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251 PHY251_M2_S01_solutions Name: PHYSICS - Final Exam Wednesday, May 16, 2001 Show all work for full credit! I. Short answer problems (120 points) 1. Calculate the momentum p and the wavelength l of: (Use eV/c and nm units!) i. a photon of 10 eV energy, Solution: p=E/c=10 eV/c; l=h/p=1240 eV/c nm/10 eV/c = 124 nm. ii. an electron of 10 eV energy, Solution: p=(2mE) = (20.51110-610) = 3.2 keV/c; l=h/p=1240 eV/c nm/3.2 keV/c = 0.388 nm. iii. an electron of 10 GeV. Solution: p=E/c=10 GeV/c; l=h/p=1240 eV/c nm/10 GeV/c = 12410-9 nm = 0.124 fm. 2. A laser emits a beam of 1017 photons per second, of wavelength 310 nm, which falls on a piece of blackened foil (mass 20 mg). Assume the photons are fully absorbed. i. Calculate the power P of the laser beam. Solution: E = hc/l = 4.0 eV; P = NE = 10174.0 eV/s = 64 mW ii. Calculate the initial acceleration of the foil. Solution: p = E/c; a = F/m = Ndp/(mdt) = NdE/(mcdt) = P/(mc) = 6410-3 Nm/s /(210-5 kg 3108 m/s) = 1.0710-5 m/s. 3. Four indistinguishable particles (with masses m = 5.0105 eV/c) are inside a cubical box, with sides equal to 1.0 nm. There is no magnetic field. i. If the particles have spin s = 0, calculate the ground state energy of the total system. Solution: The lowest energy level is at E111 = 3 /8 The next level is E211=E121=E112 = 6 /8 h mL h mL = 30.384 eV = 1.15 eV. = 60.384 eV = 2.30 eV. In the ground state, all four bosons are in the first level: Etot = 4E111 = 4.61 eV. ii. If the particles have spin s = 1/2, calculate the ground state energy of the total system, and list the possible spin orientations of the particles. Solution: Only two fermions may occupy the lowest energy level with opposite spins. The other two, with opposing spins, must go into the first excited state; thus: Etot = 21.15 eV + 22.30 eV = 6.91 eV. iii. If the particles have spin s = 3/2, calculate the ground state energy of the total system, and list the possible spin orientations of the particles. Solution: Now all fermions can go into the ground state, differentiated by their ms quantum number: ms = +3/2, +1/2, -1/2, -3/2. The total ground state energy is as in (i): 4.61 eV. 4. Compton scattering of 12.4 keV X-rays: i. Under what condition is Compton scattering observed? Solution: It assumes that the scattering is off quasi-free electrons! ii. Calculate the wavelength of X-rays Compton-scattered at a 60 degree angle with respect to the incident X-rays. Solution: l = 1240 eVnm /12,400 eV = 0.1 nm; l' = l - h/mc(1 - 1/2) = 0.100 + 1240/0.51110-6 = 0.1012 nm. file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (1 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions 5. X-rays bombarding heavy atoms can be used to eject electrons from the 1s shell in atoms; indeed, this is the starting point for Moseley's experiments. Estimate the maximum wavelength of photons required to eject an electron from the 1s shell of copper, for which Z=29. Solution: Eg = -E1(Cu) = -( Zeff /1)(-13.6 eV) @ 28 13.6 eV = 10.7 keV. Thus: l = 1240 eVnm / 10.7 keV = 0.116 nm. 6. Describe the H2+ molecular bonding (i.e. p-e-p bonding): i. Sketch both the electron wavefunction and the probability distribution as function of the coordinate along the p-p axis that leads to stable bonding (the bonding state). Solution: | *|* / | \ /* | *\_ _/* __/** ____/ ** | | | **\_ relative plus - sign | *|* / | \ _/* | *\ _/** | | | *\_ **\__ ** \____ **** probability ____ wavefunction **\___/** ***** _*/__*__*_______|_________________|_______*__*__\*____ x-> P<-- separation-->P ii. Sketch both the electron wavefunction and the probability distribution as function of the coordinate along the p-p axis that does not lead to bonding (the anti-bonding state). Solution: | *|* / | \ /* | *\_ _/* __/** ____/ ** | | | **\ *\_ *\ P **** probability relative minus -sign ____ wavefunction _*/__*__*_______|________\_________|__________________ x-> P \ | **__*_/* file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (2 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions \* \* \* | | | **_/ *_/ */ \* | */ \ | / *|* | 7. Bonding: define/describe in short, and give an example of : i. Ionic bonding. Solution: Bonding between a very electro-negative and a very electro-positive atom: one atom has a loosely bound electron, which is strongly attracted by an atom that has an almost-filled shell. Rather than shared equally as in co-valent bonding, the electron is mostly belonging to the electro-negative partner. Coulomb attraction between the partners leads to bonding. Example: NaCl. ii. ss Covalent bonding. Solution: Sharing of outer s-shell electrons between the atomic partners. The s-wavefunctions add up in between the atoms, leading to significant probability for the electrons to be found in-between the atoms, leading to a net bonding effect. Example: H2 (see also previous problem). iii. (sp) hybridization (describe the process that leads to an electronic (sp) hybrid state). Solution: Up to two s and six p electrons are possible. If an element has 4 outer-shell electrons, like C or Si, then typically one will go into the s sub-shell, and three into the p sub-shell. The s electron will have a spherically symmetric wave functions, whereas the p electrons will have elongated two-lobed distributions. Now, because electrons are all negatively charged, they repel each other in multielectron atoms. Therefore, and because of the Pauli exclusion principle, the p electrons will orient themselves in mutually perpendicular directions: we call them px, py and pz electrons. Often a process called sp hybridization occurs in this case (4 outer-shell electrons); it is based on the fact that any linear combination of wavefunctions is again a valid solution of the atomic Schrdinger equation. The lone s electron can be combined with various signs with the three p electrons, to give four independent and similar (sp) hybrid solutions. These solutions look again elongated and two-lobed, but with the crucial difference that one of the lobes is much more pronounced than the other, giving a strong directionality to the distribution: the sp electron is mostly on one side of the atom, in a specific direction compared to the other three sp electrons. It is like a person with four hands, all reaching out in different directions. Mutual repulsion between the electrons makes this a lower energy configuration than the original s plus three p states. 8. Consider the NaCl molecule, in which the atomic weights of Na and Cl are 27 and 37, respectively. The internuclear separation is 0.236 nm, and the "spring constant" for vibrations is k = w2 = 300 eV/nm. i. Calculate the energy difference between adjacent vibrational levels. Solution: = [2737/(27+37)] u = 15.6931.5 MeV/c = 14.54 GeV/c DE = hw = h(k/) = hc(k/ ) = 197 eVnm (300 eV/nm/14.5109 eV) = 0.0282 eV. c ii. Calculate the rotational energy level for angular momentum quantum number L=1. Solution: Erot(L) = | |/2 = L(L+1)( /2 I L h R ) = L(L+1)( hc / 2cR) = file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (3 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions = L(L+1)((197 eVnm)/214.5109 eV (0.236 nm)) = = L(L+1)((197 eVnm)/214.5109 eV (0.236 nm)) = L(L+1) 2.4010-5 eV. For L=1: Erot = 4.8110-5 eV. 9. At room temperature (300 K), calculate the probability that an electronic state with energy E = 5.25 eV in a material with Fermi energy EF = 5.00 eV is occupied. Solution: FFD = (e E-E F/ kT + 1) -1 = (e0.25 eV/0.025 eV + 1) -1 = 1/(e10 + 1) e -10 = 4.510-5. 