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Experiment 5 1 Photoelectric Effect Physics 2150 Experiment No. 5 University of Colorado1 Introduction In quantum physics, light is emitted in the form of photons and the energy distribution of photons is not continuous, which cannot be explained by the Maxwell's electromagnetic theory. A photon has unit energy of "ℎ𝜈", where ν is the frequency of light and ℎ is a constant. By illuminating a metal surface with light, the free electrons of the metal will absorb the photon’s energy. If the photon’s energy is higher than the barrier energy of the metal, electrons could escape from the metal surface. This effect is called the photoelectric effect. The kinetic energy of the escaped electron (i.e. photoelectron) will be: !
!
𝐸 = ℎ𝜈 − 𝑊! or ! 𝑚𝜈!
= ℎ𝜈 − 𝑊! (1) -­‐34 Where ℎ is the Planck’s constant (6.626×10
J·s), ν is the frequency of the illuminating light, 𝑚 is the mass of an electron, 𝑣𝑚 is the initial speed of the photoelectron at the metal surface, and 𝑊𝑠 is the escape energy or the work function of the metal. Equation (1) gives the maximum kinetic energy of the photoelectron without any obstruction in space. The higher the frequency of the illuminating light is, the larger the maximum kinetic energy of the photoelectron. Considering the certain initial kinetic energy of the photoelectron, there may be photoelectrons that escape from the metal surface (cathode) to form a photocurrent in the absence of positive voltage applied between the anode and the cathode. In the presence of sufficient negative voltage between the anode and the cathode, photoelectrons can no longer reach the anode leading to zero photocurrent. This negative potential 𝑈𝑆 is known as the cutoff voltage of the photoelectric effect, as described by: !
!
𝑒𝑈! − ! 𝑚𝜈!
= 0 (2) Substitute (2) into (1) to obtain: !
!
𝑒𝑈! = ℎ𝜈 − 𝑊! or ℎ𝜈 = ! 𝑚𝜈!
+ 𝑊! (3) Equation (3) is called the Einstein's equation, which states that no photocurrent is given when the photoelectron energy ℎ𝜈 is less than the work function 𝑊𝑠, as electrons cannot escape from the metal surface under such condition. For a given metal material, the minimum frequency of the illuminating light to create a photoelectric effect is 𝜈0 = 𝑊𝑠/ℎ, called the cutoff frequency of the photoelectric effect. It should be noted that work function 𝑊𝑠 is the inherent property of a metal material, which is independent of the frequency of 1 Experimental apparatus and instructions come from Lambda Scientific: www.lambdasys.com Experiment 5 2 the incident light. Equation (3) can be rewritten as: !
𝑈! = ! 𝜈 −
!!
!
!
= ! (𝜈 − 𝜈! ) (4) Equation (4) shows that the cutoff voltage 𝑈𝑠 is a linear function of the frequency of incident light, ν. When 𝑈𝑠 = 0 when 𝜈 = 𝜈0, there is no photocurrent. The slope of the straight line as described by equation (4) is a constant: 𝑘 = ℎ/𝑒 thus, ℎ = 𝑒𝑘 (5) -­‐19 Where e is the electron charge (1.602×10
Coulombs). Therefore, the Planck’s constant can be calculated by measuring the cutoff voltage 𝑈𝑠 versus the frequency of illumination light, plotting the 𝑈𝑠~𝜈 curve, and acquiring the slope 𝑘. Apparatus Figure (1) shows the experimental schematic of the photoelectric effect for determining Planck’s constant using a photoelectric tube. When a light beam of frequency 𝜈 and power 𝑃 illuminates the cathode of a phototube, photoelectrons escape from the cathode. If a positive potential is applied to the anode relative to the cathode, photoelectrons will be accelerated; if a reverse potential is applied to the anode, photoelectrons will be decelerated. The photocurrent will decrease with an increase in the reverse potential, 𝑈𝐾𝐴. Finally, the photocurrent will be zero when 𝑈𝐾𝐴 = 𝑈𝑠. Figure (1) also shows the 𝐼 ~𝑉 characteristic curve of a photoelectric tube. By illuminating a phototube with different light frequencies ν, corresponding 𝐼 ~𝑉 curves of the phototube can be acquired, so that the corresponding cutoff voltages 𝑈𝑠 of the phototube can be 3 obtained. By plotting 𝑈𝑠~𝜈 curve, an approximately linear line should be obtained as predicted by Einstein’s photoelectric equation. Hence, the Planck’s constant h can be calculated from the slope k of the line using equation (5). In addition, the cutoff voltage 𝑈0 of the cathode material can be found from the intersection of the 𝐼~𝑉 curve with the horizontal axis of the plot. Thus, the cutoff frequency 𝜈0 can be achieved from 𝑈0, which equals the electron escape potential 𝛷! . Figure 1: Experimental Schematic and I~V Characteristic Curve of Phototube Experiment 5 3 Procedure
