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Class Meets Mon + Wed + Fri   11:00 - 11:50pm in Science 277 ... Thurs at 11 in Sci.100
My office:   Science 159 (below ramp to 3rd Ave)     my e-mail :   foltzc@marshall.edu     office phone :   (304) 696-2519

Topic Ten - Atomic Nuclei Character & Stability ; Nuclear Reaction Energetics & rates

Quiz 10 was Fri.Dec.08 . . . here is my answer key

Exam 4 will be Tue.Dec.12 @ 10:15 am (or later day, by appointment)

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Readings for Topic 10: Ch.43 Home-Work PRACTICE QUESTIONS for topic 10       (NOT for grading; numbers refer to Young & Freedman's University Physics, 14th ed.)

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Physics II Topic 10 Summary

Idea #20: atomic Nucleus made of Z protons and N neutrons ; Z + N = A - - - - - -

Mass Number: the number of "heavy things" ("baryons") in each object (protons + neutrons)
    written as a SUPER script to the left of the symbol.
=> the total mass number IS the same after a reaction, as it was before the reaction
. . . "mass" is roughly conserved ... (see below for an enlarged idea of mass)

The number of fermions ... things with "Spin Angular Momentum" s = ½ h/2π
      . . . short-cut notation: write h/2π as new symbol ℏ , pronounced "h - bar"
    angular momentum is sometimes written as a superscript to the right of the chemical symbol
=> number of fermions must be even (or odd) after the reaction, if it was even (or odd) beforehand
. . . the number of fermions is conserved (↑ spin + ↓ spin = 0)
      photons have spin 1 (h/2π) , so are bosons, not fermions ... boson number is not conserved .
=> a nucleus can emit a photon (gamma ray   γ ) if one of its nucleons flips its spin
      (if final − initial = ↑½ − ↓½ = ↑1 , photon carries ↓1 h/2π)
. . . the nucleus must have started with excess angular momentum , having extra Energy in an "excited" state
      ... symbolized with a superscript star to right of the Chemical symbol , or a lower-case "m" after it

notation examples:

Hard-sphere model of a nucleus - - - - -

neutron radius ≈ proton radius =   R1 = 1¼ E−15 [m] = 1¼ [femto-meter] . . . more nucleons => bigger radius :   RA = R1 A1/3   => essentially incompressible volume Strong Nuclear Force : every baryon attracts every other nearby baryon => Deep negative Potential Energy . . . but the Strong Force weakens steeply with distance , as a negative exponential / r²       a nucleon obtains U = −2 MeV from its direct neighbors , only about −0.37 MeV from 2nd neighbors       negligible amount from any of the others Electric Force : protons repel protons => positive Electric Potential Energy . . . weaker, but fairly long-range : only 0.6 MeV from each neighbor proton,       but 0.3 MeV from 2nd neighbors, and 0.2 MeV from 3rd , 0.15 MeV from 4th ... . . . this positive PE tends to push all the protons apart ! protons can reduce their Electric PE by being farther apart, so neutrons mingle among them . . . big nuclei need to add almost 2n for each new p   (Z > 20)

Charge Distribution in a Proton Charge Distribution in a Neutron

Magnetic Force : nucleons pair up magnetically => negative Magnetic Potential Energy for each pair
. . . a lot weaker than the other nuclear Energies , and only for that pair : −0.15 MeV if they pair up correctly ( +0.15 MeV if they are oriented wrong)
      ( Gravitational Energies are about 10^−30 eV , so totally ignoreable still )
empirically, nuclei seem to favor having even Z and even N ... about 10× as much as the small magnetic energy indicates
. . . that is actually the result of Pauli's exclusion principle allowing 2 protons in their lowest available Energy level, or 2 neutrons in their lowest available Energy level
      if a nucleus has one odd proton and one odd neutron, the higher-level one (especially if it is a neutron) will be able to lower its energy by changing into the other kind of nucleon, to pair with it.
they also seem to prefer having the same number of neutrons as protons, especially in the lighter nuclei ;
. . . the energy cost of being unequal seems to be about   27[MeV] (N−Z)/A ,

Negative Potential Energy reduces the mass of the collection of nucleons
. . . if a neutron hits a proton, it will bounce off (returning to infinity) unless it loses a lot of KE
      (it was pulled Magnetically from far away, and then Strongly as it approached, so KE as it hits is ~ 3.15 MeV)
. . . they lose Energy by emitting a photon (gamma) which carries the Energy away (also momentum and angular momentum)
      the Energy lost in this manner is what "binds" the neutron to the proton
. . . the Energy loss shows up as the bound-pair having less mass than the two did separately
      Δm = ΔE/c² . . . less Energy congealed in that region, than had been there.

