Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.26
NCERT Solutions for Class 12 Physics Chapter 11 Question-26
Additional Exercise
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?
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Wavelength of ultraviolet light, λ = 2271 Aº = 2271 x 10⁻10 m
Stopping potential of the metal, Vo = 1.3 V
Planck’s constant, h = 6.6 x 10⁻34 J
Charge on an electron, e = 1.6 x 10-19 C
Work function of the metal = φO
Frequency of light = v
We have the photo-energy relation from the photoelectric effect as:
φO= hv – eVo
= (6.6 x 10⁻34) x ( 3 x 10⁸ ) /(2271 x 10⁻10) – ( 1.6 x 10-19) x 1.3
= 8.72 x 10-19 – 2.08 x 10-19
= 6.64 x 10-19 J
= (6.64 x 10-19 ) /(1.6 x 10-19) = 4.15 eV
Let vo be the threshold frequency of the metal .Therefore , φO=hv0
=>v0 = φO/h = (6.64 x 10-19)/(6.6 x 10⁻34) = 1.006 x 10¹⁵ Hz
Wavelength of red light,
λr = 6328 Aº = 6328 x 10⁻10
Therefore ,the frequency of red light ,
vr = c/λr = (3 x 10⁸) /(6328 x 10⁻10) = 4.74 x 10¹4 Hz
Since vo> vr, the photocell will not respond to the red light produced by the laser.