Ch., Problem 3 After acceleration through 0 MV both the electron and proton will each have a kinetic energy of 0 MeV. For the electron, E K + mc 0:5 M

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1 Physics 3304 Assignment solutions Grading: The problems that will be graded in detail for this assignment are Ch., number 35 and Ch. 3, number 9 èeach worth 5 points totalè. You will receive point for each of the other problems if you have made a reasonable attempt at a solution. ètotal points for this assignment8è Ch., Problem 5 aè Use the relativistic velocity transformation to determine the velocity v 0 as measured in the frame O 0 : Then the energy becomes: E 0 æ 0 mc v 0 v, u, uv c mc, èv 0 è c mc r èv,uèc,uvc and the momentum is: p 0 æ 0 mv 0 h mv 0 m r, èv 0 è c i èv,uè,vuc èv,uèc,uvc E 0, p 0 c m c 4, m c h i èv,uè,uvc èv,uèc,uvc m c 4 so the uantity E 0, p 0 c is an invariant, meaning it has the same constant value èsuare of the particle's rest energyè in any inertial reference frame. The rest mass m is sometimes referred to as the invariant mass. Ch., Problem 3 aè The ærst stage of acceleration takes the electron from rest èe 0 mc 0:5 MeVè up to the energy: E æmc 0:5 MeV, è0:99è 3:6 MeV so the ærst stage adds 3:6, 0:5 3: MeV to the energy of the electron. The second stage takes the electron up to an energy: E æmc 0:5 MeV, è0:999è :4 MeV so the second stage adds :4, 3:6 7:8 MeV to the energy of the electron.

2 Ch., Problem 3 After acceleration through 0 MV both the electron and proton will each have a kinetic energy of 0 MeV. For the electron, E K + mc 0:5 MeV p E, èmc è 0:5 MeVc c while the classical relation between kinetic energy and momentum gives: p p mk cp mc K 3:0 MeVc Since the electron is very relativistic èpc éé mc è the classical expression fails badly. For the proton, E K + mc 0:0 MeV + 938:3 MeV 948:3 MeV p E, èmc è 37 MeVc c while the classical relation between kinetic energy and momentum gives: p p mk cp mc K 37 MeVc Since the proton is non-relativistic èpc éé mc è the classical expression works well in this case. Ch., Problem 35 E p + E p 9700 MeV Since the proton and antiproton have the same speed èand rest massè, their energies are eual: E p E p æmc " mc, v c 938 MeV, v c è9700 MeVè v 0:98c è so Ch., Problem 36 Since the pion is moving at the speed 0:98c, its energy and momentum measured in the laboratory are: E æmc p c 35 MeV, è0:98è 678 MeV E, èmc è 664 MeVc

3 The problem says that the decay gamma rays have eual energies so they share the 678 MeV eually: E æ 339 MeV. The angle of the two æ-rays can be obtained from the conservation of momentum: p x 664 MeVc p æ x p æ cos èe æ cè cos :7 æ Ch., Problem 37 As in Example.6, the total energy of the kaon is 83 MeV. The speed of the kaon relative to the lab frame can be found from the expression for the energy: E æmc 83 MeV æè498 MeVè æ :656 vc 0:796 This is also the transformation speed u that is reuired to transform from the lab frame to a frame where the initial kaon is at rest. When the decay of the kaon is observed in its rest frame, the two pions share the decay energy eually and they move in opposite directions with eual and opposite velocities. So we have: E E m Kc 49 MeV æm c m c, v c Solving gives that the speed of each of the pions is v 0:87c. Now we transform back to the lab frame by making a Lorentz velocity transformation using the speed u 0:796c which is the relative velocity between the lab èoè and kaon rest èo 0 è frames. We assume in the kaon rest frame one pion moves along the +x 0 axis with speed v 0 +0:87c and the other moves along the,x 0 axis with speed v 0,0:87c. So we then have for the speeds of the pions in the lab frame: v v 0 + u +uv 0 c æ 4:873 v v 0 + u +uvc 0 æ :004 0:87c +0:796c +è0:796èè0:87è 0:9787c,0:87c +0:796c +è0:796èè,0:87è,0:0907c Then the kinetic energies of the pions in the lab frame are given by: K èæ, èm c è4:873, èè40 MeVè 54 MeV K èæ, èm c è:004, èè40 MeVè 0:6 MeV Ch. 3, Problem 3

4 aè d sin n è0:347 nmèèsin 34:0æ è 0:388 nm The spacing between the planes oriented at 45 æ is: p d sin 45 æ è0:347 nmè 0:45 nm sin 0:388 nm 0:79 or 5:æ d è0:45 nmè This is the angle measured with respect to the crystal planes. Referring to ægure 3.6 in the book, the angle of incidence measure with respect to the crystal surface is, 45 æ 7: æ, while the emerging beam makes an angle of +45 æ 97: æ with respect to the surface èmeasured from the opposite side of the surfaceè. Ch. 3, Problem 6 aè E :00 æ 0 4 ev 0:4 nm :00 æ 0 6 ev :4 æ 0,3 nm cè 350 nm : E 700 nm : E so the range is.8 ev to 3.5 ev. Ch. 3, Problem nm 3:5eV 700 nm :8eV The stopping potential èv s è is related to the work function and wavelength through the relation: ev s K max, 0:65 ev :69 ev, 40 nm, 30 nm so

5 Subtracting the two euations allows us to solve for Planck's constant, h, using only the data given: :04 ev 30 nm 40 nm yielding h 4:0 æ 0,5 ev-s Now either of the above two euations can be used to obtain the work function :, :69 ev :8 ev 30 nm Ch. 3, Problem 3 aè 4:3 ev for zinc The largest wavelength corresponds to the minimum energy photon reuired to extract a photoelectron. That minimum energy is simply the work function. E æ min max E æ min 4:3 ev 88 nm ev s K max V s :33 volts, 0:0 nm,4:3 ev :33 ev

Physics 111 Homework Solutions Week #9 - Thursday

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