Radioactivity_1 PDF

RADIATION BIOPHYSICS Objectives To explain; • Some properties of nuclei • Why some of the nuclei are instable? – – – – Forces at work Nuclear binding energy Binding energy per nucleon Instability curve • Disintegration • Radioactivity • Half life NUCLEUS: Proton Neutron Z: Atomic number, defin...

RADIATION BIOPHYSICS Objectives To explain; • Some properties of nuclei • Why some of the nuclei are instable? – – – – Forces at work Nuclear binding energy Binding energy per nucleon Instability curve • Disintegration • Radioactivity • Half life NUCLEUS: Proton Neutron Z: Atomic number, defines the number of protons in a nucleus N: neutron number A: Mass number, defines the number of particles (i.e. neutrons and protons) in a nucleus. Any nucleus: Z is also defined as a characteristic number because it defines the element itself. Example: Z=6 is a carbon atom. In a neutral atom, the number of protons is equal to the number of orbital electrons. Atoms having the same number of atomic number (Z) but different mass number (A) are called isotopes. FORCES AT WORK: Nuclear Force Electrostatic Force Atoms having the same number of atomic number (Z) but different mass number (A) are called isotopes. UNITS: 1 a.m.u. =1/12 of the mass of 12C 1 a.m.u.= 1.6605x10-24 g =1.6605x10-27 kg Energy equivalent of 1 a.m.u.: E=mc2 E=1.6605 x 10-27 kg x (3x108 m/s)2 E= 1.49 x 10-10 kg.m2/s2 = 1.49 x 10-10 Joules 1 eV is the energy acquired by an electron in falling through a potential difference of 1 volt. Energy (eV) =1 volt x electron charge = 1 V x 1.6x10-19 Coulombs 1 eV = 1.6x10-19 Joules What is the equivalent of 1 a.m.u in eV ? 1 a.m.u. = (1.49 x 10-10 J) / (1.6 x 10-19 J/eV) 1 a.m.u.= 0.931 x109 eV =931 MeV Nuclear Binding Energy: The total mass of a nucleus calculated from its constituents is larger than the measured mass of that nucleus. The difference between the calculated and measured values is called mass defect (missing mass) and it is the source of the energy that holds the particles in the nucleus together. Nuclear binding energy is the energy equivalent of the mass defect. Measured mass of an Helium atom 4.002604 a.m.u. Calculated mass of an 4He nucleus 2 protons + 2 neutrons 4.031884 a.m.u. Example 1: 4He nucleus contains two protons and two neutrons. Measured mass of an Helium atom = 4.002604 a.m.u. mass of a proton: mp = 1.007277 a.m.u. mass of a neutron: mn = 1.008665 a.m.u. mass of an electron: me = 0.00055 a.m.u. Electrons do not contribute to the n. binding energy. Therefore, to find the measured value for the nuclear mass, we directly subtract the total electron mass from the atomic mass Measured value for the mass of an Helium nucleus = 4.002604 - 2x me = 4.002604-2x0.00055 = 4.001504 a.m.u. Calculated mass of an 4He nucleus =2x mp +2x mn = 2x1.007277+2x1.008665 = 4.031884 a.m.u. Mass defect = 4.031884-4.001504 =0.03038 a.m.u. N. Binding energy = 0.03038a.m.u.x931 MeV/a.m.u.= 28.3 Mev N. Binding energy/nucleon = 28.3 Mev/4 = 7.07 MeV/nucleon Example 2: Hydrogen isotope 2H nucleus contains 1 proton and 1 neutron. mass of a proton: mp = 1.007277 a.m.u. mass of a neutron: mn = 1.008665 a.m.u. No electrons: Measured mass mm = 2.014102 a.m.u. Mass defect (Δm) = (mp + mn ) – mm = 1.007277+1.008665-2.014102 =0.00184 a.m.u. Eb= 0.00184 a.m.u.x 931 MeV/a.m.u.