Problem Set 4: Whither, thou turbid wave SOLUTIONS

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1 PH 253 / LeClair Spring 2013 Problem Set 4: Witer, tou turbid wave SOLUTIONS Question zero is probably were te name of te problem set came from: Witer, tou turbid wave? It is from a Longfellow poem, Te Wave, wic you can find ere: ttp:// com/autor/henry_wadswort_longfellow/te_wave.tml. 1. Te wavefunction of a transverse wave on a string is ψ(x, t) = (30.0 cm) cos [(6.28 rad/m) x (20.0 rad/s) t] (1) Compute te frequency, wavelengt, period, amplitude, pase velocity, and direction of motion. Solution: Te wavefunction as te form ψ(x, t) = A sin (kx ωt + δ) (2) wic lets us immediately identify ω = 2πf = 20 rad/s = f = 2π ω = 10 π s 1 (3) k = 2π λ = 6.28 rad/m = λ = 2π k 1.0 m (4) A = 30 cm (5) v = ω k 3.18 m/s (6) Since te argument as te form kx ωt, te wave is traveling along te +x direction. 2. Wat is te uncertainty in te location of a poton of wavelengt 300 nm if tis wavelengt is known to an accuracy of one part in a million? Solution: Te momentum of te poton is p = λ (7) Te uncertainty in te poton momentum can be related to te uncertainty in wavelengt (as we ave done before wit wavelengt and energy uncertainty). p = p λ λ = p λ λ = (8) λ2 λ

2 Using te uncertainty principle, x 2 p = λ 23.9 mm (9) 2p λ Tis is a uge distance for a poton, but keep in mind tat a poton takes only about 80 ps i to cover tis distance! 3. Find te potential difference troug wic electrons must be accelerated (as in an electron microscope, for example) if we wis to resolve: (a) a virus of diameter 12 nm, (b) an atom of diameter 0.12 nm, (c) a proton of diameter 1.2 fm. Sow your work, and do not forget about relativity. Solution: Te resolution of electron waves is rougly teir wavelengt. Wavelengt we can find from momentum, from wic we can get kinetic energy. Conservation of energy dictates tat an electron accelerated from rest troug a potential difference V, tereby acquiring potential energy e V, will ave a kinetic energy K = e V. Given tat te smallest distance is well below te electron s Compton wavelengt, we expect relativity may be important for at least tat resolution, and we sould perform a relativistic analysis. An energy balance using te relativistic form for kinetic energy gives K = e V = p 2 c 2 + m 2 c 4 mc 2 (10) Solving for p = /λ, we can relate wavelengt and potential difference ( e V + mc 2 ) 2 = p 2 c 2 + m 2 c 4 (11) p 2 = 1 c 2 ( e V + mc 2 ) 2 m 2 c 2 = 2 λ 2 (12) We can now find te potential difference in terms of te desired wavelengt, wic sets te microscope s resolution 2 c 2 λ 2 + m 2 c 4 = ( e V + mc 2) 2 (13) (c ) 2 e V = + (mc λ 2 ) 2 mc 2 relativistic (14) For simplicity in te resulting calculations, we can note c 1240 ev nm, mc kev, and 1 fm=10 6 nm. Tis way, putting te wavelengt in nm we obtain e V in electron volts, and te i p = pico =

3 same numerical value is V in volts. We can also note te classical result, wic makes use of K=p 2 /2m: e V = ( ) c 2 1 λ 2mc 2 classical (15) wavelengt (nm) V (relativistic) V (classical) V 0.01 V V 100 V V V Even at 0.12 nm resolution te classical result is just fine, but for fm resolution, it breaks down badly. 4. In order to study te atomic nucleus, we would like to observe te diffraction of particles wose de Broglie wavelengt is about te same size as te nuclear diameter, about 14 fm for a eavy nucleus suc as lead. Wat kinetic energy sould we use if te diffracted particles are (a) electrons? (b) Neutrons? (c) Alpa particles (m = 4 u)? Solution: We can make use of te result of te previous problem ere all we need to do is skip te step were we set K=e V and solve for kinetic energy instead. Tat gives: (c K = λ ) 2 + (mc 2 ) 2 mc 2 (16) particle rest energy mc 2 K (relativistic) K (classical) electron 511 kev 88 MeV 7.7 GeV neutron 940 MeV 4.2 MeV 4.2 MeV α 3.73 GeV 1.1 MeV 1.1 MeV Te electron is clearly deeply in te relativistic limit, since te classical expression fails badly. One way to see tis is tat te kinetic energy implied far exceeds te rest mass of te electron, wile for te oter particles te kinetic energy is well below te rest mass. Wen te particle s energy is of order its rest mass, we expect relativistic effects to be important. Anoter rule of tumb is tat te Compton wavelengt λ c = /mc for te particle in question is rougly te scale at wic one as to worry about relativistic and quantum effects. 5. Te speed of an electron is measured to witin an uncertainty of m/s. Wat is te size of te smallest region of space in wic te electron can be confined? Solution: Since v/c 10 4, we need not worry about relativity. Momentum is ten just p = mv,

