHY232 - Remco Zegers - nuclear physics 12
nuclear reactors use fission of heavy elements like uranium fusion reactor: work in progress: ITER PHY232 - Remco Zegers - nuclear physics 13
Now consider 226 Radium 226 Ra: 226.025402 u 222 Rn: 222.017571 u 4 He: 4.002602 u Sum: 226.020173 u difference: 0.005229 u (or 4.87 MeV) in favor of the separate system It is energetically more favorable for the 226 Ra to emit an alpha particle ( 4 He: 2p+2n). 4.87 MeV in energy is gained. Where does this energy go to??? PHY232 - Remco Zegers - nuclear physics 14
question 226 Ra can decay into 222 Rn (Radon) and an alpha particle. what happens to the energy that is released in the decay? a) into the excitation of electrons orbiting the nuclei. b) into the motion (kinetic energy) of the decay products c) into creating of additional particles PHY232 - Remco Zegers - nuclear physics 15
Radioactivity Spontaneous emission of radiation by unstable nuclei. Note that the binding energy for unstable particles is positive which tells you that the nucleus would not fall apart in its separate constituents. However, as we just saw it can be favorable to break up into several chunks. 4 He (α particles) β particles (electrons or positrons) γ rays (energetic photons) PHY232 - Remco Zegers - nuclear physics 16
α-decay and tunneling Coulomb barrier 234 Th α Because of tunneling, the quantum mechanical wavefunction for the α-particle is not zero outside the nucleus. It has a small probability to be there and thus a small probability it can escape. PHY232 - Remco Zegers - nuclear physics 17
β-decay β - -decay electron emission β + -decay positron emission (electron with + charge) The atomic number is changed by 1, but the mass number remains constant β-decay is governed by the weak force which is short range, but very weak. PHY232 - Remco Zegers - nuclear physics 18
fundamental forces strong force holds nucleons/quarks together strength 1 range: 1 fm electromagnetic force: electric & magnetic forces strength 1/137 range: infinity weak force describes neutrino interactions strength 10-6 range: 10-18 gravity describes force between massive objects strength 6x10-39 range: infinity PHY232 - Remco Zegers - nuclear physics 19
types of β-decay β - decay an electron is emitted A Z X turns into A Z+1Y β + decay a positron is emitted A Z X turns into A Z-1Y electron capture A Z X turns into A Z-1Y PHY232 - Remco Zegers - nuclear physics 20
not the complete story Besides the electron (positron) also an anti-neutrino (neutrino) is produced. Look at the spectrum resulting from the beta decay of many nuclei Number of events no neutrinos if neutrinos would not exist all electron would have fixed energy, because of momentum and energy conservation in the decay. This is not observed! Another particle is present: the neutrino! neutrinos (ν) are very light and don t interact well with material and thus hard to detect. Kinetic energy of the electron (MeV) PHY232 - Remco Zegers - nuclear physics 21
γ-decay Just like in the case of electrons, the nucleus has different energy levels and going from one to another is associated with the release of a photon (MeV) PHY232 - Remco Zegers - nuclear physics 22
An example ( 226 U) PHY232 - Remco Zegers - nuclear physics 23
radioactivity PHY232 - Remco Zegers - nuclear physics 24
chart of nuclei PHY232 - Remco Zegers - nuclear physics 25
Decay chain The half life indicates how long it takes for half of the nuclei in a sample to decay. PHY232 - Remco Zegers - nuclear physics 26
decay ΔN Δt ΔN R N N = λnδt ΔN = = λn Δt λt = N e = N 0 0 e t / τ N: number of particle remaining R: decay rate or Activity N 0 : initial number of particles λ: decay constant R: the decay rate τ: decay time (=1/λ) Half-life: T 1/2 =ln2/λ=0.693/λ PHY232 - Remco Zegers - nuclear physics 27
