A `12.9 eV` beam of electrons is used to bombard gaseous hydrogen atom at room temperature. Up to which energy level the hydrogen atoms would be excited?
Calculate the wavelength of the first member of Paschen series and first member of Balmer series.

Explanation

Energy of the electron in the `n^(th)` state of an atom `=-13.6z^(2)n^(2)eV`
For hydrogen atom, z = 1
Energy required to excite and an atom from initial state `(n_(i))` to final state
`(n_(f))=-13.6n_(f)^(2)+13.6n_(f)^(2)eV`
This energy must be equal to or less than the energy of the incident electron beam.
`-13.6n_(f)^(2)+13.6n_(i)^(2)eV`
This energy must be equal to or less than the energy of the incident electron beam.
`therefore" "-13.6n_(f)^(2)+13.6n_(i)^(2)=12.9`
Energy of the electron in the ground state
`=-13.612=-13.6eV`
`therefore" "-13.6n_(f)^(2)+13.6=12.9` ltbr. `13.6-12.9=13.6n_(f)^(2)`
`n_(f)^(2)=13.60=19.43`
`rArr" "n_(f)=4.4`
State cannot be a fraction number.
`therefore" "n_(f)=4`
Hence, the hydrogen atom would be excited up to 4th energy level.
Rydberg's formula for the spectrum of the hydrogen atom is given by :
`lambda =R_(1)n_(1)^(2)-n_(2)^(2)`
Here, `lambda` is the wavelength.
Rydberg's constant,
`R=1.097xx10^(7)m^(-1)`
For the first memmer of the Paschen series:
`n^(1)=3n^(2)=4`
`lambda =1.097xx10^(7)xx132-142`
`lambda=18761A`
For the first member of Balmer series :
`n^(1)=2n^(2)=3`
`lambda = 1.097xx10^(7)xx122-132`
`lambda = 6563A.`