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Home , Alpha decay, Binding energy, Decay energy, Nuclear binding energy, Particle decay, Radioactive decay

Radioactivity

Lecture 4 The Physics of and

Nature of

• Deflection in • α radiation is positively charged • β radiation is negatively charged • γ radiation is neutral • Stopping in material • α radiation has small range and large • 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 or too many ) transitioning into a more stable form. • What represents a stable or unstable configuration • Too many protons deflective Coulomb • Too many Neutrons reduce the strong binding • Too many cannot be bound together by strong force • Too much deformation increases instability of nucleus. • Heavy deformed nuclei will spontaneously fission Radiation as Release

A nucleus emits radiation of certain energy (E=hν for gammas and

E=Ekin for ) to reach the lowest possible energy configuration!

Energy release in Energy release in gravitational potential nuclear potential

1 2 Ekin = mv 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 of the Nucleus

Occurs mainly for very heavy nuclei which are not stable against alpha emission α = 4He

226 222 4 88 Ra ⇒ 86Rn + 2He Nucleus conversion through α-decay

Z

N Determine the end-product of the ‘yellow’ a-emitter: A A−4 Z N ⇒ Z −2 X N −2 + α of the Nucleus

β decay is the emission of an e- or e+ to convert a to a 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 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 m12C = 12.00000 · 1.66·10 g = 1.992·10 g -24 -22 m208Pb = 207.797665 · 1.66·10 g = 3.449·10 g -24 -22 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

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 A! Calculating the

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

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

(MeV/nucleon) B/A Binding energy normalized to mass number B/A

Mass number

Example: What is the binding energy of the 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 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 at Paris had observed the first nuclear reaction. 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 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 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