half-life N 0 N 0 /2 N 2 t 1/ 2 = N 0 1/ 2 o e t = τ ln 2 = / τ 0.693τ 1 Curie (Ci) = 3.7x10 10 decays/s 1 Bq = 1 decay/s N=N 0 e -tln2/t 1/2 t 1/2 PHY232 - Remco Zegers - nuclear physics 28
example a) Polonium-210 has a half-life of 140 days. If a sample of this material has an initial activity of 1mCi, what is the activity after 20 days and 2000 days? b) what is the activity after 3 half-lives? N=N 0 e -tln2/t 1/2 c) if after 100 seconds 1/3 of a sample of unknown material has decayed, what is it s half-life? a) N=1x10-3 e -0.693t/140 note that I can use days for the half-life. The answer will then also be in days. 20 days: N=1x10-3 e -0.693x20/140 =9.05x10-4 Ci (0.905 mci) 2000 days: N=1x10-3 e -0.693x2000/140 =5.02x10-8 Ci (0.0502 μci) b) 3 half-lives means that the original sample has decayed to ½ x ½ x ½ = (½) 3 =1/8 of its original activity, so 1/8 mci. c) N=N 0 e -0.693t/T 1/2 1/3 has decayed, so 2/3 is left over. T 1/2 = (-0.693t)/ln(N/N 0 ) =(-0.693x100)/ln(2/3)=171 seconds PHY232 - Remco Zegers - nuclear physics 29
14 C dating 14 C is produced from 14 N by Cosmic rays. While alive, organisms have a fixed 12 C/ 14 C ratio (1/1.3x10-12 ) (Carbon in CO 2 ). After dying, no more 14 C is absorbed and it decays away and the ratio of 12 C/ 14 C can be used for dating. Shroud of Turin Found to be from 1320±60 AD PHY232 - Remco Zegers - nuclear physics 30
Natural occurring radioactivity Many elements that occur in nature are radioactive. They have a long life time and so can still be found even if they were made a long time ago. fiesta ware some can be harmful, like Radon gas in basements some are man-made, like fall-out from nuclear explosions some are useful, for example when used in a smoke detector or for treatment of cancer, or for determining the age of an object. Radiation is harmful because it can damage DNA/body tissue (the particles get stopped and deposit energy in the body). For the same reason they can be used to destroy tumors PHY232 - Remco Zegers - nuclear physics 31
stopping of radioactive material different types of radiation have different ranges when passing through material. Gammas (and neutrons) have long ranges (not charged!) Beta and Alpha radiation have shorter ranges PHY232 - Remco Zegers - nuclear physics 32
nuclear reactions Besides natural radioactive decay, nuclear reactions can also be created, for example in a cyclotron lab, such as the NSCL on MSU campus, to study the properties of nuclei. cyclotrons to accelerate particles to experimental devices fragment separator to make clean beams speed of particle beams 0.4c S800 spectrometer to detect particles after reaction PHY232 - Remco Zegers - nuclear physics 33
nuclear reactions Many nuclear reactions are studied, for example: 34 P + 7 Li 34 Si + 7 Be usually written as: 34 P( 7 Li, 7 Be) 34 Si Note that mass number and atomic number must be conserved before and after the reaction takes place: (A=34, Z=15) + (A=7, Z=3) (A=34,Z=14) + (A=7, Z=4) A sum =41, Z sum =18 A sum =41, Z sum =18 PHY232 - Remco Zegers - nuclear physics 34
examples which element is missing? a) 12 C(α,γ)? b) 56 Ni( 2 H,?) 57 N a) before reaction: 12 C: Z=6, A=12 α: Z=2, A=4 (α is 4 He) Total Z=8, A=16 after reaction: γ: Z=0, A=0 So the unknown particle is has Z=8 and A=16, which is 16 O b) before reaction 56 Ni: Z=28, A=56 2 H: Z=1, A=2 Total: Z=29, A=58 after reaction: 56 Ni: Z=28, A=57 So the unknown particle is has Z=1 and A=1, which is 1 H (proton) PHY232 - Remco Zegers - nuclear physics 35
where do all these nuclei come from? stable nuclei remnant of an exploded star (supernova) PHY232 - Remco Zegers - nuclear physics 36