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Unformatted text preview: PHYSICS 13 FINAL EXAM FALL 2003 NAMﬁ___:QL_U_ [ﬂALS ______________ n:s Schrodinger equation: (h=h/21c) ,2 d2
_ i W w(x) +V(x)\v(x) = E‘lKX) For two interacting bodies the m is replaced by reduced mass u and x is the separation.
Quantum mechanical angular momentum lLl= [l(l+1)]‘/Zh and L2 = m,h Note the hbar or more generally HI = [j(j+1)]"2h and J2 = mj h with m, = j, H, ..., j Addition of orbital (l) and spin (1/2) angular momenta yields j=l1/2, (+1/2. Addition of J1 and J2 yields sum J with quantum numberj = lj1j2l, lj1jzl+1, ..., j1+j21, j1+j2 Atomic electron spectroscopic notation: n 1‘ °' 2 °‘" , where n is principle quantum number, 1 is orbital angular
momentum quantum number, l=0,1,2,...,n1, with notation t=0>s, 1>p, 2>d, 3>f,... Avogadro's Number: NA = 6.0 x 1023 number of atoms, molecules or nuclei in a MOLE (or gram molecular wt) Nonrelativistic Kinetic Energy =mvz/2 Magnetic dipole interaction energy = #sz Lorentz Transformation frame 3’ moves in the +x direction with V as seen in S: x=y(x’+Vt’), y=y’, z=z’, t=y(t’+Vx’/c2) where y=1 /‘/(1v2/c2)
Doppler effect A' =x~l(1+v/c)/\/(1v/c) ' Time dilation t' = y t
Relatvistic Energy E= m0 c2 /‘](1v2/c2) ; Rest energy = moc2 ; Relativistic Kinetic Energy = E  moc2 Photon: E=hf ; c=7tf ; k=21c/7l. ; (o =2nf Electric dipole moment p51: Q r (Q=charge, r=separation) Average Thermal Energy at Temperature T(Kelvin) = k(Boltzmann)T/2 per degree of freedom
Decay N(t) = N(0)exp(7tt), 7l.=1/ (mean lifetime) = ln2/T1,2 where T1,; is the halflife Nuclear Binding Energy Eb = (Zm(p) + Nm(n)  MA) c2 Coulomb Potential kEMQ1Q2/ r
Newton Law of Gravitation : potential energy = —Gm,m2/r Constants: Velocity of light: c = 3.00x 108 meter/sec electron charge: e=1.60x10“9 Coul
Newton gravitational constant G = 6.67x10'”J*m/Kg2 k(Boltzmann) = 1.38x10‘23J/°K = 8.62x10‘5eV/°K Planck’s constant: h = 6.63 x 10‘34 J sec: 4.14 x 10'15 eV sec 71 = h/21c = 1.05 x 10‘“ J sec = 6.58 x 1045 eV sec kEM = 1/41t80 = 8.99x109 NmZ/Coul2 hc = 1.24 x 10’5 eV m Rest energy of electron = mec2 = 511 keV = 0.511 MeV me=9.1x10'3.‘Kg
Uniﬁed mass units 1 u = 1.66x10‘27 Kg (1 u)c2 = 931.5 MeV
umts; Energy: 1 J(oule) = 1 Kg mZ/secZ ; 1 eV(electron Volt):= 1.6x10“9 J
H E T F LA T PR BLE V VI VII i. r n ** XTRA REDlT Pr b.Vll PHYSICS 13 gﬁ) L FINAL EXAM , FALL 2003 ‘ Grades: points /possible points I / 100 II /50 Ill /40 IV /40 V / 3 0* WW
Vl / 30* VII / 30* Total /290 ——_—_—_—___—————_—________——_—__—_—______—___———_— I. Multiple choice  circle the one hesI answer. 1) The “Curve of Binding Energy” is a graph of the magnitude of the binding_enetng.er
nucleon (BE/A) of the neutrons and protons plotted against mass number A. The
lightest nuclei have smaller: BE/A than a. deuterons, which makes fission possible.