= 1.713 MeV This energy represents the energy that would be released in the synthesis of one atom of H from 1 proton and 1 neutron Example 3:Measured nuclear mass of 6028Ni is 59.930789 a.m.u. A)What is the calculated mass? B)What is mass defect of an6028Ni nucleus? C)What is the binding energy per nucleon ? C) Is this a tight or a weak binding ? Compared to what? Calculated mass = 60.481036 amu Mass defect (Δm) = 0.565647 amu Eb= 0.565647 a.m.u. x 931 MeV/a.m.u.= 526.6173 MeV Binding energy/nucleon = 526.61MeV/60 nucleon = 8.77 MeV/nucleon, A tight binding Binding energy / nucleon ( MeV) Region of very stable nuclides ~60>A>50 Mass number (A) -An attractive nuclear force is postulated to exist between neutrons, between neutrons and protons and even between protons and protons. -Nuclear force is “short range”. It can felt only at a distance of one nuclear radius. -The total mass of a nucleus is always less than the sum of the masses of the individual masses of nucleons which constitute it. According to Einstein mass-energy relation (E=mc2), we obtain the binding energy of nucleons (in eV). -Binding energy is an indicator of stability but does not predict stability against radioactive decay alone. Example: binding energy is higher for But is stable Binding energy/nucleon Line of Stability N N=Z 8.4 7.8 Mev N2 N1 A=55 Mass number, A Z1 Z In the nucleus, the only charged particles are the protons. Therefore positively charged protons will repel each other. This repelling force is counteracted with the binding energy.When the two forces are equal in magnitude the nucleus will be in the stable condition. Binding energy saturates for high A numbers. For low atomic number elements, proton number is almost equal to the neutron number and Binding energy balances proton’s Coulomb force (stable isotopes are on the line for Z=N). But for high atomic number elements, Binding energy cannot balance Coulomb force. In order to have a stable condition, more neutrons are tried to keep in the nucleus (as seen by green point on the figure) and the curve for stable isotopes lie above the line for Z=N. • What is nuclear binding energy? Mass defect? • Stability of nuclei Nuclear binding energy/nucleon (A tight binding is more stable) Mass number region of very stable nuclides Mass number greater than 83 is unstable n/p=1-1.5 stable (there might be exceptions) n/p>1.5 in-stable Even numbers are more stable compared to odd numbers Magic numbers of nucleons (exceptionally stable nuclides) Line of Stability Neutron number , N alpha emitters Unstable nuclides lie in the shaded regions β- emitters Stable nuclides β+ emitters Proton number, Z If an isotope is not stable, then it disintegrates to go to a stable position: Modes of disintegration: 1. Negatron Decay 2. Positron Decay 3. Alpha Decay 4. Electron Capture Negatron emission: β- decay: Consider an atom present above the stable condition (dark blue). N=Z N N/Z> stable condition n → p+ + e- +ν (antineutrino) Z Z One neutron in the nucleus is transformed into one proton, one electron and one antineutrino. After the decay, the