1) Turn on the mercury lamp and measurement unit and allow the mercury lamp to preheat for 20 minutes. During the preheating stage cover the entrance to the photoelectric tube box. During the preheating stage cover the output of the mercury lamp source. 2) Unplug the photocurrent input and zero the measurement by setting the “Current Range” to 10^-­‐13 and adjusting the “Current Zero” knob to achieve zero readout. Plug the photocurrent cable back in after zeroing. Do not touch the “Current Zero” knob through the remainder of the experiment. 3) While you are waiting for the lamp to heat up, measure the “dark current” of the photoelectric tube. a. Set the aperture on the photoelectric tube to “B” by rotating the wavelength selector. b. Set the “Current Range” switch to 10^-­‐13 and “Voltage Range” button to the “in” position (-­‐2 to +20 V) and set the “Voltage Adjust” knob fully counterclockwise. c. Slowly rotate the “Voltage Adjust” knob clockwise to increase the voltage. Record the dark current value under different voltage readings. 4) Make a plot of Dark Current vs. Voltage. Do you notice anything interesting about your graph? Why might it be important to measure the dark current? 5) After the 20 minute pre-­‐heating period we are able to collect data to determine ℎ/𝑒. a. Set the phototube approximately 30cm from the light source. b. Set the “Current Range” to 10^-­‐13 and “Voltage Range” to -­‐2 to +2 (the “out” position) and turn the voltage knob fully counter-­‐clockwise (the lowest voltage). c. Select the 365.0 nm filter and the aperture size to 8mm. d. Slowly increase the voltage until the current-­‐meter reads zero. Record this voltage and the corresponding filter wavelength. e. Repeat the previous steps for 404.7 nm, 435.8 nm, 546.1 nm, and 577.0 nm wavelengths, but keep the aperture the same size. 6) Make a plot of negative “stopping” voltage vs. frequency and perform a linear regression on your data, which will yet again yield h/e. Quantitatively assess your data. You may find linfit.nb useful to analyze your data. If your errors or discrepancies are high, think about possible sources of error (if necessary) and how to resolve them. Do you think the dark current has an effect on your measurements? What about the presence of an anode or cathode current? 7) Next, one I~V characteristic curve of the phototube will be measured. a. Make sure the phototube is placed 30cm from the light source, and set “Current Range” to 10^-­‐13 and “Voltage Range” to -­‐2 to +2 V. b. Select the 435.8 nm filter and set the aperture size to 8mm. c. Starting with the lowest voltage, increase the voltage slowly (allow the current to “relax” for a few seconds before recording data) while observing the change in photocurrent. Make a data table of current and voltage. Use a fairly small step size. (In total, gather roughly 20 data points across a wide range of currents from around -­‐40*10-­‐13 to 50*10-­‐13 A.) Experiment 5 4 8) Plot your I~V curve. You may find combining ListPlot[] and ListLinePlot[] and nice way to visualize your data e.g. Show[ListPlot[{data1}], ListLinePlot[{data1}]]. 9) You’ll notice that your graph will begin flat then rise very sharply. Describe why this behavior is observed, in particular what is going on in the phototube along various regimes on the I~V curve. What similarities and difference might occur for different wavelength I~V curves? Make a sketch (by hand is okay) of the other four curves for the other four wavelengths. 10) Determine the relationship between photocurrent and light energy by filling out the following table (note: keep 𝑈!" fixed at a reasonable value and take note of it): Aperture Size 2 mm 4 mm 8 mm I365.0nm I404.7nm I435.8nm I546.0nm I577.0nm 11) Choose a fixed aperture size, wavelength, and 𝑈!" value and vary the distance between the light source and the phototube, i.e. make a plot of photocurrent vs. distance and photocurrent vs. 1/distance2 to determine if a relationship exists.