A hydrogen atom in a having a binding of `0.85eV`makes transition to a state with excited energy `10.2eV`(a) identify the quantum number n of theupper and the lower energy state involved in the transition (b) Find the wavelength of the emitted radiation

Answer

A, B, D

Explanation

(a) From the energy data, we see that the 'H' atom transists from builing energy of `0.85 eV` to excitation energy of `102 eV`
=Binding Energy of `-3.4eV`
`So, `n=4` to `n=2`
(b) We know
`1/lambda = 1.097 xx 10^7 xx 10^7((1)/(4)-(1)/(16))`
implies `lambda = (16)/(1.097 xx 3 xx 10^7)`
`=4.8617 xx 10^-7`
`=487 nm`.