4 and since m is a constant for an electron, we can say p=m v. We are given v= m/s, so we can just apply te uncertainty principle. x p = x (m v) (17) 4π = x = 4πm v = 2.8 nm (18) 2m v 6. Use te distribution of wave numbers A(k) = A o e (k k o) 2 /2( k) 2 k R (19) and equation 4.23 from your text, viz., y(x) = A(k) cos kx dk (20) to obtain equation 4.25 from your text, viz., y(x) = A o k 2πe ( k x)2 /2 cos k o x (21) Solution: Te matematics will be far easier if we write te function y(x) as a complex exponential and take te real part later: y(x) = A(k) cos kx dk = A(k)R { e ikx} dk = A o e (k k o) 2 /2( k) 2 R { e ikx} dk (22) Remembering tat we will take te real part in te end, integrating over all possible k gives y(x) = A o e (k k o) 2 /2( k) 2 e ikx dk (23) Collect te exponential terms and expand, ten factor out te parts tat don t contain k (since tey are not integrated)

5 y(x) = = A o A o e (k k o) 2 /2( k) 2 e ikx dk = A o e 1 k2 o e 1 2( k) 2 [k 2 2kk o +k 2 o 2ik( k) 2 x] dk (24) 2( k) 2 [k 2 2k(k o +i( k) 2 x)] e 2( k) 2 dk (25) Let a = k o + i ( k) 2 x for convenience, and note tat te second exponential term can be pulled from te integral. y(x) = A o e k2 o 2( k) 2 e k2 2ak 2( k) 2 dk (26) Te part remaining to be integrated is almost a Gaussian, wic we would know wat to do wit. We can make it a Gaussian by noting tat k 2 2ak=(k a) 2 a 2. y(x) = A o e k2 o 2( k) 2 e (k a)2 a2 2( k]) 2 dk = A o e k2 o 2( k) 2 e 2a2 2( k) 2 e (k a)2 2( k]) 2 dk (27) Once again te second exponential term doesn t depend on k, so we may remove it from te integral. We will also substitute back in a=k o + i ( k) 2 x: y(x) = A o e k2 o (ko+i( k) 2 x) 2 2( k) 2 e 2( k) 2 e (k a)2 2( k) 2 dk (28) Te integral is now just a Gaussian, wic evaluates to k 2π. Expand wat remains and simplify: y(x) = A o k 2π e 1 2( k) 2 [ k 2 o+k 2 o+2ik o ( k) 2 x ( k) 4 x 2 ] = Ao k 2π e x2 ( k) 2 /2 e ik ox (29) Now recall we only want te real part of te expression. Te only complex quantity left is e ik ox, and R { e ik ox } =cos k o x. Tus we obtain te desired result: y(x) = A o k 2π e x2 ( k) 2 /2 cos k o x (30) 7. A particle of mass m is confined to a one-dimensional line of lengt L. From arguments based on te uncertainty principle, estimate te value of te smallest energy tat te body can ave.

