Question

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.]

Answer 1

Einstein’s photoelectric equation is given as:

eV₀ = hv − `phi_0`

`"V"_0 = "h"/"e" "v" - phi_0/"e"` .............(1)

Where,

V₀ = Stopping potential

h = Planck’s constant

e = Charge on an electron

v = Frequency of radiation

`phi_0` = Work function of a material

It can be concluded from equation (1) that potential V₀ is directly proportional to frequency v.

Frequency is also given by the relation:

`"v" = "Speed of light (c)"/"Wavelenght (λ)"`

This relation can be used to obtain the frequencies of the various lines of the given wavelengths.

`"v"_1 = "c"/lambda_1 = (3 xx 10^8)/(3650 xx 10^(-10)) = 8.219 xx 10^14  "Hz"`

`"v"_2 = "c"/lambda_2 = (3 xx10^8)/(4047 xx 10^(-10)) = 7.412 xx 10^14  "Hz"`

`"v"_3 = "c"/lambda_3 = (3 xx 10^8)/(4358 xx 10^(-10)) = 6.884 xx 10^14  "Hz"`

`"v"_4 = "c"/lambda_4 = (3 xx 10^8)/(5461 xx 10^(-10)) = 5.493 xx 10^14  "Hz"`

`"v"_5 = "c"/lambda_5  = (3xx10^8)/(6907 xx 10^(-10)) = 4.343 xx 10^14  "Hz"`

The given quantities can be listed in tabular form as:

Frequency × 1014 Hz	8.219	7.412	6.884	5.493	4.343
Stopping potential V0	1.28	0.95	0.74	0.16	0

The following figure shows a graph between νand V_0.

It can be observed that the obtained curve is a straight line. It intersects the ν-axis at 5 × 10¹⁴ Hz, which is the threshold frequency (v₀) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the λ₅ line, and therefore, no stopping voltage is required to stop the current.

 Slope of the straight line = `"AB"/"CB" = (1.28 - 0.16)/((8.214 - 5.493) xx 10^14)`

From equation (1), the slope `"h"/"e"` can be written as:

`"h"/"e" = (1.28 - 0.16)/((8.214 - 5.493) xx 10^14)`

∴ `"h" = (1.12 xx 1.6 xx 10^(-19))/(2.726 xx 10^(14))`

= 6.573 × 10⁻³⁴ Js

The work function of the metal is given as:

`phi_0` = hv₀

= 6.573 × 10⁻³⁴ × 5 × 10¹⁴

= 3.286 × 10⁻¹⁹ J

= `(3.286 xx 10^(-19))/(1.6 xx 1^(-18))`

= 2.054 eV