b. deuterons, which makes fusion possible.
c. medium mass number nuclei, which makes fission possible. @medium mass number nuclei, which makes fusion possible.
e. electrons in atoms. 2) The hindingmmy of the neutron and proton in a Uranium nucleus is larger than the
typical atomic or molecular binding energy by a factor of a. 32 b. roughly 6x1 023 (90%th 106
. d. roughly 100 e. roughly 1/6xTO' 3 3) In Grand Unified Theories of particles the proton is unstable. In one such theory the mean lifetime is 1032 years (age of universe is ~ 10‘0 years). That being the
case, which statement is most true? a. Protons will not be seen to decay in our lifetime. b. A single proton is very likely to decay in 10 years. N U 3 “(6%
h t c. proton in a sample ofége, 2 will probably decay in a year.
d. All protons in a sample 0 32 protons will probably decay in t e m an lifetime. Sam) M L? /0 3&
e. No protons have decayed yet in the universe. W 00. MW M 4) The nucleons are held together in a nucleus by W a. gravitational interaction b. electromagnetic interaction
c. electroweak interaction d. weak interaction @trong interaction 5) Consider the element 1oNe. Its ground state is a. 1521p62$2 b. 1522523523p4 {? SZZSZZDG
d. 1522522p42d2 e. 1523 8 6) A free particle of kinetic energy (KE) 3 eV, traveling in a straight line, approaches a
constant higher potential of +2.5 eV. The particle will a. have a 100% probability to continue along at KE =O.5 eV. b. be reflected back at the boundary with 100% probability. c. have a 100% probability to decay exponentially in the barrier. d. lose energy in the 2.5 eV region until it stops.
,«f e. be partially reflected with KE=3 eV and partially transmitted with
KE=O.5 eV. 7) A particular crystalline solid has an energy_gap_of 5.0 eV between the valence band
and the nearest conduction band. At TEOK the valence band is filled and the conduction band is empty. At room temperature (TE3OOK, kTE0.03 eV) a. many electrons will be in the conduction band making this a semiconductor. b. many electrons will be in the conduction band making this an insulator.
) c. very few electrons will be in the conduction band making this a semiconductor.
”@few electrons will be in the conduction band making this an insulator. e. the valence and conduction bands will overlap making this like a metal. 3 8) Quarks are elementary particles that a. only interact gravitationally b. only interact electromagnetically
c. are constituents of electrons d. have the charge of the electron and positron
Ore constituents of protons and neutrons 9) The current theory of elementary particles, the Standard Model, includes a. 3 colored quarks and 8 colored leptons.
‘3 colored quarks and 8 colored gluons. »
c. gravitational interactions of the quarks with the leptons.
d. the unification of all the forces of nature.
e. only the weak interaction. 10) The Big Bang theory of the origin of the universe is in agreement with many
observations and predictions. Which of the following isnnt a true observation or
prediction that confirms the theory? a. The most distant observed galaxies are in much earlier stages of formation than the
Milky Way. b. The universe is filled with blackbody radiation at temperature 2.7K.
c. The universe is nearly flat. d. Galaxies recede from us with velocities proportional to their distance. instein showed that a stationary universe is the only solution to his field equations
f General Relativity. ***************************************************************** ll. Hubble’s Law relates the receding velocity of a galaxy to its distance from Earth via v =(dR/dt)= HR where the Hubble constant H = (23Km/sec)/(106 light years) at the
current time. a. What is the speed of the galaxy G40 that is 5 billion light years away?
:: 93K? 2< §x [061/5977  gxm x9 3:20 Vs
7” mgr/0 4g “,0353c b. What will be the wavelength measured on Earth of the 590.0 nm Sodium line emitted
from stars in G40? ’ H34; (333 ’ . 5"“ ' '
I '2: I : O'OMM X“ 3'24, €5QO,ON"«XII‘+?7 4 2383,2MN II. c. Suppose the mass of a nearer galaxy at distance R is m and the total mass of all
matter within a sphere of radius R (centered on the Earth) is M. ln the nonrelativistic
Newtonian theory of gravity, the gravitational potential energy U(R) of the galaxy is ( GmM/R) . Using Hubble’s Law express the galaxy’s nonrelativistic kinetic energy K(R) in terms of its distance R. ,
mi?) 4 M W“ ﬁW—JV’ZrifT[EB ‘ , ”2 g».