6 Solution: Te particle must be somewere witin te given segment, so te uncertainty in its position cannot be greater tan L. If we say x=l, te uncertainty principle implies a momentum uncertainty of p /4πL. Wit maximum position uncertainty, we ave a minimum momentum uncertainty. Tis in turn implies a minimum energy, since K=p 2 /2m. If we assume tat te minimum momentum is just alf of te uncertainty (i.e., te momentum may be zero plus or minus alf te uncertainty), ten p min = 1 2 p=/8πl. Tus, K min = p2 2m = 2 128π 2 ml 2 (31) Given ow crude our arguments are, te dependence on mass and lengt agrees reasonably well wit te minimum energy for a particle in a one-dimensional potential well we derived in class, E 1 = 2 8mL 2 (32) 8. (a) Find te de Broglie wavelengt of a nitrogen molecule in air at room temperature (293 K). (b) Te density of air at room temperature and atmosperic pressure is kg/m 3. Find te average distance between air molecules at tis temperature and compare wit te de Broglie wavelengt. Wat do you conclude about te importance of quantum effects in air at room temperature? (c) Estimate te temperature at wic quantum effects migt become important. Solution: (a) Te de Broglie wavelengt depends on te momentum, wic depends on mass and velocity. Te mass of a nitrogen molecule is 28 u or kg. Wat its its average velocity? Neglecting vibrations of te molecule, eac degree of freedom (axis of motion) gets on average 1 2 k BT wort of energy, for a total of 3 2 k BT average termal energy. Tis sows up as kinetic energy of te molecule, so on average: 1 2 m v 2 = 3 2 k BT 3kB T = v = m (33) (34) Te de Broglie wavelengt is ten, on average, λ = m v = 3kB Tm (35) At room temperature, tis is about 0.03 nm.

7 (b) We don t even really need te density in fact. We know tat at standard temperature and pressure, one mole of gas is about molecules and takes up 22.4 L (note 1 L = m 3 ). Tis lets us figure out ow many molecules tere are per cubic meter: N = molecules/m 3 = molecules 22.4 L Tat means, on average, eac molecule takes up V = m 3 /molecule 1 L m molecules/m 3 (36) 1 m molecules m 3 /molecule (37) If te molecules are distributed evenly, ten tere augt to be about 3 N molecules along eac side of a 1 m cube. Put anoter way, eac one must on average occupy a little cube wic is 3 V on a side, giving te average spacing as d 3 V 3 nm (38) One significant figure is plenty for an analysis of tis nature. Using te given density, we could arrive at te average volume per molecule anoter way, using te molecular mass of nitrogen (28 g/mole): ( kg V = m 3 ) ( 1 mole 28 g ) ( 1000 g 1 kg ) ( ) molecules molecules/m 3 (39) 1 mole (c) We would expect quantum effects to be important wen te average de Broglie wavelengt is about te same as te distance between molecules for a given density. Presuming we keep te density fixed, tis means λ = = T 3kB Tm d (40) K (41) 3k B md2 9. Refer to te video lecture by Feynman you watced. (a) In te two slit experiment, is it possible to design an experiment wic allows you to tell wic slit an electron goes troug witout spoiling te interference pattern? (b) A particle can proceed from point A to point B along two indistinguisable pats. Pat A as an amplitude of a, pat B as an amplitude of b. Wat is te overall probability for te particle to go from point A to point B? Solution: (a) No, it is not possible to design an experiment wic allows one to tell wic slit te electron went troug witout spoiling te interference. (b) Wen te pats are indistinguisable, we must add amplitudes and ten square to get te probability, so P =(a + b) 2.

8 10. X-ray potons of wavelengt nm are produced by a copper source. Suppose tat of tese potons are absorbed by te target eac second. (a) Wat is te total momentum p transferred to te target eac second? (b) Wat is te total energy E of te potons absorbed by te target eac second? (c) If te beam sines perpendicularly onto a perfectly reflecting surface, wat force does it exert on te surface? Recall F = p/ t. (d) For tese values, verify tat te force on te target is related to te rate of energy transfer by dp dt = 1 de c dt Solution: (a) We know eac poton as a momentum of /λ, so te net momentum transfer in one second is te momentum per poton times te number of potons arriving in one second : p net t = ( potons/s ) ( ) kg m/s (42) λ (b) Te energy of eac poton is E = c/λ, so te energy in one second is just te energy per poton times te number of potons arriving in one second: E tot t = ( potons/s ) ( ) c = c p net λ t J/s = W (43) (c) Since te potons are reflected, teir momentum canges from p to p, and te target must ten acquire momentum 2p per poton tat its it to conserve momentum. Tat means te net force is just twice te net momentum transfer per second: F net = 2 p net t = N (44) (d) Actually, tis as been verified in part b already, but let s do it more formally. Let N be te number of potons striking te target, suc tat te rate of potons arriving is dn/dt = s 1. Tis gives potons arriving in one second as te problem specifies, we ave only assumed tat tey arrive at a uniform rate. dp net dt de net dt = dp dt = 1 de c dt = d ( N ) = dt λ λ = d ( N c ) dt λ dn dt = c dn λ dt (45) (46) (47)

9 We can also verify te relationsip numerically, de/dt dp/dt = J/s kg m/s = m/s = c (48)