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Question

The work function of a metal is 4 eV. To emit a photoelectron of zero velocity from the surface of the metal, the wavelength of the incident light should be:
  1. 3105 A
  2. 2500 A
  3. 5600 A
  4. 1600 A


Solution

The correct option is A 3105 A
W=4 eV
 hν=hν0+12mv2
Given, v=0
So, hν=hν0
hcλ=(4 eV)(1.6×1019) J/eV
λ=(6.626×1034 Js)(3×108 m/s)(4×1.6×1019) J
λ=3105×1010 m
λ=3105 A

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