30 KEG?) =Q%WRZ}U>W_ _. .IMHER. d. For a flat universe the kinetic energy of the distant galaxy is exactly balanced by the
potential energy. Use that condition to get an expression for the critical density of matter in the universe, Pcritical Express this in terms of H and G, the Newtonian gravitational constant. :2, e. Evaluate that critical density in units of protons per cubic meter. The mass of a
proton is 938 MeV/c2 or 1.67x10'27Kg. g 48
H .. RIBFIDtFM/é _ "2*” 1/0 ,vl
IDEXCX3E0iS .V “J6 .
G=e.e:xlo‘"f%‘ 2 ism/W ’32: = 6'7 l :7an Ill. The quantum mechanical explanation of many properties of metallic filaments Can be modelled by a system of electrons confined to a one dimensional tube or wire of length L.
Treat the electrons as non—interacting fermions. a. w(x) = (Z/L)“2 sin(kx) is a normalized solution to the ldimensional stationary
Schrodinger equation for a Mum with kinetic energy h2(k2)/ 2m. Under what condition on k will this w(x) be a wavefunction for one electron in the tube of‘
length L? (You will need to introduce a quantum number for this system, call it n.) rm)  o ”1&me M.
> k L
b. What are the allowed energies E(n) for a single electron in terms of n? EﬂgoffDZ/Bf: bimu c. Suppose electrons are added to the tube, one at a time, by putting each electron in
the lowest allowed energy state. When there are N electrons in the box, what is the
WWW (assume N is a very large number and
remember that electrons have spin 1/2)? Call this energy E "near. Express your answer in
terms of h, m (mass of the electron), L and N. d. Aluminum has a free electron number density of 1.8lx1023/cm3. Taking the 1/ 3 power of this gives 5.66xlO7/m (or 56.6 electrons per micron) as the linear number density. What is the Fermi Energy EJW’ of an Al filament in this 1—dimensional model? Use units of
eV. , EM: ”E: (N, 7*: hqcf (géé x/07M_'>2. F 32m L 3JMCQ .___ (latix/ evMMW/M
W
31% S/lx/ﬂe V = emu/Nev 6 IV. The fission process n + 23592U > 9°37Rb + ‘44?CS + ?n
releases energy that you will calculate. a. HowmanLnemLQns must be released? What is Z, the atomic number, for Cs (cesium)? 2 Na . Z(_"C<>) == {7 ,9 b. The masses of the nuclei are given below. What is the that disappears
_(i.e. gets converted to other forms of energy) in the reaction? (Be sure to account for the neutrons.) M6“) +41 (U) 2 ,4 (tb) + Mains) +3Wm)%
322%) == 0.0167x10'25 Kg A": M01)" M Kb) .__/I{C3).— MM) 3.9184x10'25 Kg 7
M(9°Rb) = 1.4925x1025 Kg _. 2 [0 ‘3
M(”4Cs) = 2.3892x1025 Kg “ :0 X IE7 c. The mass that disappears is converted to how much energy? Express your answer in
Joules or MeLunits. q
;; 2:. AM co“ : 1,5! x/D"'0j : [4/ x/Oev
EM Q : l.lX/03/1€V
M d. How muchenergy would be produced if about 1/2 Kg of the Uranium undergoes fission, i.e. every nucleus in 1/2 Kg undergoes fission? You’ll need to find the number of
U nuclei in that mass first. Express your answer in Kilotons. (1 Kiloton = 4.2x10‘2J = 2.6x1025MeV. 1 Kiloton is the equivalent of one thousand tons
of TNT explosives.) lMMM £1,137}? , 3'73 §1§§§= x/3W. V. Consider the potassium flouride KF molecule. Suppose it is primarily ionic. a. The ionization energy (energy to remove one electron) for K is 4.34 eV. The electron
affinity (energy released when one electron is added) for F is 3.45 eV. When the neutral atoms K and F are separated to infinity (or a very large distance R), how much energy
does it take to create the ions K+ and F‘ ? $34evﬁgzq§§fv z: 0.%7€\/ "39M b. Calculate the Coulomb potential energy for K+F‘ for which the equilibrium
separation is 0.217 nm. Assume the two ions are point charges.  8
' x e“!  .... ‘l ‘ [Ibo/[0 M
EM an J’IixlD—IDA "' 8!?7X/0 A/ﬂ. ( ) W? 2.!73/0"€,.1 :Mélés €\/' m c. The measured dissociation energy (to pull apart the molecule into neutral K and
F atoms) is 5.07 eV. Given your answers to parts (a) and (b), how much energy is
associated with repulsion of the ions at equilibrium separation? VI. Top quarks are the most massive elementary particles with mass = 175
GeV/cz. Suppose many top quarks are produced in an accelerator laboratory with an
average total energy of 245 GeV by special reactions. Top quarks decay with a half life of about 7.0x10‘26 sec.
