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Question
Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.
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Solution
Work function (or threshold energy) (W0): The minimum energy of incident radiation required to eject the electrons from the metallic surface is defined as the work function of that surface.
W0 = hv0 = `(hc)/λ_0` Joules; v0 = Threshold frequency; λ0 = Threshold wavelength
Work function in electron volt, W0(eV) = `(hc)/(eλ_0) = 12375/(λ_0(Å))`
Einstein's photoelectric equation is E = W0 + Kmax
Maximum energy = `hv - phi`
According to the problem for the first condition wavelength of light λ = 600 nm and for the second condition, the wavelength of light λ' = 400 nm
Also, the maximum kinetic energy for the second condition is equal to twice the kinetic energy in the first condition.
i.e., K'max = 2Kmax
then K'max = `(hc)/λ - phi`
⇒ 2Kmax = `(hc)/λ^' - phi`
⇒ `2(1230/600 - phi) = (1230/400 - phi)` ......[∵ hc = 1240 eV nm]
⇒ `phi = 1230/1230` = 1.02 eV
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