Application : any Nucleus Will eventually Decay if its PE is too high - - - - -

The nucleus that is most stable has protons and neutrons with the lowest (most deeply negative) Energy .
. . . this is called "binding Energy" per nucleon ... maximum is around ⁵⁶₂₆Fe ... ⁶⁰₂₈Ni , ⁶⁴₃₀Zn , ⁵²₂₄Cr   are close
      you can think of individual free protons and neutrons being assembled into a nucleus :
          as they fall into the (strong negative) PE well, they radiate away (as γ-ray photons) their excess Energy , and become trapped .
      => the Energy that they have lost while forming the nucleus is the negative PE that they reside at while in there;
          we would need to give them that much Energy, to "unbind" them and set them free again ... so it is called their Binding Energy .

      the amount of Energy radiated by each new constituent (as the nucleus is being assembled) is typically 5 to 8 MeV or so
      ... 8 MeV is 8 E6 * e [Volt] = 1.28 pico-Joule ... for each nucleon!   From E = m c² , the mass effectively radiated away is
          roughly equivalent to m = E/c² = 14 E-30 kg each!
      ... each nucleon is only about 1673 E-30 kg , so the lost Energy shows up as a "mass defect" in the actual nucleus
          (or when you measure the mass of the atom, which is a lot easier to actually do in the lab).
      86 proton masses + 136 neutron masses + 86 electron masses = 86 (1 672.622 E-30 kg) + 136 (1 674.927 E-30 kg) + 86 (0.910938 E-30 kg) = 371 713.910 E-30 kg
      . . . but the radon-222 atom actually has mass   368 668.835 E-30 kg . . . so its "binding mass-Energy" is   Δm = 3045 E-30 kg ... Δm c² (/e) = 1712.8 MeV
      that means its PE = −1712.8 MeV ... but dividing it among the 222 nucleons => − 7.715 MeV/nucleon ... find it on Kartunnen's graph !
      . . . the stable nuclei about that size have binding Energies roughly 7.8 MeV/nucleon ... that is, 20 MeV (total) deeper .

      the decay product masses don't add up , either ... this time I'll use atomic mass units (1 u = 1660.539 E-30 kg) :
          helium-4 mass + polonium-218 mass = 4.002 602 u + 218.008 973 u   =   222.011 575 , rather than the radon's 222.017 571 u
      . . . the excess mass (0.005 996 u) in the parent radon , becomes KE in the products : (.005 996 u)*(1660E-30 kg/u) * c² (/e) => 5.6 MeV
      on this very small size scale, you can think of "mass" as being "coagulated Energy" that is well-localized

. . . 3) it has too many protons, too close together ... the protons cause (positive) Electric Potential with long range (1/r)
      the (negative, attracting) potential due to the strong Force goes to zero very nearby (e^−r/r )!
      => the ratio N/Z needs to become larger than "1" for nuclei with a lot of protons Z .
. . . it makes most sense to treat proton number Z as the "independant" variable, with all of the various neutron numbers N vertically, as on the first chart
      (rt-click to enlarge) ... but most isotope charts arrange Z ↑ and N→ , as chart 2 below . . .

. . . here is a link to Brookhaven National Lab's Interactive Chart ... very interesting, don't spend too much time browsing in it!

Half-life and Mean Life ; Activity and Decay Likelihood - - - - -

if you had a million Radon atoms you would expect 2 to decay each second ... ΔN / Δt = − 2/sec
. . . after 3 days (quarter million seconds) you would only have half million Radons left, so only 1 would decay each second
      the negative simply means that the number of Radons decreases ... as the number of Poloniums (and Heliums) increases .