a. What is the average speed of a top quark in the accelerator laboratory? (Hint:
Use the relativistic energy vs. the rest energy.) MC; v‘=.L —_. v1
(—5; X2. big/0 % 51‘0,l+7& 5—— 0,10 m v=2./0x/0gm{3 b. What is the half life measured in the accelerator laboratory? alga“ 1%,; MW mm"; WW 3% c. Those top quarks travel in the lab and decay as they go. Suppose 1/2 of them
have not decayed after traveling a certain distance. What is that distance? . v 93 _ 3M —>
095V LAB 2./px/o§x7$x/0 s ; 2IOé X/[n/M VII. Carbon 14 (14C) is radioactive and undergoes beta decay. It has a half life of 5730 Years.
a. What is its mean lifetime (1M)? 4t&_ i 8 ' :1,
§ 21.2.13” 1942:2647“
{I
Nitrogen 14 . Neutron capture .Carbon 14 ‘1— ‘ﬁ V; ————r lost protoV i
All three iséopes X m of carbon (common
5‘ C12.rareC138.
radioactive C14 are , absorbed by _.
living organism s. "" ﬂ Following death a burial. mod & bones lose
C14 as it changes to N 14 by beta decay Carbon 14 .———b Bela decay .Nilrogenﬂ Beta ;
particle GP roton ONeulron b. A piece of wood from a recently cut tree shows 12.4 14C decays per minute. A
sample of the same size from a tree cut thousands of years ago shows
3.5 decays per minute. What is the age of the sample? Am (0: AweRt * (amok/am 2%: SW :72? 394??? /:2/2 c. How many atoms K] ”C are in the sample from the old tree? ‘63 [,2é)‘ 8’35? Aal=w A) M’W?) ﬁe ~21“: W
= may” :2 No?) ”a WP 79—.
/\/(t) : W% :15 Wow ************ EXTRA CREDIT** Try this ONLY if you have finished the others.****?*** VIII. Consider a diatomic molecule, like NaCl.
a. For small oscillations x, about the equilibrium separation, the Schrodinger equation _ applies, with the mass being the reduced mass I1 and the potential being the simple
harmonic oscillator potential V(x) = (1/2) p (DOZXZ. For the classical oscillator the resulting motion has angular frequency 030. Show by substitution that
\I0(x) = exp[(p. (non)/2h] can be a solution to the appropriate Schrodinger equation. A ‘3”) 78 —.‘_MA) 4);
We?” ‘ 1’07“?
1 —( )7? °‘ 1 \ * ‘ 2
W 7—2” iW i M ‘ ETI’
? . 11;. 9: 0 +¥M 0 .. o
b. From the substitution In (a), find the correspondmg energy E, for this vibrational state
in terms of (Do and h? Em=£§wlp E0 : Q’Kmo ll ...
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 Fall '04
 GaryR.Goldstein
 Physics

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