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

It is given that the wavelengths, λ 1 =3650  A o , λ 2 =4047  A o , λ 3 =4358  A o , λ 4 =5461  A o

λ 5 =6907  A o , the stopping voltages corresponding to wavelengths, V 01 =1⋅28 V,

V 02 =0⋅95 V, V 03 =0⋅74 V, V 04 =0⋅16 V, V 05 =0 V

The formula of photo energy relation can be written as,

ϕ=hυ−e V 0 V 0 = h e υ− ϕ e (1)

The formula of frequency is,

υ= c λ

Calculate the values of frequencies for given wavelengths respectively.

υ 1 = c λ 1 = 3× 10 8 3650× 10 −10 =8⋅219× 10 14  Hz

υ 2 = c λ 2 = 3× 10 8 4047× 10 −10 =7⋅412× 10 14  Hz

υ 3 = c λ 3 = 3× 10 8 4358× 10 −10 =6⋅884× 10 14  Hz

υ 4 = c λ 4 = 3× 10 8 5461× 10 −10 =5⋅493× 10 14  Hz

υ 5 = c λ 5 = 3× 10 8 6907× 10 −10 =4⋅343× 10 14  Hz

Draw the table as given below.

Frequency × 10 14  Hz	8⋅219	7⋅412	6⋅884	5⋅493	4⋅343
Stopping potential V 0	1⋅28	0⋅95	0⋅74	0⋅16	0

Draw a graph between stopping potential and frequency as per given values.

The graph intersects the frequency axis at 5× 10 14  Hz, so this frequency will be the threshold frequency.

From the graph, the slope can be calculated as,

AB CB = 1⋅28−0⋅16 ( 8⋅214−5⋅493 )× 10 14

This slope will be equal to slope of equation (1) as h e .

h e = 1⋅28−0⋅16 ( 8⋅214−5⋅493 )× 10 14

Substitute the value of e in the above equation.

h= 1⋅12×1⋅6× 10 −19 ( 8⋅214−5⋅493 )× 10 14 h=6⋅6× 10 −34  Js

The formula of work function is,

ϕ=h υ 0

Substitute the values in above expression.

ϕ=6⋅6× 10 −34 ×5× 10 14 =3⋅286× 10 −19  J = 3⋅286× 10 −19 1⋅6× 10 −19 =2⋅1 eV

Hence, the value of Planck’s constant is 6⋅6× 10 −34  Js, the threshold frequency is

5× 10 14  Hz and the work function is 2⋅1 eV.