Name ______Pd_____
AP Physics B - Photoelectric Effect
Determination of Planck’s Constant, Work Function, & Threshold Frequency
Purpose:
To determinePlanck’s constanth, Work Functionandthreshold frequencyusing the measured stopping potential Vo of photoelectrons and incident light of known wavelength .
Discussion:
The photoelectric effect was explained by Einstein as being due to quantized light energy (photons) either being enough to excite electrons so that they can both break their bonds with their metal atoms and then move across a vacuum to complete a circuit or being not enough to do so. He expressed the energy of each incident light photon as:
E = hor, since c= E = hc/
The photoelectric effect can be expressed in terms of the incoming energy hc/being equal to the energy for the electron to break its bond with its metal atom plus havingenough kinetic energy left over to jump the vacuum gap qVo . Put all together,
hc/qVo
Rearranging this slightly, we get a linear function whose y-intercept is -and whose slope is h:
qVoh(c/
(corresponds to y = mx – b)
The slope h of this function can be found from a graph of qVo(y-axis) versus (c/(x-axis). Using several known wavelengths and their corresponding measured stopping potentials Vo , a line of best fit should yield Planck’s constant.
The accepted value for Planck’s constant is 6.626 x 10-34 Js. See how close you can get!
Other accepted values are: c = 2.998x108 m/s and qelectron = 1.602 x 10-19 C
Materials:
Photocurrent detector with amplifier, voltmeter, high intensity fluorescent or tungsten lamp with known wavelength filters or monochromatic sources of known wavelengths. (Wavelength can be determined with a spectroscope as well)
Precautions:
Be careful not to look directly at the filament of the high intensity light source, and be sure flammable materials are not near the lamp, as it may get very hot.
Procedure:
Attach the photodetector to a voltmeter. Zero the photodetector with the aperture shielded by a metal plate while the voltage on the detector is set to zero (dial completely counter-clockwise). Place a filter of known wavelength over the aperture and turn on the light source. Adjust the distance of the light source so that the current on the photodetector reads 10 nA. Gradually increase the voltage on the photodetector until the 10nA current goes to zero. Take the voltage reading from the voltmeter at that point. This reading is the stopping potential Vo. Repeat the procedure with the other known wavelength filters. Plot the stopping potentials times the charge on an electron qVo(y-axis) versus (c/known(x-axis). Find the slope of your graph. Determine the percent error of your graph slope.
Data
known (m) / Vo(J/C) / qVo (J) / c/known (Hz) / Slope (Js) / % errorSample Calculations:
Quantity / Substitution / Answer with UnitsqVo
c/known
Slope
% error
Graph (attach to lab write up)
You must plot your graph on graph paper for it to be valid.
Do not truncate your graph axes (both x and y axis should start at zero)
Draw the line of best fit through your plotted points.
Circle the points on your line of best fit that you used to find your graph slope
Questions:
- What did the slope of your graph represent? ______
- Try a different line of best fit and compare that slope to the one you found originally. How much error does this one change generate? SHOW your new line of best fit on your graph as a dotted line and SHOW your calculation for your NEW SLOPE and the %DIFFERENCE below:
- What does the y-intercept of your graph represent?______
- Calculate the y-intercept for your graph, showing all work in the space below.
- What does the x-intercept of your graph represent?______
- Calculate the x-intercept for your graph, showing all work in the space below.
- Name one source of error other than the slope angle variation you already discussed that may account for the percent error you got for your slope versus the accepted value, and thoroughly describe and demonstrate (using calculations) the effect of a reasonable amount of change in that source of error. Show clearly how much change there would be in your percent error as a result of the change in your source of error.