Physics 220 Homework #2 Spring 2015 Due Wednesday 4/15/15 1
Transcription
Physics 220 Homework #2 Spring 2015 Due Wednesday 4/15/15 1
Physics 220 Homework #2 Spring 2015 Due Wednesday 4/15/15 1. Photons with a wavelength λ = 410nm are used to eject electrons from a metallic cathode (the emitter) by the photoelectric effect. The electrons are prevented from striking the anode (the collector) by applying a stopping potential of 0.88V . What is the work function of the metal and what is the likely composition of the metal surface? From the photoelectric effect, we have hc hc eVstop = −φ →φ = − eVstop λ λ ⎡⎛ 6.6 × 10 −34 Js × 3 × 10 8 ms ⎞ ⎤ 1eV φ = ⎢⎜ × ⎟⎠ 1.6 × 10 −19 J ⎥ − 0.88eV = 2.14eV 410 × 10 −9 m ⎣⎝ ⎦ Looking up the work function at http://hyperphysics.phyastr.gsu.edu/hbase/tables/photoelec.html, the metal could be cesium at φ = 2.1eV . 2. Photons with energy E are incident on a photocathode (metal surface) and produce electrons with a maximum kinetic energy of 2eV . When the energy of the photons is doubled, the maximum kinetic energy of the electrons increases to 8eV . What is the work function of the metal surface? From the first piece of information, we can calculate the energy of the incident photon. We have 2eV = E − φ → E = 2eV + φ . From the second piece of information we have 8eV = 2E − φ = 2 ( 2eV + φ ) − φ = 4eV + φ → φ = 4eV . 3. When radiation of wavelength λ = 410nm is incident on a cathode with an unknown work function, and electrons are ejected by the photoelectric effect. The minimum voltage necessary to prevent the electrons from reaching the anode, the “stopping potential” is unknown. Experimenters change the wavelength of the radiation and measure the change in the stopping potential. What wavelength of radiation would correspond to an increase in the stopping potential of 1V ? Let λ and λ ' be the original and new wavelengths respectively. We have originally, hc hc eVstop = − φ and now eVstop + 1eV = − φ . Solving this for λ ' we get λ λ' hc 6.6 × 10 −34 Js × 3 × 10 8 ms λ'= = = 308nm . −19 −34 8 m hc ⎛ ⎞ 1.6 × 10 J 6.6 × 10 Js × 3 × 10 s 1eV + 1eV × ⎟⎠ + λ ⎜⎝ 1eV 410 × 10 −9 m 4. Radiation with a wavelength λ = 300nm of is directed at a metallic surface. Estimate the radiation flux (power per unit area) necessary to have an average of 1 photon per second strike each atom on the surface of the metal. Assuming the atom has a rough diameter of 0.3nm , we can calculate the area over which the radiation is spread. We have A = π r 2 = π ( 0.15 × 10 −9 m ) = 7.07 × 10 −20 m 2 . 2 Next we want 1 photon per second to strike an atom, so this corresponds to an energy (per second) of hc hc 6.6 × 10 −34 Js × 3 × 10 8 ms 1 photon λ 300 × 10 −9 m Etotal = λ × = = = 6.6 × 10 −19 Js . Thus the photon s s s intensity is S = Energy power 6.6 × 10 −19 Js = = = 9.3 mW2 . time × area area 7.07 × 10 −20 m 2 5. Make a plot of eVstop versus frequency for cesium, magnesium, and platinum on the same axis. Provide for positive values of Vstop for 0 to 5 Volts and indicate on your graphs the visible portion of the spectrum. Looking up the work functions at http://hyperphysics.phyastr.gsu.edu/hbase/tables/photoelec.html, we find φCs = 2.1eV , φ Mg = 3.7eV , and φPt = 6.4eV . The plot is generated using eVstop = hf − φ for frequencies from the infrared ( λ IR = 10 µ m ) to the ultraviolet ( λuv = 200nm ). The plot is below, where the blue line is for cesium, the red is for magnesium, and the yellow is for platinum. eV 6 4 2 1.0¥101 5 -2 -4 -6 1.5¥101 5 fHs- 1 L 2.0¥101 5 6. Photons with kinetic energy equal to the mass energy of an electron collide with an π electron at rest and scatters at an angle of . Calculate the energy of the electron 2 after the collision. Express your answer in terms of mc 2 . Starting from the Compton wavelength formula, divide both sides by hc . We have h λ + (1− cos φ ) 1 1 (1− cos φ ) λ' mc . Next we evaluate the expression at = → = + hc hc E' E mc 2 mc 2 0 . Now the kinetic φ = 90 and determine the scattered photon energy to be E ' = 2 energy of the recoiling electron is the difference between the incident photon energy mc 2 mc 2 = and the scattered photon energy and we have KE = E − E ' = mc 2 − . The 2 2 total energy is the sum of the electron’s rest energy and its kinetic energy. Thus the 3 energy of the recoiling electron is Ee = KE + mc 2 = mc 2 . 