Photoelectric Effect with an Unknown Metal
A student shines monochromatic light on a metal cathode in a vacuum tube and measures the stopping voltage needed to reduce the photocurrent to zero. With light of wavelength 400 nm the stopping voltage is 0.90 V. The student then switches to a source of wavelength 250 nm and finds a stopping voltage of 2.80 V. The student suspects the apparatus is miscalibrated because the numbers do not match the tabulated work function of the metal they think they are using. A classmate points out that the two measurements can be checked against each other without knowing the metal's identity. Planck's constant is 6.63 x 10^-34 J·s, the speed of light is 3.00 x 10^8 m/s, and the elementary charge is 1.60 x 10^-19 C.
1. Use the two data points to determine whether they are mutually consistent, and estimate the work function of the cathode surface.
Read the worked answer
Use the photoelectric equation in the form eVs = hc/λ − W, where Vs is the stopping voltage.
For λ = 400 nm: energy of photon = hc/λ = (6.63e-34)(3.00e8)/(4.00e-7) = 4.97e-19 J = 3.11 eV. Then W = 3.11 − 0.90 = 2.21 eV.
For λ = 250 nm: photon energy = (6.63e-34)(3.00e8)/(2.50e-7) = 7.96e-19 J = 4.98 eV. Then W = 4.98 − 2.80 = 2.18 eV.
The two values agree to within about 1 percent (2.21 eV vs 2.18 eV), which is within the rounding of the stated constants and voltages. The data are therefore mutually consistent, and the work function is about 2.2 eV. The student's mismatch with the tabulated metal is likely a wrong identification of the cathode material, not a calibration failure, because an internal inconsistency would show up as two different work functions.
2. The student wants to observe a photocurrent using light of wavelength 600 nm. Predict what happens and justify your answer quantitatively.
Read the worked answer
Photon energy at 600 nm = hc/λ = (6.63e-34)(3.00e8)/(6.00e-7) = 3.32e-19 J = 2.07 eV.
The work function is about 2.2 eV, so 2.07 eV is below the threshold. No electrons can be emitted regardless of intensity, because a single photon carries less energy than the escape energy; increasing intensity only increases the number of sub-threshold photons.
The threshold wavelength is λmax = hc/W = (6.63e-34)(3.00e8)/(2.2 × 1.60e-19) ≈ 5.6e-7 m, or about 560 nm. Since 600 nm exceeds 560 nm, the prediction is zero photocurrent. Note the reasoning depends on the work function derived from the data, not on recall of a table.