Consider a hydrogen atom with its electron in the `n^(th)` orbital. An electromagnetic radiation of wavelength `90 nm` is used to ionize the atom. If the kinetic energy of the ejected electron is `10.4 eV`, then the value of `n` is `(hc=1242 eV nm)`

Answer

2

Explanation

`E_(Ph)=(hc)/lambda=1242/90=13.8 eV`
`E_(Ph)=DeltaE+(K.E.)`
`13.8=DeltaE+10.4`
`DeltaE=3.4 eV`
so electron initially was in `n=2`