daughter nucleus contains one less neutron, but one more proton . Therefore atomic number of the daughter nucleus has increased by one unit. A X→ A Z Z+1Y +β-+ν A A -+ν X→ Y + β Z Z+1 Example: 14C Stable isotope of C is 12C. 14C β− 14 For 12C N/Z = 6/6=1 7N Ζ For 14C N/Z = 8/6 > 1 14C → 147N + β - + ν Mass defect for Negatron emission (βdecay): A X→ A Z Z+1Y +β-+ν Nuclear Mass of AZX : (remember on the charts, atomic masses are given) define: ma(X), atomic mass of AZX mn(X), nuclear mass of AZX then: mn(X) = ma(X) - Zx me Mass defect = mass difference between the before and after reactions Δm = mn(X) -mn(Y)- me = ma(X) - Zx me - {ma(Y) - (Z+1)x me }- me Δm = ma(X) - ma(Y) This energy is shared by a negatron and antineutrino if the decay occurs to ground state. Example 1: Δm = ma(X) - ma(Y) 20F ma(F) = 19.999987 amu ma(Ne) =19.992440 amu 7.02 MeV β− 1.63 MeV 20 10Ne Δm=19.999987-19.992440 = 0.007547 amu E= 0.007547 amu X 931 MeV/amu = 7.02 MeV Maximum energy = 7.02-1.63=5.39 MeV This energy is shared by negatron and antineutrino. Particles are emitted over a continuous range of energies # negatro n Average Energy Emax ( 5.39 MeV) Energy Energy of Antineutrino # negatro n Energy of Negatron Emax ( 5.39 MeV) Energy Energy of Negatron Energy of Antineutrino Example 2: Δm = ma(X) - ma(Y) 41Ar 2.5 MeV β− γ 1.29 MeV 41K ma(Ar) = 40.98108 amu ma(K) = 40.97840 amu Δm= 40.98108−40.97840 = 0.00268 amu E= 0.00268 akb X 931 MeV/amu = 2.4951 MeV Maximum energy = 2.50-1.29=1.21 MeV Example 3: 1.71 MeV 32 Δm = ma(X) - ma(Y) 15P β− 32 16S ma(P) = 32.0678 amu ma(S) =32.066 amu Δm=32.067832.066 = 0.00184 amu E= 0.00184 amu x 931 MeV/amu = 1.71 MeV Positron emission:β+ decay: If the isotope has fewer neutron (dark blue) N=Z than the stable condition, then one proton N is transformed into one neutron, one positive electron and one neutrino. Hence, the daughter nucleus has one less proton. N/Z< stable condition p+ → n + e+ + ν (neutrino) A X→ A Y Z Z-1 +β++ν Example: 11C Stable isotope of C is 12C For 12C N/Z = 6/6=1 For 11C N/Z = 5/6 < 1 If N/Z< stable condition (decay occurs either by positron emission or/and electron capture) Z 11C β + 11 5 B 11 → 5 B+ β + + ν 11C Z Positron emission: β+ decay: A X→ A Y Z Z-1 +β++ν Nuclear Mass of AZX : ma(X), atomic mass of AZX mn(X), nuclear mass of AZX then: mn(X) = ma(X) - Zx me Mass defect = mass difference between the before and after reactions Δm = mn(X) -mn(Y)- me = ma(X) - Zx me - {ma(Y) - (Z-1)x me }- me Δm = ma(X) - ma(Y) -2x me The energy equivalent of the mass defect is shared by a positron and neutrino but sharing is not in a given proportion. A ZX→ A Z-1Y + 11C +β +ν β + Example 1: 11C Stable isotope of C is 12C. For 12C N/Z = 6/6=1 For 11C N/Z = 5/6 < 1 11C → 115B+ β + + ν 11 B 5 Ζ %90 β+ decay A 22 A Y +β++ν X→ Z Z-1 11Na β+ 1.27 MeV Example 2: 22Na Stable isotope of Sodium is 23Na. For 23Na N/Z = 12/11>1 For 22Na N/Z = 11/11 = 1 22 Ne+ β + + ν Na → 11 10 22 γ 22 10Ne (stable) Ζ Electron Capture: If the condition (N/Z< stable condition ) is satisfied by electron capture. Electron and proton combines to produce one neutron. Therefore after the reaction the daughter nucleus contains one less proton, i.e. Z number decreases