Consider a hydrogen atom with its electron in the `n^(th)` orbital An electomagnetic 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 eVnm)

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`