Wave motionHard
Question
When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5 V0. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V0. Then the work function of the metallic surface is :
Options
A.
B.
C.
D.
Solution
= 5 eV0 + φ
= eV0 + φ ⇒ 
= 4eV0 ⇒ φ = 
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