10. Consider a free-electron gas at a temperature T such that kT<< EF. Calculate an expression for the electron number density N/V for electrons that have an energy in excess of EF. Show, by making the change of variables (E-EF)/kT = x, that the electron number density is proportional to T. Solution: ne(E>EF) = (Ne/V) = = (8p2 me3/2/h) F E EF ne(E)dE = -1 EF g(E)dE FFD(E;T) = EF E dE (ex + 1) (8p2 me3/2/h)EF kT dx (ex + 1) -1 = = (8p2 me3/2/h)kT EF ln2 = [2 me3/2/(ph)]kT EF ln2 = (6.85 nm-3) kT EF ln2. 11. Sketch a typical energy band structure for a the following types of crystalline solids. Mark the position of the Fermi level in the graphs, and clearly label the valence and conduction bands. Give the typical size of the band-gap (in eV), and name an example of each type of solid. i. Insulator, ii. Conductor, iii. Semi-conductor. Solution: | cond.| | band | |______| (few eV) EF ****** ______ | | | cond.| | band | | | | | | cond.| | band | | | ^ | | E |______| |******| | val. | | band | |______| | | ****** (0.2-0.6 eV) | | | val. | | band | SemiConductor (Si, Ge) | val. | | band | Insulator Conductor (Diamond-C) (metals-Cu) file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (4 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions 12. A current of 3.0 A flows in an aluminum wire 2.0 mm in diameter. Use the table at the end for your calculations: i. Show that the electron density in Aluminum is 181028 m 3. Solution: ne = (#electrons/atom)NA[#atoms/mole]rD[g/m]/M[g/mole] = ii. Calculate the drift speed of the conduction electrons. Solution: j = I/A = enevd vd = I/Aene = = 3 C/s / (p10 6 m 1.610 19 C 181028 m 3) = 3.310 iii. Calculate the resistivity of Aluminum. Solution: r E/j = (mvd/ ) / (enevd) = / et m enet - - - -5 - m/s. = -3 = 0.511106 (eV/c) / (e 1.610-19 C 181028 m 0.8010-14 s) = - = 5.6810-12 V / (1.610-19 A 181028 m-1 0.8010-14 ) = 2.4510 8 m 13. A one-dimensional quantum well of width 12 nm is created in a solid, and measurements show that the energy difference between the ground state and the first excited state is 0.16 eV. What is the effective mass of the electron in the well? Solution: The energy for an electron occupying the ground state in a one-dimensional box is E1 = nh/ (8mL) = h/(8mL). The energy of the first excited state is E2 = 4h/(8mL). The difference between these two states is DE=3 h/(8mL) = 0.16 eV (this difference is the same as for a 3-dimensional box!). Thus, the effective electron mass is: m =3 h/(8DE L) = 3(1240 eV/cnm)/(80.16 eV 144 nm) = 2.5104 eV/c = 0.049 me (= 4.510 -32 kg)! 14. Give the electronic configurations of the elements Phosphorus (P), Aluminum (Al), and Iron (Fe). Solution: P = [Ne](3s)2(3p)3 = [(1s)2(2s)2(2p)6](3s)2(3p)3; Al = [(1s)2(2s)2(2p)6](3s)2(3p)1; Fe = [Ar](4s)2 (3d)6 = [(1s)2(2s)2(2p)6(3s)2(3p)6](4s)2(3d)6 15. Doping and semiconductors: i. Describe the process of doping in silicon; how does it lead to enhanced conduction? List a typical donor material, and of a typical acceptor material. Solution: Doping is the controlled addition of "impurities" to pure silicon. The dopants are either from group 3 or group 5; i.e. they either lack a fourth p electron and are likely to bind a free-roaming electron relatively tightly (acceptor impurities like Al or Ga) thus providing a hole for conduction. Impurities that instead have five p electrons, have one left over once they covalently bind with surrounding silicon atoms, i.e. their fifth electron is only lightly bound, and is easily transferred into the conduction band (donor impurities like P or As). Silicon doped with donor impurities has electrons as the majority charge carriers (n-silicon), whereas silicon doped with acceptor impurities has holes as majority charge carriers, and is called p-silicon. ii. Describe the steps that lead to formation of the