Q. 284.3( 3 Votes )

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 :

Given:



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,



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.


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