When radiation of wavelength A is used to illuminate a metallic surface, the stopping potential is V . When…

When radiation of wavelength A is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3A, the stopping potential is V4. If the threshold wavelength for the metallic surface is  then value of n will be :

Solution

hcλ=ϕ+eV ...........(1)

hc3λ=ϕ+eV4      ---------(2)

from (1) & (2)

hcλ1-13=34ev

hcλ23=34eV

eV=89hcλ

hcλ=ϕ+89hcλ

ϕ=hc9λ=hcλth

λth=9λ

k=9

Asked in: JEE Main 2020 (02 Sep Shift 1)

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