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An electron and a proton are separated by a large distance. The electron starts approaching the proton with energy \( 2 eV \). The proton captures the electron and forms a hydrogen atom in first excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength \( 4600 A \). The maximum KE of the emitted photoelectron is (take, \( h c=12420 eV A \) ). |
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Answer» Intial Energy = 2 + 0 = 2 ev Final energy = \(\frac{13.6\times1}1\) 2 = -13.6 + Ep Ep = 15.6 ev ɸ = \(\frac{12420}{4600}\) ɸ = 2.7 then E = hv - ɸ E = 15.6 - 2.7 E = 12.9 ev |
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