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प्रश्न
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.
उत्तर
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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