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Wave Optics

Question
CBSEENPH12038474

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 volt, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~ 105Wm–2) red light of wavelength 6328 Å produced by He-Ne laser?

Solution

Here,
Sopping potential,  V0 = 1.3 V 
Wavelength of ultraviolet light, λ = 2271  Å                                                            = 2271 × 10-10m

Now, using Einstein's photoelectric equation, 

        hv = hv0+12mv2max = ϕ0+eV0  

   ϕ0 = hcλ-eV0 
where, ϕo is the work function of the metal. 

Taking,      h = 6.62 × 10-34 Js
                 e = 1.6 × 10-19C
                 c = 3 × 108 ms-1 
We have,

ϕ0 = 6.62 × 10-34× 3 × 1082271 × 10-10-1.6 × 10-19 × 1.3

    ϕ0 = 8.745 × 10-19-2.08 × 10-19     = 6.665 × 10-19J   

i.e.,  ϕ0 = 6.665 × 10-191.6 × 10-19 = 4.166 eV. 

Threshold wavelength is given by, 

      λ0 = hcϕ0       = 6.62 × 10-34 × 3 × 1086.665 × 10-19      = 2.98 × 10-7m 

i.e., λ0 = 2980 Å 

As calculated above, wavelength of red light is greater than threshold wavelength λ0. Therefore, photocell will not respond when red light of wavelength 6328 A produced by He-Ne laser is incident on the photocell.

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