depletion region in a p-n junction diode. Solution: When a piece of p-silicon is brough in contact with n-silicon, holes and electron charge carriers cross the junction and start to "annihilate" one another: electrons from the n-silicon get trapped in the holes on the p-silicon side (leading to a net negative charge there), and vice-versa , until quickly a potential is established that prevents any more electrons and holes from crossing the junction. The region depleted of charge carriers thus established is called the depletion region. iii. In what direction will the diode conduct? What is the approximate magnitude of the depletion voltage in silicon? Solution: The forward voltage that allows conduction is applied positive to the p-side and negative to the n-side; the voltage has to be larger than the depletion voltage (about 0.5-0.7 V) for an appreciable current to develop. iv. Sketch the I-V characteristic curve for a p-n diode (label the axes!). file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (5 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions Solution: 100 mA-+ | Forward Current [mA] | | | | 10 mA-+ | | | | | _/ _/ _/ / / / | / | | / | ----------------------+---__/-------+------> Forward ______________________|__/ reverse or leakage | current (<<0.1mA) | ~0.6 V Voltage [V] II. Consider the normalized wave function of the form y(x) = C exp{- /2 }. a x (50 points) 1. Show that the normalization constant C equals (ap)-. Solution: The value of C is such that 1 = - |y(x)|2 dx = |C|2- exp{-x/ a }dx = |C|2 ap ; thus C = (ap)-. 2. Show that the average of x is zero, i.e. the expectation value x = 0. Solution: x = - x|y(x)|2 dx = - x|C|2 exp{-x/ }dx , which is an integral of an odd function of x, a over an interval symmetric with respect to x=0: therefore, the result is 0: x = 0. The result makes sense physically: the wavefunction is perfectly symmetric around x=0, and therefore the average should be zero as well. file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (6 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions 3. Calculate the value of Dx, i.e. the root-mean-square of x. Solution: First, we need to calculate x = - a. Then: Dx = RMS (x - x) = x = /2 . 4. Show that the average of px is zero, i.e. the expectation value px = 0. Hint: Use the proper definition of the expectation value, and the operator identity: (px)op = -ih(/ ). x a x|y(x)|2 dx = |C|2- x exp{- / }dx = |C|2 ap. = a x Solution: p = - = |C|2- y*(x) [-ih(/ )y(x)] dx = |C|2- x a x x a exp{-x/ h 2a} [-ih(/ )exp{- /2 }] dx = x x a x exp{-x/ } 2a [-ih( / )exp{- /2 }] dx = -i( / )|C|2- a exp{- / } dx = 0; again because a x the integrand is an odd function. The particle represented by the wavefunction is equally likely to travel left or right, and the vector average is therefore zero. 5. Calculate the value of Dpx, i.e. the root-mean-square of px. Solution: First, we calculate the value: p = - y*(x) [{-ih(/ )}y(x)] dx = -h|C|2- x [(x/ a exp{-x/ 2a} [( / ) exp{- /2 }] dx = x a x = -h|C|2- h a ) - 1](1/ )exp{- / }] dx = a a x x = -( / )|C|2[(1/ )- a exp{- / } dx - - a h a2 x exp{-x/ }dx] a = -( / )[1/2 - 1] = /2 . a a h h Then: Dp = RMS (p - p) = p = / 6. Check your result for the product DxDpx against the Heisenberg uncertainty relationship and comment. Solution: Thus: DxDp = /2 . For this particular wavefunction the product DxDp reaches the minimum allowed by the Heisenberg uncertainty relationship! In many other examples - see the book - the product comes out larger. III. Consider the hydrogen atom z f h (60 points) 1. The angular momentum z-component in operator form is given by (L )op -ih(/ ). Show that the wavefunction y21+1 is an eigenfunction of (L )op, and calculate the eigenvalue for this wavefunction. z Solution: )opy21+1 (L -ih(/ )y21+1 = -ih(/ )[1/ z f f (8pa0) ( /2 0)e a r -r/2a 0 sinq e+if] = = -ih(+i)[1/ (8pa0) ( /2 0)e a r -r/2a 0 sinq e+if] = h y21+1. Thus, the function is an eigenfunction of the Lz operator, with eigenvalue +h. 2. Calculate the probability that a 1s electron in hydrogen (see the corresponding wavefunction y100 in the formula table) is found inside the volume of the nucleus. Take the nucleus to be a sphere with radius R = 1.0 fm. Solution: P100(0rR=10 -15 m) = 0 R |y 100(r)| 2 rdr sinq dq df = 0 R |R10(r)Y00|2 rdr sinq dq df = file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (7 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions = [ 0 R |R10(r)|2 rdr][ |Y00|2sinq dq df ] = [0 -15 R |R10(r)|2 rdr][1]. Over the very small distance under consideration here: r R = 10 m << a0, the exponential in the Radial function R(r) is very close to unity, and we may set it to 1 without loss of precision! Then the integral reduces to: P100(0rR=10 -15 m) = 0 R |R10(r)|2 rdr = 4a0-3 0 R rdr= 4a0-3 (R/3) = 4/3(R/a0) = 4/3(210 5) 10-14. - 3. To map the orbits with some reasonable precision, it is necessary to localize the electrons with a precision Dx = 0.5 a0. What amount of kinetic energy (in eV) is likely to be transferred to the electron that is under observation with a probe that can localize the electrons with this precision? Solution: Dp ( /2 ) = (h/a0) = 197/0.053 eV/c = 3.7 keV/c. This is the spread in momentum; because the Dx h average momentum is zero (see the end of problem VI.2 above!), the average p = (Dp) = 14 (keV/c). The corresponding average kinetic energy is then: K = p/2m = 14 (keV/c) / 1.022 MeV/c = 14 eV. This is a substantial energy compared to the typical binding energy of electrons in hydrogen, and the atom will be significantly perturbed. 4. A hydrogen atom is in an excited 4d state. Show in a diagram all possible transitions that can occur and that lead to emission of photons. Solution: Note: the selection rule dictates that only transitions with Dl = 1 are allowed (to first order): 4s-- 4p---- 4d==== 4f-/ / / / V / 3s-- 3p====/ 3d-/ / / / V / V / / 2s====/2p==== / / V 1s==== V / / 5. In the atomic quantum theory it is generally not possible to specify all three components of the angular file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (8 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions momentum simultaneously. Is the l=0 angular momentum state an exception to this rule? Explain! Solution: Yes, for l=0 the magnitude of the angular momentum is zero, and hence all three components are trivially zero as well. However, this state does not violate the Heisenberg uncertainty relationship because the position and the momentum of the electron in a l=0 state are still uncertain as required by the rule DpxDx h/2. 6. An unknown magnetic field produces a set of lines for a transition from a state with l=2 to one with l=1, between which there is a maximum energy difference of 6.110 4 eV. List all allowed emission lines. Calculate the magnitude of the magnetic field. Solution: The l=2 state has a total of 2l+1 = 5 magnetic sub-states, and the l=1 state has three. Transitions may occur between the following levels (a total of 9 allowed transitions): l=2, m=2 l=1, m=1 DEB = -BB l=2, m=1 l=1, m=1 DEB = 0 l=2, m=1 l=1, m=0 DEB = -BB l=2, m=0 l=1, m=1 DEB = +BB l=2, m=0 l=1, m=0 DEB = 0 l=2, m=0 l=1, m=-1 DEB = -BB l=2, m=-1 l=1, m=0 DEB = +BB l=2, m=-1 l=1, m=-1 DEB = 0 l=2, m=-2 l=1, m=-1 DEB = +BB The transition emission line is thus split into three (assuming the splitting of the l=2 and the l=1 levels is the same), with energies DE21, and DE21 BB; with BB = 5.7910-5 eV/T B = 6.110 - -4 - eV. Therefore B = 6.110 4/5.7910-5 T = 10.5 T. 7. Positronium is a short-lived bound state in which an electron e- and a positron (anti-electron) e+ form a sort of atom in which the two particles circle each other around their common center of mass. Using the fact that the ground state energy of hydrogen is -13.6 eV, quickly calculate the ground state energy of positronium. (Hint: positronium is like hydrogen, but with the massive proton replaced by the positron which has a mass exactly equal to the mass of the electron.) Solution: The reduced mass of positronium is m/2; therefore E = -1/2 mc = -1/4 mc = -13.6 eV /2 = - 6.8 eV. IV. Consider a particle of mass m = 0.100 MeV/c and energy E0 = 20.0 eV incident from the left (x<0) on a potential step V(x<0) = 0, V(x>0) = V0 (not a barrier!): (50 points) a. Calculate the momentum p1 and the corresponding wave number k of the incident particle. Solution: The incident momentum is: p1 = (2mK) = (20.10106 eV 20 eV/c) = 2000 eV/c. The wave number: k = p1/h = 2000 eV/c / 197 eVnm/c = 10.2 nm 1. b. Derive that the reflection probability is R = [ k-q - / k+q ] for the case V0<E0. file:///C|/Documents%20and%20Settings/Linda%20Grau...20HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (9 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions Solution: Continuity at the boundary: y1(x=0) = y2(x=0) A + B = C, y1'(x=0) y1/x| Thus: B/A = ( k-q x=0 = y2'(x=0) ikA - ikB = iqC A - B = (q/k)C, = (q/k)(A + B). k-q / k+q ), and: R = [ / k+q ] c. For a potential step of V0 = 15 eV height, calculate the wave number q of the particle on the right-hand side (x>0) of the potential step and calculate the probability for reflection. Solution: q = [2m(K-V0)] / h = 1000 eV/c / 197 eVnm/c = 5.1 nm-1. |R| = [ k-q / k+q ] = [(1/2k)/(3/2k)] = [1/3] = 0.111 d. If instead the potential step is V0 = 25.0 eV (i.e. 5.0 eV larger than the incident particle's energy!) show that the penetration probability as function of penetration depth x (x>0) is proportional to e-2 , with k = 5.1 nm-1. Solution: In this case the wave number q in the region behind the step will be purely imaginary: q = [2m(K-V0)] / h = i[2m(V0-K)] / h = 5.1i nm-1. kx Thus, the wave function behind the step will be completely real because iq is real: y2=Ceiqx + De -iqx . Here, the first exponential will be an exponentially growing function of x, and cannot remain finite as the probability interpretation of the wavefunction-squared requires! Therefore the coefficient C must vanish. That leaves: y2=De -iqx . = De-5.1 . x -2iqx The barrier penetration probability P as function of x is thus: P(x>0) = |y2(x)| = |D|e = |D|e -10.2x . e. For this last case, calculate the mean penetration depth inside the barrier. Solution: The mean penetration is simply the mean of the falling exponential, namely the value where it reaches 1/e down from its maximum: x = 0 |D|xe-x/a dx / 0 |D|e-x/a dx = 0 xe-x/a dx / 0 e-x/a dx = a/a = a = 1/10.2 nm-1 = 0.10 nm. V. The negatively charged tau particle (t-, with mass = 1800 MeV/c and mean life of 0.30 ps). (30 points) a. If the tau has a relativistic energy of 36 GeV as measured in the laboratory, calculate its mean life as observed in the laboratory. Solution: g = / E mc = 36,000 MeV/1,800 MeV = 20. Thus: tlab = g t0 = 200.30 ps = 6.0 ps. b. Calculate the speed of the tau (with 36 GeV energy). Solution: The speed is b = (1-1/g) 1 - 1/800 = 0.99875, and v = 0.99875 c. c. Calculate the mean travel distance of the tau in the laboratory, and compare this to the distance it would travel in a purely Newtonian world. Solution: The mean distance traveled in the lab: d = v tlab = 0.99875 c 6.0 ps = 0.99875 0.30 mm/ps 6.0 ps = 1.8 mm. Without relativity we would have found: v = (2K/m) = (2K/m) = c (72,000/1,800) = 6.3c. And the corresponding distance in the lab would have been: d = v t0 = 6.3 0.30 mm/ps 0.30 ps = 0.57 mm only. Alternatively, if we were given the