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Find the de Broglie wavelength of photoelectrons ejected with maximum kinetic energy, when light ofwavelength 100 nm is incident on a cesium surface.(Work function of cesium = 3.4 eV) |
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Answer» Broglie wavelength of photoelectrons EJECTED with maximum kinetic energy, when light of wavelength 100 nm is INCIDENT on a CESIUM surface is λ = 4.08 A⁰The de-broglie wavelength is represented as: λ = h/mvAs per the QUESTION, wavelength = 100 nm = 100 x10⁻⁹ mNow, we know that, (Einstein's equation)Thus, upon putting the VALUES, we get, 12.4 eV = 3.4 eV +KEKE = 9eVNow, the de-broglie wavelength will be :λ = 4.08 A⁰ |
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