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Radioactive Decay and Nuclear Reactions

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Radioactive Decay and Nuclear Reactions Radioactivity Lecture 4 The Physics of radioactive Decay and Nuclear Reaction Nature of Radiation • Deflection in electric field • α radiation is positively charged • β radiation is negatively charged • γ radiation is neutral • Stopping in material • α radiation has small range and large mass • β radiation has deep range and small mass • γ radiation has largest range and no mass – electromagnetic radiation Origin of Radiation • Radioactive decay of unstable nucleus configuration (too many protons or too many neutrons) transitioning into a more stable form. • What represents a stable or unstable isotope configuration • Too many protons deflective Coulomb forces • Too many Neutrons reduce the strong binding force • Too many nucleons cannot be bound together by strong force • Too much deformation increases instability of nucleus. • Heavy deformed nuclei will spontaneously fission Radiation as Energy Release A nucleus emits radiation of certain energy (E=hν for gammas and E=Ekin for particles) to reach the lowest possible energy configuration! Energy release in Energy release in gravitational potential nuclear potential 1 2 Ekin = mv Particle decay to excited states 2 with subsequent de-excitation by emission of gamma radiation β-decay γ-spectrum γ-decay + β decay + _ p ⇒ n+e +ν β- decay n ⇒ p+e-+ν Alpha Decay of the Nucleus Occurs mainly for very heavy nuclei which are not stable against alpha emission Alpha particle α = 4He 226 222 4 88 Ra ⇒ 86Rn + 2He Nucleus conversion through α-decay Z A A−4 N Determine the end-product of the ‘yellow’ a-emitter: Z X N ⇒ Z −2 X N −2 + α Beta Decay of the Nucleus β decay is the emission of an electron e- or positron e+ to convert a neutron to a proton or a proton to a neutron inside the nucleus 6 · e+ 7 · e+ +1 · e- Too many protons The β decay always converts along isobars Too many neutrons Gamma Decay of Nucleus excited states in nucleus Excitation of nucleus γ emission with subsequent l< 10-15 m characteristic γ emission Excited states correspond Ex to vibration, rotation or quantum state excitation Nucleus conversion through β+,--decay Z N + + A A Determine the end-product of the β -emitter: Z X N ⇒ Z −1X N +1 + β A A − - Determine the end-product of the β -emitter: Z X N ⇒ Z +1X N −1 + β Energy in Nuclei According to Einstein’s formula, each nucleus with a certain mass m stores energy: E=mc2 -24 -24 Proton mp = 1.007596 · 1.66·10 g = 1.672·10 g -24 -24 Neutron mn = 1.008486 · 1.66·10 g = 1.674·10 g -24 -23 Carbon m12C = 12.00000 · 1.66·10 g = 1.992·10 g -24 -22 Lead m208Pb = 207.797665 · 1.66·10 g = 3.449·10 g -24 -22 Uranium m238U = 238.050783 · 1.66·10 g = 3.952·10 g http://amdc.impcas.ac.cn/web/masseval.html www.nndc.bnl.gov/chart/reCenter.jsp?z=42&n=53 1 -24 1 amu= /12(m12C)=1.66 · 10 g Breaking up nuclei into their constituents requires energy Some unavoidable unit considerations kg ⋅m2 1 J =1 s2 1eV =1.6022⋅10−19 J A g ≡ 6.022⋅1023 particles c = 300000 km / s 1 amu =1.66⋅10−27 kg = 931.49 MeV / c2 http://en.wikipedia.org/wiki/Electronvolt Nuclear binding energy The mass M of the nucleus is smaller than the mass of its proton and neutron constituents! 