A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ1 = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4 = 5461 Å, λ5 = 6907 Å,
The stopping voltages, respectively, were measured to be:
V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V
Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10–19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Given:
Sl | Wavelength (λ) | Stopping voltage (vs) |
1 | 3650 Å | 1.28 v |
2 | 4047 Å | 0.95 v |
3 | 4358 Å | 0.74 v |
4 | 5461 Å | 0.16 v |
5 | 6907 Å | 0 v |
The frequency, v of a wave with wavelength λ is given by,
v = c/λ …(1)
(c = speed of light = 3 × 108ms-1)
By using equation (1) we can find frequency for each case,
Sl | Wavelength (λ) | Stopping voltage (vs) | Frequency (v) |
1 | 3650 Å | 1.28 v | 8.2 × 1014Hz |
2 | 4047 Å | 0.95 v | 7.412 × 1014Hz |
3 | 4358 Å | 0.74 v | 6.88 × 1014Hz |
4 | 5461 Å | 0.16 v | 5.49 × 1014Hz |
5 | 6907 Å | 0 v | 4.3 × 1014Hz |
The equation for photo-electric effect is given as,
Φ0 = hv- eV0
We can rewrite this equation as,
V0 = 
…(2)
Where,
h = Planck’s constant
c = speed of light = 3 × 108m
λ = wavelength of light
e = charge on each electron = 1.6 × 10-19C
We find that the slope of the graph remains same,
Slope of the line = 
From equation (2),
We can write,
Slope = h/e
h = e × slope
h = 
h = 6.58 × 10-34 Js
From the same graph, we can infer that, threshold frequency of the metal is 5 × 1014 Hz
i.e. v0 = 5 × 1014Hz
Work function, Φ0 = hv0 …(2)
Where,
h = Plank’s constant = 6.57 × 10-34Js
v0 = threshold frequency
By plugging the data in equation (2), we get,
Φ0 = 6.57 × 10-34Js × 5 × 1014Hz
→ Φ0 = 3.28 × 10-19 J
→ 
→ Φ0 = 2eV
Hence, the work function of the metal is 2eV.