2 7. A photon with energy E collides with an electron at rest. Calculate the maximum amount of kinetic energy transferred to the electron. Make a graph of KE versus E, where the units for the energies are in electron volts. The kinetic energy of the electron is given by KE = E − E ' , where 1 1 (1− cos φ ) Emc 2 . Combining these two equations we = + → E' = E' E mc 2 mc 2 + E (1− cos φ ) Emc 2 have KE = E − E ' = E − 2 . The maximum energy transferred to the mc + E (1− cos φ ) electron is when the scattered photons energy is the least and this occurs for complete backscattering, or at an angle of φ = 180 0 . The maximum kinetic energy of the Emc 2 2E 2 = electron is therefore KE = E − 2 . The plot of the kinetic energy mc + 2E mc 2 + 2E versus the incident photon energy is shown below. KEHMeVL 1.5 1.0 0.5 0.5 1.0 1.5 2.0 EHMeVL 8. From the three equations in class we have from conservation of total energy (1) and conservation of momentum (2 & 3) hc hc hc 1: + mc 2 = + Ek + mc 2 = +E λ λ' λ' h h 2 : = cos φ + p cosθ λ λ h 3 : 0 = sin φ − psin θ λ To derive the Compton shift in wavelength, square (1) and simplify terms to get 2 ⎛ hc hc ⎞ ⎛ hc hc ⎞ 2 4 : E = ⎜ − ⎟ + 2 ⎜ − ⎟ mc 2 + m 2 c 4 . ⎝ λ λ'⎠ ⎝ λ λ'⎠ Next, square (2) and (3) and add the results. Multiply the result by c2. This produces (5) h2c2 h2c2 h2c2 2 2 5: p c = 2 + 2 − 2 cos φ . λ λ' λλ ' Solve (4) for the expression p2c2 and set the two results equal to one another and you obtain (6). 2 h2c2 h2c2 h2c2 ⎛ hc hc ⎞ ⎛ hc hc ⎞ 6 : E 2 − m 2 c 4 = p 2 c 2 → ⎜ − ⎟ + 2 ⎜ − ⎟ mc 2 = 2 + 2 − 2 cos φ . ⎝ λ λ'⎠ ⎝ λ λ'⎠ λ λ' λλ ' Solve (6) for the scattered wavelength h λ' = λ + (1 − cosφ ) . mc ( ) 9. Show that the scattering angle of the electron scattered in a Compton effect λ sin φ experiment is given by tan θ = . In 1950 two scientists Cross and h ⎞ ⎛ ⎜⎝ λ + ⎟ (1− cos φ ) mc ⎠ Ramey were able to detect the electrons from 2.6MeV gammas scattered to φ = 30 0 . At what angle were they able to detect the scattered electrons? Using conservation of momentum (2) and (3), we have dividing (3) by (2): psin θ hsin φ hsin φ tan θ = = = p cosθ ⎛h h ⎞ ⎛ ⎞ λ ' ⎜ − cos φ ⎟ ⎝ λ λ' ⎠ ⎛ ⎜ ⎟ h h h cos φ ⎟ (1− cos φ )⎞⎟⎠ ⎜ − ⎜⎝ λ + h mc ⎜ λ ⎛⎜ λ + ⎟ (1− cos φ )⎞⎟⎠ ⎝ ⎝ ⎠ mc tan θ = sin φ λ sin φ = h ⎞ h ⎞ ⎛ (1− cos φ ) ⎛⎜⎝ 1+ ⎟⎠ (1− cos φ ) ⎜⎝ λ + ⎟ λ mc mc ⎠ Using the result, we find the angle of the detected electrons, where the wavelength of hc the incident photons is determined from E = . Thus, λ λ sin φ tan θ = h ⎞ ⎛ ⎜⎝ λ + ⎟ (1− cos φ ) mc ⎠ tan θ = 4.7596 × 10 −13 m × sin 30 = 0.6145 ⎛ ⎞ 6.6 × 10 −34 Js −13 ⎜⎝ 4.7596 × 10 m + 9.11× 10 −31 kg × 3 × 10 8 m ⎟⎠ (1− cos 30 ) s θ = tan −1 ( 0.6415 ) = 31.6 0 Cross and Ramey actually detected them at θ = 31.30 . 10. A 1.0MeV photon collides with an electron at rest and scatters at an angle of φ = 45 0 . What are the energy of the scattered photon and the kinetic energy of the scattered electron? Starting as in problem #6, we’ll calculate the energy of the scattered photon and then the kinetic energy of the scattered proton. The energy of the scattered photon is 1 1 (1− cos φ ) 1 (1− cos 45 ) → E ' = 0.6357MeV . Therefore the = + = + 2 E' E mc 1MeV 0.511 MeV × c2 c2 energy of the scattered proton is KE = E − E ' = 1MeV − 0.6357MeV = 0.3643MeV .