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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 26
Total: 37
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–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 =
hc
λ
 =
.63×10−34×3×108
2271×10−10×1.6×10−19
 = 5.47 eV
Maximum kinetic energy of emitted electron can be judged by stopping potential of 1.3 volt.
1
2
mvmax2 = 1.3 eV
Using Einstein’s equation hu =W0+
1
2
mvmax2
5.47 eV = W0 + 1.3 eV
W0 = 4.17 eV Red light of wavelength 6328 Å will have energy of each photon
E =
hc
λ
 =
6.63×10−34×3×108
6328×10−10×1.6×10−19
 = 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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