Physics 0 Homework # Spring 06 Due Monday 4//6. Photons with a wavelength λ = 40nm 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 ev stop = hc λ φ φ = hc λ ev stop 6.6 0 34 Js 3 0 8 m φ = s 40 0 9 m ev.6 0 9 0.88eV =.4eV J Looking up the work function at http://hyperphysics.phyastr.gsu.edu/hbase/tables/photoelec.html, the metal could be cesium at φ =.ev.. When radiation of wavelength λ = 40nm 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 V? Let λ and λ ' be the original and new wavelengths respectively. We have originally, ev stop = hc λ φ and now ev stop +ev = hc φ. Solving this for λ ' we get λ ' hc 6.6 0 34 Js 3 0 8 m λ ' = ev + hc = s ev.6 0 9 J λ ev + 6.6 = 308nm. 0 34 Js 3 0 8 m s 40 0 9 m
3. Make a plot of ev stop versus frequency for cesium, magnesium, and platinum on the same axis. Provide for positive values of V stop for 0 to 5 Volts and indicate on your graph the visible portion of the spectrum. K (ev) 6 5 4 Cs Mg 3 Pt 0 4 6 8 Energy (ev)
4. Starting with the work-kinetic energy theorem, show that the relativistic kinetic energy is given as K = ( γ )mc and show that this reduces to the classical result in the limit that the speed of the object becomes much less than the speed of light. Hint: Use the relativistic form of the momentum, p! = p = γ mv to show K = ( γ )mc, and use the binomial expansion ( + x) n = 0! nx n(n )x!! ( )x3 3! n(n ) n... to show the classical limit. The work-kinetic energy theorem states that: W = F! d r! d p = (! dt ) d r! = vdp = K f assuming the mass starts from rest. The speed is related to the magnitude of the momentum through p = γ mv = mv v c. Take the derivative of the momentum with mv m respect to the velocity and we get by the chain rule: dp = c ( ) + 3 ( v ) v c c dv. Thus the work done is given as: v mv K = m vdp = v c ( ) + 3 ( v ) v c c dv = 0 v c mc = ( γ )mc. The integral was evaluated on Mathematica as shown below. To show the relativistic kinetic energy reduces to the classical result, expand the factor gamma using the binomial expansion. We have ( ) γ = ( v c )! 0! = + v. Where we ve dropped! c n v the higher order terms since v << c v c << ; c <<<<<<. Substituting this ( x) n = 0! nx ( )( v c )
expression for gamma into the expression for the relativistic kinetic energy gives K = ( γ )mc + v c mc = mv, which is the desired result. 5. Starting with the total relativistic energy E = γ mc, show that E = p c + m c 4. The total relativistic energy of a particle is the sum of its kinetic and rest energies. We have, E = γ mc. Since the relativistic momentum is p = γ mv γ m = p v, E = p v c. Solving this for the ratio of v which is in the expression for gamma, we c have v c = pc E E = γ mc =. Substituting this into the expression for the total energy we have mc E = p c + m c 4. pc E. Square both sides and cross multiply we get 6. Using the ideas from class, show that the Compton shift in wavelength is given as Δλ = h ( cosφ ) mc. From the three equations in class we have from conservation of total energy () and conservation of momentum ( & 3) : hc λ + mc = hc λ ' + ( E + k ) mc = hc λ ' + E : h λ = h cosφ + pcosθ λ 3: 0 = h sinφ psinθ λ To derive the Compton shift in wavelength, square () and simplify terms to get 4 : E = hc λ hc λ ' + hc λ hc λ ' mc + m c 4. Next, square () and (3) and add the results. Multiply the result by c. This produces (5) 5 : p c = h c + h c h c λ λ ' λλ ' cosφ. Solve (4) for the expression p c and set the two results equal to one another and you obtain (6).
6 : E m c 4 = p c hc λ hc λ ' + hc λ hc λ ' mc = h c + h c h c λ λ ' λλ ' cosφ. Solve (6) for the scattered wavelength λ ' = λ + h ( cosφ mc ). 7. Photons with kinetic energy equal to the mass energy of an electron collide with an electron at rest and scatters at an angle of π. Calculate the energy of the electron after the collision. Express your answer in terms of mc. Starting from the Compton wavelength formula, divide both sides by hc. We have h λ ' λ + cosφ hc = mc ( ) hc E ' = + ( cosφ ). Next we evaluate the expression at E mc φ = 90 0 and determine the scattered photon energy to be E ' = mc. Now the kinetic energy of the recoiling electron is the difference between the incident photon energy and the scattered photon energy and we have KE = E E ' = mc mc = mc. The total energy is the sum of the electron s rest energy and its kinetic energy. Thus the energy of the recoiling electron is E e = KE + mc = 3 mc.
8. 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 K 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 E ' = E ( ) cosφ + E ' = mc Emc. Combining these two equations we mc + E( cosφ) Emc have KE = E E ' = E. The maximum energy transferred to the mc + E( cosφ) electron is when the scattered photons energy is the least and this occurs for complete backscattering, or at an angle of φ = 80 0. The maximum kinetic energy of the electron is therefore KE = E Emc mc + E = E. The plot of the kinetic energy mc + E versus the incident photon energy is shown below. K (MeV).5.0 0.5 0.5.0.5.0 Energy (MeV)
9. Show that the scattering angle of the electron scattered in a Compton effect λ sinφ experiment is given by tanθ = λ + h. In 950 two scientists Cross and mc ( cosφ) Ramey were able to detect the electrons from.6mev gammas scattered to φ = 30 0. At what angle were they able to detect the scattered electrons? Using conservation of momentum () and (3), we have dividing (3) by (): tanθ = psinθ pcosθ = hsinφ hsinφ h λ ' λ h = λ ' cosφ λ + h cosφ mc ( ) h λ h λ + h cosφ cosφ mc ( ) tanθ = sinφ cosφ ( ) + h = λmc λ sinφ cosφ ( ) λ + h mc Using the result, we find the angle of the detected electrons, where the wavelength of the incident photons is determined from E = hc λ. Thus, λ sinφ tanθ = λ + h mc ( cosφ) 4.7596 0 3 m sin 30 tanθ = 6.6 0 34 Js 4.7596 0 3 m + 9. 0 3 kg 3 0 8 m s cos 30 θ = tan ( 0.645) = 3.6 0 Cross and Ramey actually detected them atθ = 3.3 0. ( ) = 0.645