Mean Lifetime T_mean = 1/λ . . . λ is likelihood to decay per some time interval [/sec].
      any ²⁰₁₁Na is 65% likely to decay in 1 second . . . (the odd proton captures an atomic electron, becoming ²⁰₁₀Ne + ⁰₀ν )
      . . . that is, 2 out of 3 will capture an electron during the first second ... more than half of them.
      . . . half of them do it in less time than that 1 sec : the half-life is shorter than the expected life.
half life = T_½ = T_mean· 0.693   . . .
. . . with a large sample   N₀ of items at time t = 0 ,
=>   ΔN = − λ N₀   will become something else during one second .
      a high decay likelihood clearly means that lots of them will decay ... with λ > ½ , has the majority not surviving the first second
. . . but each surviving radon has the same likelihood to decay during the next second , as they did during the first second. no memory that they are old
. . . since there are fewer remaining after that , the expected number to decay will also be less .
      the number remaining (that haven't done it yet) decreases exponentially :
=> − ΔN/Δt   =   Decay Rate   =   N₀ e^(−λ t)   . . . this   e   ≈ 2.71828... is base of natural log)
. . . half-life   τ_½   is how long it takes for half a sample to be decay , so half is left ; (rather than 1/2.718 = 36.8 % to remain)
      τ_½   = (0.693147...)/λ   . . . is shorter than the mean lifetime ;
=> R = − ΔN/Δt   =   N₀ (½)^(− t/τ)   . . . the exponent is just "how many half-lives has it survived?"
      a sample's decay rate is also called its radioactive Activity .

Because the number of undecayed parents decreases (exponentially) with time,
. . . 1) the mass of those undecayed parents also decreases exponentially with time: m = m_o e^{−λ t}
. . . 2) the activity of those undecayed parents decreases exponentially : A = A_o e^{−λ t}
. . . 3) the rate of Energy release (Power) from them decreases exponentially : P = P_o e^{−λ t}

Example 1 : neutron decays into proton and electron . . . mean life-time 881 [s]
      ¹₀n   →   ¹₁p   +   ⁰₋₁β   +   ⁰₀ν  
. . . the   beta β⁻ is an electron ... needed to conserve charge .
      the   neutrino ν , predicted by Pauli, evidenced by Cowan ... needed to conserve spin ;
      . . . this is really an anti-neutrino , with negative fermion number , so the total number of fermions are strictly conserved

total momentum (3-d vector!) is also conserved in every reaction (relativistic kinematics often required)
      in the neutron's c.o.m. frame, the decay products must be traveling in "opposite" directions . . . the 3 p vectors add to zero .
. . . if the neutron decays from rest, all 3 momentums have similar magnitude (but different directions) ... the proton's p = mv ...
      the proton has 2000× as much mass as the electron, so the electron is going nearly 2000× as fast as the proton
      ... since KE ≈ ½p·v   , the electron has nearly 2000× the KE as the proton ... and the neutrino carries even more than the electron!

. . . total Energy is also conserved , but only if you include "mass Energy"
      mass: 939.565 [MeV/c²]   →   938.272 [MeV/c²]   +   0.511 [MeV/c²]   +   "0"
      so KE: 0 [MeV]   →   0.782 [MeV] KE . . . more KE ~ ½p·v in lower-mass pieces !

Half-life of a free neutron is 881 sec × ln2 = 881 s × 0.693 = 610 seconds ... about 10 minutes , so after an hour   (=6 half-lives)   only 1.6 %   =(½)^6   will survive .

Example 2 : proton fuses with proton , making deuterium . . . initial (Threshold) Kinetic Energy is required !

  ¹₁p^+½   +   ¹₁p^−½   +   KE   →   ²₁H⁰   +   ⁰₊₁β^+½   +   ⁰₀ν^−½   +   ⁰₀γ^±1 .
. . . initial KE is needed so that the 2 protons can get very close together (2E−15[m] = 2 [femto-meter]) , which requires high PE :
      PE = qV = q (kQ/r) ~ 0.72 [MeV] , to get +e to be 2[fm] from the other +e .
. . . anti-electron   ₊β⁺ , fondly called a "positron" , has mass identical to an electron's , but has positive electric charge .
      we've found antiparticles for protons and neutrons , and almost all the mesons , etc.
      when positron and electron collide , all that is emitted are 2 gamma rays (opposite directions, opposite angular momentums)