2 2 2 M · c < Z mp · c + N mn · c Z=3 protons E=m·c2 N=3 neutrons The mass difference is the binding energy B The binding energy is the energy that needed to dissociate a nucleus into its single constituents. It is released when N neutrons and Z protons fusion together to form a nucleus with the mass number A! Calculating the Nuclear Binding Energy Binding energy B B = (Z · m + N · m - M) · c2 of nucleus p n 12 = ⋅ + ⋅ − ⋅ 2 B( C) (6 mp 6 mn m12C ) c 2 = (6⋅1.672⋅10−24 g + 6⋅1.674⋅10−24 g −1.992⋅10−23 g)⋅(3⋅108 m / s) = 2 2 2 m m =1.56⋅10−25 g ⋅9⋅1016 (m / s) =1.404⋅10−8 g =1.404⋅10−11 kg =1.404⋅10−11 J s s 12 B( C) − J =1.17⋅10 12 A nucleon 238 = ⋅ + ⋅ − ⋅ 2 B( U ) (92 mp 146 mn m 238U ) c 2 = (92⋅1.672⋅10−24 g +146⋅1.674⋅10−24 g − 3.952⋅10−22 g)⋅(3⋅108 m / s) = 2 2 2 m m = 3.03⋅10−24 g ⋅9⋅1016 (m / s) = 2.725⋅10−7 g = 2.725⋅10−10 kg = 2.725⋅10−10 J s s 238 B( U ) − J =1.145⋅10 12 A nucleon Nuclear Potential and internal forces (A − 2Z )2 B = a ⋅ A − a ⋅ A2/3 − a ⋅ Z ⋅(Z −1)⋅ A−1/3 − a ⋅ +δ v s c sym A −4/3 −4/3 δ = +a p ⋅ A (Z, N even); δ = −a p ⋅ A (Z, N odd); δ = 0 (A = Z + N odd); av =15.5MeV; as =16.8MeV; ac = 0.72MeV; asym = 23MeV; a p = 34MeV http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/liqdrop.html#c2 1 MeV = 1.602·10-13 J Nuclear Binding Energy Components 1 MeV = 1.602·10-13 J B/A (MeV/nucleon) B/A Binding energy normalized to mass number B/A Mass number Example: What is the binding energy of the oxygen isotope 18O? -24 mp= 1.007596 · 1.66 · 10 g -24 mn= 1.008486 · 1.66 · 10 g M(18O) = 17.99916 · 1.66 · 10-24 g Z=8, N=10, A=18 2 B = (Z · mp+ N · mn- M) · c B(18O) = 1.382 · 105 keV = 2.21·10-11 J B(18O)/A=1.23·10-12 J 18 11 1g O contains 7.41·10Atomic Mass Unit: J (W·s) 1 amu=1/12(M 12C)=1.66 · 10-24 g Nuclear Decay Processes A ⇒ B+b A(b)B 14C(β-ν)14N 234U(α)230Th Decay energy is released in kinetic energy of emitted particles or in electromagnetic gamma radiation energy 2 2 −19 Qd = (mB + mb )⋅c − mA ⋅c 1eV =1.6⋅10 J = + − = + − Qd BB Bb B A Q234U B234U B4 He B230Th = + − = Q234U 1779.325 MeV 28.296 MeV 12755.128 MeV 52.5 MeV = = ⋅ 7 = ⋅ −12 Q234U 52.5 MeV 5.25 10 eV 8.4 10 J Nuclear Reactions and Energy Release Frederic Joliot and Irene Curie at Paris had observed the first nuclear reaction. Enrico Fermi showed the existence of neutron induced reactions, which produce artificial radioactivity. Nuclear reactions can produce energy Q > 0 exothermic or need energy Q < 0 endothermic Nobel Prize 1938 A(a,b)B 2 Q = (mA+ ma- mB- mb)·c Q = BB+ Bb- BA-BB a Difference of masses in entrance A and exit channel determines Q Q value of nuclear reaction process projectile A+a ⇒ B+b a product A(a,b)B b B 2 2 A Q = (mB + mb )⋅c − (mA + ma )⋅c recoil Q = (BB + Bb )− (BA + Ba ) target Q > 0 exothermic reaction Q < 0 endothermic reaction Example: Nuclear Reaction energy budget Conversion of nuclei through fusion or fission leads to release of energy! 2 2 4 1 H1 +1H1⇒2 He2 + Q isotope B (J) Q = B(4He) − 2⋅(B(2H ) + B(2H )) 2H 3.34131·10-13 Q = 2⋅3.34131⋅10−13 J − 4.53295⋅10−12 J = 3.8647⋅10−12 J 4He 4.53297·10-12 12C 1.47643·10-11 238 ⇒ 119 + 119Pd 1.59643·10-10 92U146 2 46Pd73 Q 119 119 238 238U 2.88631·10-10 Q = B( 46 Pd73 ) + B( 46 Pd73 ) − B( 92 U146 ) Q = 2.88631⋅10−10 J − 2⋅1.59633⋅10−10 J = 3.06542⋅10−11 J http://www.nndc.bnl.gov/exfor/endf00.jsp http://www.nndc.bnl.gov/exfor/exfor.htm .
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