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:
$\lambda_1=3650\ \mathring{\text{A}},\ \lambda_2=4047\ \mathring{\text{A}},\ \lambda_3=4358\ \mathring{\text{A}},\ \lambda_4=5461\ \mathring{\text{A}},\ \lambda_5=6907\ \mathring{\text{A}},$
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

Einstein's photoelectric equation is given as:

$\text{eV}_\circ=\text{hv}-\phi_\circ$

$\text{V}_\circ=\frac{\text{h}}{\text{e}}\text{v}-\frac{\phi_\circ}{\text{e}}....(1)$

Where,

$\text{V}_\circ$ = Stopping potential

h = Planck's constant

e = Charge on an electron

$\phi_\circ$ = Work function of a material

It can be concluded from equation (1) that potential $\text{V}_\circ$ is directly proportional to frequency v.

Frequency is also given by the relation:

$\text{v}=\frac{\text{Speed of light (c)}}{\text{Wavelength}\ (\lambda)}$

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

$\text{v}_1=\frac{\text{c}}{\lambda_1}=\frac{3\times10^8}{3650\times10^{-10}}=8.219\times10^{14}\ \text{Hz}$

$\text{v}_2=\frac{\text{c}}{\lambda_2}=\frac{3\times10^8}{4047\times10^{-10}}=7.412\times10^{14}\ \text{Hz}$

$\text{v}_3=\frac{\text{c}}{\lambda_3}=\frac{3\times10^8}{4358\times10^{-10}}=6.884\times10^{14}\ \text{Hz}$

$\text{v}_4=\frac{\text{c}}{\lambda_4}=\frac{3\times10^8}{5461\times10^{-10}}=5.493\times10^{14}\ \text{Hz}$

$\text{v}_5=\frac{\text{c}}{\lambda_5}=\frac{3\times10^8}{6907\times10^{-10}}=4.343\times10^{14}\ \text{Hz}$

The given quantities can be listed in tabular form as:

Frequency × 1014 Hz8.2197.4126.8845.4934.343
Stopping potential $\text{V}_\circ$1.280.950.740.160

The following figure shows a graph between vand $\text{V}_\circ.$

It can be observed that the obtained curve is a straight line. It intersects the v-axis at 5 × 1014 Hz, which is the threshold frequency $(\text{V}_\circ)$ of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the $\lambda_{5}$ line, and therefore, no stopping voltage is required to stop the current.

Slope of the straight line $=\frac{\text{AB}}{\text{CB}}=\frac{1.28-0.16}{(8.214-5.493)\times10^{14}}$

From equation (1), the slope $\frac{\text{h}}{\text{e}}$ can be written as:

$\frac{\text{h}}{\text{e}}=\frac{1.28-0.16}{(8.214-5.493)\times10^{14}}$

$\therefore\ \text{h}=\frac{1.12\times1.6\times10^{19}}{2.726\times10^{14}}$

= 6.573 × 10-34 Js

The work function of the metal is given as:

$\phi_\circ=\text{hv}_\circ$

= 6.573 × 10-34 × 5 × 1014

= 3.286 × 10-19 J

$=\frac{3.286\times10^{-19}}{1.6\times10^{-18}} $

= 0.054 eV

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