speed as 0.99875c then we would find classically: ct = c0.3 ps = 0.090 mm. d. Using the Heisenberg uncertainty relationship, calculate the uncertainty in the mass energy of the tau. Solution: Dmc hc/(2ct0) = 197 eV nm / (23.0105 nm/ps 0.30 ps) = 1.110-3 eV. file:///C|/Documents%20and%20Settings/Linda%20Gra...0HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (10 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions VI. Consider a collection of NA (Avogadro's number) non-interacting identical but distinguishable atoms. (30 points) The system has a ground state energy E0 = 0, and a single excited state at energy E1 = 3.00 eV. The system is at temperature T = 34815 K. Assume the atoms obey Maxwell-Boltzmann statistics, and take both the ground state and the excited state to be non-degenerate. a. Calculate the ratio of the number of atoms in the excited state to the number in the ground state. Solution: At 34815 K, the energy kT = 8.61710-5 34815 eV = 3.00 eV. The ratio of atoms in the excited state and those in the ground state is: NE/N0 = F = e- E E/(kT) = e-3.00/3.00 = e-1.0 = 0.368 . b. Calculate the average energy of an atom in this system. Solution: The average energy is simply: E = (N E + N00)/(N + N0) = F E/(F + 1) = (0.368/1.368) E = 0.269 E = 0.807 eV. E E E E c. Calculate the total energy Etot of the system in eV and in Joule. Solution: The total energy is simply the total number of atoms times the average energy: Etot = NA E = 6.021023 0.807 eV = 4.851023 eV = 7.82104 J. d. Calculate the heat capacity C dEtot/ (in J/K!) of this system. dT Solution: C dE / dT = / (Etot) = /dT[NA e- d d dT E/(kT) E /(e- E/(kT) + 1)] = .... Constants and Formulae Scale Prefixes: Constants c = 2.9979108 m/s = 3.001017 nm/s, Masses: p k(ilo)=103, M(ega)=106, G(iga)=109, T(era)=1012. m(illi)=10-3, (icro)=10-6, n(ano)=10-9, p(ico)=10-12, e = 1.60210-19 C f(emto)=10-15 me = 0.511 MeV/c = 9.1110-31 kg; m = 938 MeV/c; mn = 940 MeV/c, 1u = 931.5 MeV/c = 1.66010-27 kg Planck's constant: h = 1240 eV/c nm = 4.13610-15 eVs h ( /2 ) = 197 eV/c nm p h Boltzmann constant: k = 8.61710-5 eV/K,; at room temperature: T 290 K, kT 0.025 eV Coulomb's constant: kc (4pe0)-1, Fine structure constant: Compton wavelength: hc kce = 1.440 eVnm a (kce/ ) = 1.440 eVnm/197 eVnm = 1/137.02 lC / ) = 0.00242 nm, lC / ) = 386 fm mc mc h h file:///C|/Documents%20and%20Settings/Linda%20Gra...0HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (11 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions Bohr radius: a0 ( C/ ) = 0.0529 nm Avogadros number: NA = 6.0221023 /mole Bohr magneton: Laws Special Relativity: mB eh/(2m ) = 5.78810-5 eV/T e b / , c v g [ 1/ (1 - b) ] p = gmv, E = (pc + mc4) = gmc, p m K = E - mc = (g -1) mc ( = /2 Bragg scattering: 2d sinq = nl Photo-electric effect: E = hf = Kmax + W = eVstop + W Stefan's law: 0 for v<<c) R dl = sT4, with s = 5.6710-8 W/m/K4 Wien's displacement l -3 6 peakT = 2.89810 mK = 2.89810 nmK law: Planck's radiation law: R(l;T) = ( /4) (8 / 4) ( / ) [1/ l l c p hc e hc/(lkT) - 1] Compton scattering: Dl = l' - l = [h/(mec)] (1 - cosq) de Broglie wavelength: Heisenberg's uncertainty relationships: Probability: Average of f(x): l = h/p = 2p/k, with k the wavenumber DxDpx h/2, DEDt h/2 P(x) = |y(x)|, with y properly normalized, i.e. - |y(x)|dx = 1 or |y(r)|dV = 1 f(x) y (x) f(x) y(x) dV; with dV = dx dy dz = dr rdq rsinq df root-mean-square of Dx (x - x) variable x: Static Schrdinger equation: (- /2 ) / y(r) + V(r) y(r) = Ey(r) m r h Constant potential y(x) = Aeikx + Be-ikx, with wave number k = (2mK)/ , m the particle mass and K the kinetic energy h V=0: (1-dimensional) K=p/2 m file:///C|/Documents%20and%20Settings/Linda%20Gra...0HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (12 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions