Photoelectric Effect: A Straight-Line Graph Through an Impossible Point
A student uses a photocell with a potassium cathode (work function 2.30 eV) and a set of interference filters to test the photoelectric equation. Light from a mercury lamp passes through filters of wavelength 365, 405, 436 and 546 nm. For each wavelength the stopping potential is measured with a digital voltmeter: 1.10, 0.76, 0.54 and 0.13 V respectively. The 546 nm filter transmits a small amount of violet light because of a manufacturing defect. The voltmeter has a resistance of 10 MΩ, and the photocell dark current is about 0.02 nA. The student plots stopping potential against 1/λ and obtains a straight line through all four points. The accepted value of Planck's constant is 6.63 × 10⁻³⁴ J s, and 1 eV = 1.60 × 10⁻¹⁹ J. The cathode and anode are in an evacuated tube; the anode is a ring so that light reaches the cathode.
1. Using the 365 nm and 436 nm data, calculate Planck's constant and the work function. Show the unit conversions clearly and state whether the values are consistent with the accepted values.
Read the worked answer
Use the photoelectric equation in the form
[ eV_s=\frac{hc}{\lambda}-W ]
It is convenient to use (hc\approx1240\ \text{eV nm}).
At 365 nm:
[ E_{365}=\frac{1240}{365}=3.40\ \text{eV} ]
At 436 nm:
[ E_{436}=\frac{1240}{436}=2.84\ \text{eV} ]
Difference in photon energy:
[ \Delta E=3.40-2.84=0.56\ \text{eV} ]
This equals (e\Delta V_s) with (\Delta V_s=1.10-0.54=0.56\ \text{V}), so the numbers are internally consistent.
Planck's constant:
[ h=\frac{e\Delta V_s}{c\Delta(1/\lambda)} ]
[ \Delta\left(\frac{1}{\lambda}\right)=\frac{1}{365\times10^{-9}}-\frac{1}{436\times10^{-9}} =4.46\times10^{5}\ \text{m}^{-1} ]
[ h=\frac{(1.60\times10^{-19})(0.56)}{(3.00\times10^8)(4.46\times10^5)} \approx6.7\times10^{-34}\ \text{J s} ]
Work function from the 365 nm point:
[ W=3.40-1.10=2.30\ \text{eV} ]
From the 436 nm point:
[ W=2.84-0.54=2.30\ \text{eV} ]
The work function agrees well with the stated 2.30 eV. Planck's constant is about 1% above 6.63 × 10⁻³⁴ J s, which is acceptable given the rounded data.
2. Explain why the 546 nm point must not be treated as a valid stopping potential. What would including it do to the graph and to the deduced value of Planck's constant?
Read the worked answer
Photon energy at 546 nm:
[ E_{546}=\frac{1240}{546}=2.27\ \text{eV} ]
This is less than the work function of 2.30 eV. A single 546 nm photon therefore cannot eject an electron from potassium, no matter what voltage is applied. The true stopping potential for that wavelength is not 0.13 V; ideally there would be no photocurrent at all.
The 0.13 V reading is almost certainly an artefact: the defective filter transmits some violet light, and the dark current or amplifier offset can also produce a small apparent voltage. It is not a photoelectric stopping potential.
On the graph of (V_s) against (1/\lambda), this point lies above the line predicted by the other three points. Including it would pull the fitted line upwards at small (1/\lambda). That would reduce the slope and increase the intercept, giving a lower apparent Planck constant and a larger apparent work function. The point must be rejected, and the reason stated.
3. Suggest two improvements that would reduce systematic error in this experiment, other than simply using more wavelengths.
Read the worked answer
Any two of the following, with explanation:
- Use a monochromator instead of interference filters. This avoids stray light of shorter wavelength, which is especially important near the threshold where a small amount of violet light can produce a spurious photocurrent.
- Measure and subtract the dark current. Record the current–voltage characteristic with the light blocked, then subtract that background from each measurement. This removes offset and leakage effects.
- Use a high-impedance buffer or electrometer. The 10 MΩ voltmeter can draw current and distort the stopping-potential measurement; a much higher input resistance reduces this loading error.
- Use a guard-ring or non-photoemissive anode. The anode should not emit electrons when light reaches it; a ring-shaped anode reduces this, and a guard ring can reduce leakage currents.
A good answer should link the improvement to a specific systematic error, not just say 'be more careful'.