Hydrogen gas in the atomic state is excited to an energy level such that the electrostatic polential energy becomes `-3.02 eV`
Now, the photoelectric plate hoving W=4.6 eV is exposed to the emission spectra of this gas. Assuming all the transition of be possible, find the minimum de-Broglie wavelength of ejected photoelectrons.

Explanation

Potential energy `U,=3.02 eV`
We know,
Total energy, `E=("Potential energy(U)")/(2)=(-3.02)/(2)=-1.511 eV`
Using the relation , `E_(n)=(13.6)/(n^(2)) eV`
`rArr " " -1.511=(-13.6)/(n^(2))rArrn^(2)=9rArr n=3`
Three transitions are possible, `i.e.,rarr2,2rarr1,3rarr1`.
The emission will take place for those transition in which `DeltaEgt1`
Further ,using the relation, `DeltaE=13.6Z^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
For `3rarr2,`
`DeltaE=-13.6((1)/(3^(2))-(1)/(2^(2)))=-13.6xx5//36=+1.88 eV`
For `3rarr,1`
`DeltaE=-13.6((1)/(3^(2))-(1)/(1^(2)))=+13.6xx(8)/(9)=12.08 eV`
Minimum KE of ejected electron is possible in `3rarr1.`
So, according to the relation , `DeltaE=phi+K_("max")`
where, `phi=` work function `12.08=4.6+K_("max")`
`:.` Maximum de-Broglie wavelength `lambda_(d)=(h)/(sqrt2mK_("max"))`
`=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx7.48xx1.6xx10^(-19)))`
`=0.45 nm`