NEET | 2016Wave motionHard

Question

When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the samesurface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. The threshold wavelengthfor the metallic surface is:

Options

A.

4λ

B.

5λ

C.

52λ

D.

3/2

Solution

In photo electric effects
eV0 = 48 - W
eV0 = hcλ-w
eV = hcλ-w  .... (i)
ev4=hc2λ-w    ....….(ii)
From (i) and (ii)
hcλ-w=4hc2λ-w
hcλ-w=2hcλ-4w
3W =hcλ⇒W=hc3λ
hcλmax=hc3λ⇒λmax= threshold wavelength 3λ

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