When radiation of wavelength λ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3λ, the stopping potential is V/4. If the threshold wavelength for the metallic surface is nλ then value of n will be:
When radiation of wavelength λ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3λ, the stopping potential is V/4. If the threshold wavelength for the metallic surface is nλ then value of n will be:
Asked by Shiksha User
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1 Answer
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hc/λ = Φ + eV
hc/3λ = Φ + eV/4
from (1) and (2)
hc/λ (1-1/3) = 3/4 eV
eV = 8/9 hc/λ
eV = 8/9 (hc/λ - Φ)
Φ = hc/9λ
λ? = 9λ ∴k=9
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