Photoelectric Effect With Two Wavelengths
A photoelectric tube is illuminated by monochromatic light. With 400 nm light, the stopping potential is 0.62 V. With 350 nm light, the stopping potential is 1.06 V. When the intensity of the 400 nm light is doubled, the stopping potential remains 0.62 V but the photocurrent doubles. A student claims that stopping potential should increase with intensity because brighter light carries more energy, and predicts that 500 nm light will produce a stopping potential of about 0.31 V, exactly half the 400 nm value. The tube’s work function is not given. The voltage supply is accurate to ±0.02 V, and the wavelength calibration is accurate to ±2 nm. The student must determine the work function and threshold wavelength and evaluate the prediction for 500 nm.
1. Determine the work function and threshold wavelength from the two stopping-potential measurements. Show that the two wavelengths give consistent results.
Read the worked answer
Use the photoelectric equation in the form
eV_stop = hc/λ − φ.
A convenient value is hc = 1240 eV·nm.
At 400 nm, the photon energy is
E = 1240/400 = 3.10 eV.
So
φ = 3.10 − 0.62 = 2.48 eV.
At 350 nm,
E = 1240/350 = 3.54 eV,
so
φ = 3.54 − 1.06 = 2.48 eV.
The two wavelengths give the same work function within the stated voltage accuracy. The threshold wavelength is found from
φ = hc/λ_threshold,
so
λ_threshold = 1240/2.48 = 500 nm.
Light of wavelength longer than 500 nm cannot eject electrons from this surface.
2. Evaluate the student’s intensity claim and the prediction for 500 nm. Explain what doubling the intensity does change.
Read the worked answer
The intensity claim is incorrect. The stopping potential measures the maximum kinetic energy of the emitted electrons. That energy depends on the photon energy hc/λ and on the work function, not on the number of photons per second. Doubling the intensity at 400 nm doubles the number of incident photons and therefore doubles the saturation photocurrent, as observed, but each photon still has the same energy. The maximum kinetic energy and the stopping potential therefore remain 0.62 V.
The prediction for 500 nm is also wrong. At 500 nm,
E = 1240/500 = 2.48 eV,
which equals the work function. Thus K_max = 0 and V_stop = 0 V. Emission just begins at 500 nm; it does not give half the 400 nm stopping potential. At wavelengths longer than 500 nm, the photon energy is below the work function and no electrons are emitted, regardless of intensity.
3. Predict the stopping potential for 450 nm light and whether photocurrent will flow. Justify your answer with a calculation.
Read the worked answer
At 450 nm, the photon energy is
E = 1240/450 = 2.756 eV.
The maximum kinetic energy is
K_max = E − φ = 2.756 − 2.48 = 0.276 eV.
Therefore the stopping potential is
V_stop = K_max/e = 0.276 V,
or about 0.28 V.
Because 450 nm is shorter than the 500 nm threshold, the photon energy exceeds the work function and electrons can be emitted. Photocurrent will flow if the light intensity is nonzero and the emitted electrons are collected. Increasing the intensity at 450 nm would increase the current but would not change the 0.28 V stopping potential.