Danger !   Energetic Projectiles are Emitted - - -

Nuclear radiation typically carries 1-5 Million eV , so each projectile can ionize 100,000 atoms

alpha penetrate about 5 μm per MeV, so a sheet of paper (40 μm) can stop an 8 MeV &aplha;
      they actually lose Energy less readily as they become slower;
      . . . a 10 keV α penetrates 25 μm per MeV ... that is only 0.25 μm.
. . . concentrated Energy loss is good, if it happens in your shielding material (clothes),
      and is not too bad if it is in your skin (those outer layers are dead already)
      but it is bad if it occurs inside you ... say, via radon decay in your lungs.

beta lose only about .1 or .2 MeV per mm (by EM Forces) , so a 1 MeV electron penetrates about 10 mm
. . . until they start to lose momentum by collisional Energy loss
      a 10 keV electron loses Energy much more readily to the material it goes thru,
      . . . because their wavelength becomes much longer, and their wavefront much wider
      about 2.0 MeV per mm, so only goes 0.005 mm = 5 μm farther as it stops.
      so beta radiation will ignore your skin and outer layers of fat and muscle,
      . . . to deposit most of its Energy deeper inside some vital organ.
. . . All along the beta's path, it has ionized lots of atoms that it passes
      10,000 ions per mm at first, means 1 ion every 100 nm ; only hits ¼ %
      200,000 per mm at the end, means 1 ion every 5 nm ; some 5 % along the track get ionized

gamma are 0.4 % likely to be absorbed in each meter of air that they go thru
      (compare: 50 keV x-rays are 4% likely per meter of air)
. . . so half of the gammas will make it thru (0.693)/0.004(1/m) = 178 meters of air ... ¼ of them thru 356 m , etc.
      6 or 8 of those 178m thicknesses (1400m) gets rid of almost all the original gammas.
. . . density of water (we are essentially water) is 770× the density of air, so
      gammas are 0.3% likely to be absorbed in 1 mm of water (or tissue)
      gammas that enter a person will impart about half their Energy to that person's first 200mm (8") thickness
      . . . with another ¼ of their energy being deposited in the next 8" (exit-half side of the body).
. . . a typical gamma will first ricochet ("Compton scatter") from an atomic electron :
      momentum conservation means that the electron recoils when it is hit, knocked out of its atom.
      . . . the gamma has less Energy after the richochet (so longer wavelength, still speed c)
      the ejected electron might have enough KE to ionize several dozen atoms along its track
      . . . the ion which lost that electron probably emits a UV cascade as the vacancy is filled in
. . . the gamma ricochets several times ; farther apart at first, when it has its highest Energy (smallest wavelength)
      closer together as it loses Energy to these electrons (longer and wider gamma hits more of them).
. . . eventually the gamma gets absorbed by an atom as it knocks out a deep atomic electron
      that atom will emit an x-ray, as a higher-n electron fills the vacant hole ... UV cascade.

positrons (the anti-electron) find an electron almost immediately, and annihilate with it
      the Energy in the positron (KE + m c²) plus the Energy in the electron (m c²)
      is shared among 2 gammas ... each is at least 0.511 MeV (in the center-of-momentum reference)
      . . . these gammas go outward from the annihilation point , acting like gammas (see above)
      . . . the atom that had an electron annihilated fills in its vacancy via UV cascade.

Deeper than the Nucleus : the Standard Model of matter (leptons , quarks , & gluons) - - -

proton is made of 2 "up" quarks + 1 "down" quark  
. . . Z = 2 (+2/3) + 1 (−1/3) = 4/3 − 1/3 = +3/3 = +1   . . . e's of course.
. . . A = 2 (1/3) + 1 (1/3) = 1   . . . each quark is 1/3 of a baryon
. . . s = (1/2) − (1/2) + (1/2) = 1/2   . . . 3 fermions make a fermion

neutron is made of 2 "down" quarks + 1 "up" quark  
. . . Z = 2 (−1/3) + 1 (+2/3) = 0/3 = 0   . . . of course.

Each meson seems to be made of 2 quarks ;
pion . . . π⁺ = "up" + "anti-down" . . . π⁻ = "anti-up" + "down"
      verify that their charges add up correctly (all antiparticles have opposite charge)
      s = (1/2) − (1/2) = 0   . . . 2 fermions make a boson
. . . being bosons (not conserved) they can pop into existence temporarily
      anyplace there is some excess Energy . . . enough stress in space
=> typical nuclei are surrounded by a pion cloud ("dressed")
      ... but each pion only exists 10% of the time (or less).