Hydrogen gas in the atomic state is excited to an energy level such tht the electrostatic potential energy of H-atom becomes `-1.7 eV`. Now, a photoelectric plate having work function w=2.3 eV is exposed to the emission spectra of this gas. Assuming all the transitions to be possible, find the minimum de-Broglie wavelength of the ejected photoelectrons.

Hydrogen gas in the atomic state is excited to an energy level such th

1.	Hydrogen gas in the atomic state is excited to an energy level such tht the electrostatic potential energy of H-atom becomes `-1.7 eV`. Now, a photoelectric plate having work function w=2.3 eV is exposed to the emission spectra of this gas. Assuming all the transitions to be possible, find the minimum de-Broglie wavelength of the ejected photoelectrons.
Answer» Correct Answer - C `U = - 1.7 eV` `:. E = (U)/(2) =- 0.85 eV = (-13.6)/(n^2)` `:. N =4 ` Ejected photoelectron will have minimum de-Broglie wavelength corresponding to transition from n = 4 to n=1. `Delta E = E_4 -E_1 =- 0.85 - (-13.6)` = 12.75 eV `K_(max) = DeltaE - W = 10.45 eV` `:. lambda = sqrt((150)/(10.45)) Å` (for an electron)` = 3.8 Å

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