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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 26 of 37
Marks: +1, -0
Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~ 105 W m−2m^{-2}) red light of wavelength 6328 Å produced by a He-Ne laser?
Solution:  
Let us find energy of each photon of given ultraviolet light
E = hcλ\frac{hc}{\lambda} = .63×10−34×3×1082271×10−10×1.6×10−19\frac{.63 \times 10^{-34} \times 3 \times 10^8}{2271 \times 10^{-10} \times 1.6 \times 10^{-19}} = 5.47 eV
Maximum kinetic energy of emitted electron can be judged by stopping potential of 1.3 volt.
12mvmax2\frac{1}{2} m v^2_{max} = 1.3 eV
Using Einstein’s equation hu =W0+12mvmax2W_0 + \frac{1}{2} m v_{max}^2
5.47 eV = W0W_0 + 1.3 eV
W0W_0 = 4.17 eV Red light of wavelength 6328 Å will have energy of each photon
E = hcλ\frac{hc}{\lambda} = 6.63×10−34×3×1086328×10−10×1.6×10−19\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6328 \times 10^{-10} \times 1.6 \times 10^{-19}} = 1.96 eV
Thus energy of red light photons is less than work function 4.17 eV, hence irrespective of any intensity, no emission will take place.
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