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

C

Explanation

`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