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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
Questions With Solutions

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Question : 28
Total: 37
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.
Solution:  
In order to calculate Planck’s constant ‘h’ we need slope of the graph between cut off voltage and frequency.
So, let us first calculate the frequency (υ = c/λ) in each case and the table shows corresponding stopping potential.
λ υ V0
3650 Å 8.2 × 1014 Hz1.28 V
4047 Å 7.4 × 1014 Hz 0.95 V
4358 Å 6.9 × 1014 Hz 0.74 V
5461 Å 5.49 × 1014 Hz 0.16 V
6907 Å4.3 × 1014 Hz 0.0 V

V0 versus υ plot shows that the first four points lie nearly on a straight line which intercepts the x-axis at threshold frequency, υ0 = 5.0 × 1014 Hz.
The fifth point υ (= 4.3 × 1014 Hz) corresponds to υ < υ0, so there is no photoelectric emission and no stopping voltage is required to stop the current.
Slope of , V0 versus υ graph is tan θ =
ΔV0
Δυ
 =
(1.28−0)V
(8.2−5.0)×1014s−1
 =
h
e
= 4.0 × 10−15 V s =
h
e
Planck’s constant,
h = e × 4.0 × 10–15 J s = 1.6 × 10–19 × 4.0 × 10–15 J s = 6.4 × 10–34 J s.
(b) Threshold frequency, υ0 = 5.0 × 1014 Hz
∴ Work function,
W0 = hυ0 = 6.4 × 10–34 × 5.0 × 1014 J =
6.4×5.0×10−20
1.×10−19
 eV = 2.00 eV
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