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

Explanation

U=-1.7 eV
`E=(U)/(2)=-0.85 eV`
`E_(n)=-(13.6)/(n^(2)) eV`
`-0.85=-(13.6)/(n^(2)) implies n^(2)=16`
n=4

Six transitions `4 to 1,4 to 2,4 to 3,3 to 2,3 to 1,2 to 1 DeltaE gt 2.3 eV`
`4 to3, DeltaE=0.66 eV`
`4 to 2, Delta E=2.55 eV`
` 4 to 1, Delta E=12.75 eV`
Maximum K.E. of ejected electrons is possinle in ` 4 to 1 DeltaE=phi+K_(max)`
`12.75=2.3+K_(max) implies K_(max)=10.45 eV`
`lambda_(d)=(h)/(sqrt(2mK_(max)))=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx10.45xx1.6xx10^(-19)))`
=0.38 nm