by one unit. A X Z + -1e → AZ-1Y + ν proto neutron electron Essentials of Nuclear Medicine Physics and instrumentation, Powsner R.A., 2013 Alpha ( α) Decay: High atomic number elements cannot go to a stable level by emitting only one particle; instead they emit more then one type of particle. In the alpha decay two protons and two neutrons are emitted from the nucleus. A X→ A-4 Y Z Z-2 + 42He (α) Mass defect Δm = mn(X) -mn(Y)- mn(He) = ma(X) - Zx me - {ma(Y) - (Z-2)x me }- {ma(He) - 2x me } = ma(X) - Zx me - ma(Y) + Z x me- 2x me-ma(He) + 2x me } A X Z Δm = ma(X) -ma(Y)- ma(He) α A-4 Z-2Y Δm = ma(X) -ma(Y)- ma(He) Example 3: 226 0.187MeV α 4% 1 222 88Ra ma(Ra)= 226.0254 a.m.u. ma(Rn)= 222.0177 a.m.u. ma(He)= 4.0026 a.m.u. 96% α 2 4.748 MeV 86Rn Δm = ma(X) -ma(Y)- ma(He) = 226.0254-222.0177-4.0026 Δm =0.0051 a.m.u. E = 0.0051 a.m.u.x931 MeV/a.m.u. E = 4.748 MeV Εα1 =4.748−0.187 =4.56 MeV Εα2 =4.748 MeV Number of parent atoms Half Life λ: Decay constant: probability of decay per time (a characteristic value for each nuclide) N0: Number of atoms at t=0 λ: Number of atoms which disintegrate in a unit time. Disintegration Rate (Decay constant). t: time Definition: Half time: t= Thalf t=T1/2 (half time) =T N=No/2 N= No/2 by definition No/2=Noe-λThalf 1/2=e-λThalf ln 1- ln 2 = ln(e-λΤhalf) 0 – ln 2 = - λThalf λT1/2=0,7 decay of 131I Number of atoms (N) T1/2 = 8 days; Τ1/ 2Τ1/ 2 2 Time (days) N0 =20000 atoms Activity of a Radioactive Element: How do we measure radioactive activity of an isotope? in grams? Common units of radioactive activity are Becquerel (Bq) and Curie (Ci) 1 Becquerel is defined as one radioactive decay per second 1 Curie is defined as 3.7 1010 decays per second (radioactivity emitted by 1 g of Radium) 1 Ci = 3,7x1010 particles emitted /s 1 Ci= 3.7×1010 Bq = 37 GBq The units Bq or Ci provides to determine the sample’s radioactivity depending on time We can express the same equation in terms of activity. A = Ao e-λt Its unit is Becquerel or Curie. Define 1 Curie is 3x1010 disintegrations in one second. That is 3x1010 atoms decay in 1 second. Example 1: −3 hr−1 . Find the activity in T has a decay constant of 9.5 × 10 81 Becquerels of a sample containing 1010 atoms. 201 A=λN λ=(9.5 10-3 hr-1) / (3600sec/hr) = 2.6 10-6 s-1 A=(2.6 10-6 s-1) (1010 atoms) =2.6 104 atoms/s = 2.6 104 Bq Activity of a sample depends on its decay constant and half life. If the half life is short, we should be careful about the sample. Example When we order 40 mCi of 131I , its activity will become 20 mCi after 8 days. So we should take into account the decrease in its activity while we calculate the dose. Essentials of Nuclear Medicine Physics and instrumentation, Powsner R.A., 2013 Problem: What is the reason for 137 Cesium to be radioactive (its atomic number is 55)? 137 Cs 55 M = 137 A = P = 55 M=P+N 137 = 55 + N N = 137 - 55 = 82 N >>> P “That’s why 137 Cesium is radioactive”. Example : Half life of 231Pa is 3,25x104 years. What is the disintegration rate (λ) of 231Pa? What is the activity of 20 mCi of Pa after 6,5x104 years? A=Aoe-λT λT=0,7 λ = 0,7/ 3,25x104 = 0,2x 10-4= 2 x10-5 1/year A=Aoe-λt =20 e-1,4 = 20 x 0,247 =5 mCi OR t=0 t=T t=2T A= 20 mCi A = 10 mCi A= 5 mCi