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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 28 of 37
Marks: +1, -0
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\lambda_1 = 3650 Å, λ2\lambda_2 = 4047 Å, λ3\lambda_3 = 4358 Å,
λ4\lambda_4 = 5461 Å, λ5\lambda_5 = 6907 Å,
The stopping voltages, respectively, were measured to be:
V01V_{01} = 1.28 V, V02V_{02} = 0.95 V, V03V_{03} = 0.74 V,
V04V_{04} = 0.16 V, V05V_{05} = 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 V0V_0
3650 Å 8.2 × 101410^{14} Hz1.28 V
4047 Å 7.4 × 101410^{14} Hz 0.95 V
4358 Å 6.9 × 101410^{14} Hz 0.74 V
5461 Å 5.49 × 101410^{14} Hz 0.16 V
6907 Å4.3 × 101410^{14} 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\upsilon_0 = 5.0 × 101410^{14} Hz.
The fifth point υ (= 4.3 × 101410^{14} Hz) corresponds to υ < υ0\upsilon_0, so there is no photoelectric emission and no stopping voltage is required to stop the current.
Slope of , V0V_0 versus υ graph is tan θ = ΔV0Δυ\frac{\Delta V_0}{\Delta \upsilon} = (1.28−0)V(8.2−5.0)×1014 s−1\frac{(1.28-0)V}{(8.2-5.0)\times 10^{14}\,\text{s}^{-1}} = he\frac{h}{e}
= 4.0 × 10−1510^{-15} V s = he\frac{h}{e}
Planck’s constant,
h = e × 4.0 × 10−1510^{-15} J s = 1.6 × 10−1910^{-19} × 4.0 × 10−1510^{-15} J s = 6.4 × 10−3410^{-34} J s.
(b) Threshold frequency, υ0\upsilon_0 = 5.0 × 101410^{14} Hz
∴ Work function,
W0W_0 = hυ0h\upsilon_0 = 6.4 × 10−3410^{-34} × 5.0 × 101410^{14} J = 6.4×5.0×10−201.×10−19\frac{6.4 \times 5.0 \times 10^{-20}}{1. \times 10^{-19}} eV = 2.00 eV
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