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:

λ_{1} = 3650 Å, λ_{2} = 4047 Å, λ_{3} = 4358 Å, λ_{4} = 5461 Å, λ5 = 6907 Å,

The stopping voltages, respectively, were measured to be:

V_{01} = 1.28 V, V_{02} = 0.95 V, V_{03} = 0.74 V, V_{04} = 0.16 V, V_{05} = 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 × 10^{8}ms^{-1})

By using equation (1) we can find frequency for each case,

The equation for photo-electric effect is given as,

Φ_{0} = hv- eV_{0}

We can rewrite this equation as,

V_{0} = …(2)

Where,

h = Planck’s constant

c = speed of light = 3 × 10^{8}m

λ = wavelength of light

e = charge on each electron = 1.6 × 10^{-19}C

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 × 10^{14} Hz

i.e. v_{0} = 5 × 10^{14}Hz

Work function, Φ_{0} = hv_{0} …(2)

Where,

h = Plank’s constant = 6.57 × 10^{-34}Js

v_{0} = threshold frequency

By plugging the data in equation (2), we get,

Φ_{0} = 6.57 × 10^{-34}Js × 5 × 10^{14}Hz

→ Φ_{0} = 3.28 × 10^{-19} J

→

→ Φ_{0} = 2eV

Hence, the work function of the metal is 2eV.

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