Constant box potential: V = 0 inside the box (size: LxLyLz) y(r) = (2/Lx)sin(nxp x/Lx) (2/Ly)sin(nyp y/Ly) (2/Lz)sin(nzp z/Lz) with nx, ny, nz=1,2,3,... Enxnynz = (nx/Lx + ny/Ly + nz/Lz)( /8 ), with m the mass of the particle in the box m h V = outside the box Harmonic potential E = (n+) hw = (n+) h(k/ ) = (n+) hn ; n = 0,1,2,... n m V = kr: with the reduced mass m = mM / m+M and m and M the masses of the particles involved. Coulomb potential y(r) = R (r) Y (q,f) = R (r) P (q) e-imlf. ; n=1,2,3,...; l=0,1,2,...,(n-1); nl lm nl lm V = kcZe/r: ml =0,1,2,...,l En = -( / ) 1/2amc = -13.61 eV ( / ) , with the reduced n n Z Z mass m = mM / m+M , m the electron and M the nuclear mass. Ground state and some excited states of hydrogen (Z=1): yn,l,ml(r,q,f) 1s: y100 = 1/ 2s: y200 = 1/ (pa0) e -r/a 0 r -r/2a (8pa0) (1 - /2 0)e a r (8pa0) 0 2p, ml=0: y210 = 1/ ( /2 0)e a r a -r/2a 0 cosq 0 2p, ml=1: y211 = 1/ 3s: y300 = 1/ (27pa0) (8pa0) r ( /2 0)e a -r/2a sinq eif -r/3a 0 (1 - 2 /3 0+ 2r/27 0)e a Atomic quantum numbers: n = 1,2,3,4,... ( K,L,M,N,... shells) l = 0,1,2,3,...,n-1 ( s,p,d,f,g,... ) ml = 0,1,2,3,...,l. Transitions: Dl = 1 s = , ms = Atomic angular momenta: Particle Spin |L| = h[l(l+1)] , l=0,1,2,...,(n-1); Lz = mlh, ml=0,1,2,...,l; |S| = h[s(s+1)] , s=; Sz = msh, ms= (for electrons, protons and neutrons) |S| = h[s(s+1)]; Fermions: s=1/2,3/2,5/2,..; Bosons: s=0,1,2,..; Sz = msh, |ms| = s,s-1,s-2,...0 file:///C|/Documents%20and%20Settings/Linda%20Gra...0HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (13 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions Magnetic potential energy: VB = -B = -zB (z//B), = ( /2 )L, with Q the charge; me gQ for e-: VB, = ( /2 )Lz B = mBml B; e L me VB, = mBgems B = mB2ms B S Moseley's law: Rotation: DE = hn = Zeff(1/2mca)(3/4); Erot = [ /2 ], I L Zeff Z - 1 L=0,1,2,... Transitions: DL = 1 mM L = L(L+1) h , with the moment of inertia I = mR, with reduced mass m = Vibration: Statistics: Evib = , (N+) hw = (N+) h( / ) = (N+) hn, m k / m+M and separation distance R N=0,1,2,... Number of microstates: Maxwell-Boltzmann: Wj = N!/(nj,0! nj,1! nj,2! nj,3! nj,4! ... ) FMB(E;T) = Ae-E/kT F(v;T) = 4pn ( /2 (E-EF)/kT + 1)] e E/kT - 1)] Be Maxwell speed distribution: Fermi-Dirac: Bose-Einstein: m pkT )3/2 v2e- mv/(2kT) FFD(E;T) = [1/( FBE(E;T) = [1/( (for photons: B = 1) N n(E)dE = N(E)/V dE = g(E)dE Fxx(E;T), with normalization condition: 0 n(E)dE = /V Density of states for a cubic box: g(E) = [8p2 me3/2 /h] E = [2 me3/2 /(ph)]E. Solid State Drift speed of electrons: vd = eEt/m, with t is the (quantum) Drude relaxation time; l = t vF = 1/(nis) is the mean free path, E the electric field, vF the Fermi velocity (i.e the speed of electrons with energy EF), s the cross section for electron scattering, and ni the number density of ions in the solid. Electric current density: j I/A = enevd , where A is the cross sectional area, and ne the number density of conduction electrons. Resistivity: r E/j Metal Li Al Cu Nuclear physics: Nuclear decay: Atomic weight Drude relaxation Fermi velocity Mass density Conduction [g/mole] time t [s] [m/s] [kg/m] electrons/atom 6.9 27.0 63.5 ( dN(t) dt 0.8810-14 0.8010-14 2.710-14 / ) = -lN(t), 1.29106 2.03106 1.57106 N(t) = N0e- = N0 2- / tt lt 0.53103 2.70103 8.93103 1 3 1 Math formulae: 0 0 xn e-x/a dx = n! an+1 ; - e-x/a dx = ap ; - x e-x/a dx = ( /2)p ; (x 1) a (ex + 1) -1 dx = ln 2; ln(1+x) = x - [(x2)/2] + [(x3)/3] - [(x4)/4] +... file:///C|/Documents%20and%20Settings/Linda%20Gra...0HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (14 of 15) [2/4/2008 4:41:09 PM] PHY251_M2_S01_solutions file:///C|/Documents%20and%20Settings/Linda%20Gra...0HEP-H2/PHY251%20Sp01/PHY251_F_S01_solutions.html (15 of 15) [2/4/2008 4:41:09 PM] ... View Full Document