Photoelectric Effect with an Unshielded LED
In a photoelectric-effect experiment, a student illuminates a photocathode with four LEDs and measures the stopping potential with a variable DC supply and a sensitive galvanometer. The LED wavelengths and measured stopping potentials are: 470 nm, 0.64 V; 505 nm, 0.60 V; 590 nm, 0.10 V; 630 nm, no reproducible stopping potential. The accepted value of Planck’s constant is 6.63×10−34 J·s, and the speed of light is 3.00×10^8 m/s. The student plots stopping potential versus 1/λ and obtains a slope that gives h = 6.1×10−34 J·s, about 8% low. The 505 nm LED is noticeably brighter than the others and is operated without a shield; room lights are on. The voltmeter has 0.01 V resolution, and the LED wavelength tolerance is ±10 nm. The student wants to know whether the result can be improved by using the 505 nm point or by excluding it.
1. Using the 470 nm and 590 nm data, calculate Planck’s constant and the work function. State whether these two points are consistent with the accepted h.
Read the worked answer
For the photoelectric equation,
(eV_s=\frac{hc}{\lambda}-\phi), so (V_s=\frac{hc}{e}\frac{1}{\lambda}-\frac{\phi}{e}).
The slope of (V_s) versus (1/\lambda) is (hc/e). Using the two points:
(1/\lambda_{470}=2.128\times10^6\ \mathrm{m^{-1}}), (1/\lambda_{590}=1.695\times10^6\ \mathrm{m^{-1}}).
(\Delta(1/\lambda)=4.33\times10^5\ \mathrm{m^{-1}}), (\Delta V_s=0.54\ \mathrm{V}).
Slope (=0.54/(4.33\times10^5)=1.25\times10^{-6}\ \mathrm{V,m}). Thus
(h=\frac{(1.25\times10^{-6})(1.602\times10^{-19})}{3.00\times10^8}\approx6.66\times10^{-34}\ \mathrm{J,s}),
within about 0.5% of the accepted value. The intercept gives (\phi/e\approx2.01\ \mathrm{V}), so (\phi\approx2.0\ \mathrm{eV}) (about (3.2\times10^{-19}\ \mathrm{J})).
2. The 505 nm point gives V_s = 0.60 V. Using the h and work function from the first task, calculate the predicted stopping potential and decide whether the point should be excluded. Propose a specific experimental cause and a check.
Read the worked answer
At 505 nm, the photon energy is
(E=\frac{1240\ \mathrm{eV,nm}}{505\ \mathrm{nm}}\approx2.46\ \mathrm{eV}).
With (\phi\approx2.01\ \mathrm{eV}), the predicted stopping potential is
(V_s=2.46-2.01=0.45\ \mathrm{V}).
The measured 0.60 V is 0.15 V higher, fifteen times the 0.01 V voltmeter resolution. A ±10 nm wavelength error changes the photon energy by only about 0.05 eV, so it cannot explain the discrepancy. The 505 nm point should be excluded unless the cause is found. The unshielded, brighter LED and room lights can admit shorter-wavelength stray light, producing photoelectrons that require a larger stopping potential. A check is to repeat the 505 nm measurement with the room lights off, the LED shielded, and the intensity varied; the stopping potential should not change with intensity for a clean photoelectric signal.