Dual Nature of Matter

JEE Main level, numerical answer, medium

167 of 200

Question 167 of 200

When radiation of wavelength AA is used to illuminate a metallic surface, the stopping potential is VV. When the same surface is illuminated with radiation of wavelength 3A3A, the stopping potential is V4\dfrac{V}{4}. If the threshold wavelength for the metallic surface is nλn\lambda then value of nn will be :

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Answer: 9

hcλ=ϕ+eV\dfrac{hc}{\lambda }=\phi +eV ...........(1)
hc3λ=ϕ+eV4\dfrac{hc}{3\lambda }=\phi +\dfrac{eV}{4} ---------(2)
from (1) & (2)
hcλ(1−13)=34ev\dfrac{hc}{\lambda }\left(1-\dfrac{1}{3}\right)=\dfrac{3}{4}ev
hcλ23=34eV\dfrac{hc}{\lambda }\dfrac{2}{3}=\dfrac{3}{4}eV
eV=89hcλeV=\dfrac{8}{9}\dfrac{hc}{\lambda }
hcλ=ϕ+89hcλ\dfrac{hc}{\lambda }=\phi +\dfrac{8}{9}\dfrac{hc}{\lambda }
ϕ=hc9λ=hcλth\phi =\dfrac{hc}{9\lambda }=\dfrac{hc}{{\lambda }_{th}}
λth=9λ{\lambda }_{th}=9\lambda
∴k=9∴k=9

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