A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited? Calculate the wavelengths of the first member of Lyman and first member of Balmer series.

#### Solution

Energy of the electron in the *n*^{th} state of an atom `=-(13.6z^2)/n^2eV`

Here, z is the atomic number of the atom.

For hydrogen atom, z is equal to 1.

Energy required to excite an atom from the initial state (*n*_{i}) to the final state (*n*_{f}) = `-13.65/(nf^2)+13.6/n_(i^2)eV`

This energy must be equal to or less than the energy of the incident electron beam.

`:.-13.6/(nf^2)+13.6/n_(i^2)=12.5`

Energy of the electron in the ground state = `-13.6/1^2=-13.6V`

`:.-13.6/nf^2+13.6=12.5`

`=>13.6-12.5=13.6/(nf^2)`

`=>nf^2=13.6/1.1=12.36`

`=>n_f=3.5`

State cannot be a fraction number.∴ n_{f} = 3

Hence, hydrogen atom would be excited up to 3rd energy level.

Rydberg formula for the spectrum of the hydrogen atom is given below:

` 1/lambda=R[1/n_(1^2)-1/n_(2^2)]`

Here, λ is the wavelength and *R* is the Rydberg constant.

R = 1.097 × 107 m^{-1}

For the first member of the Lyman series:

n_{1} = 1

n_{2} = 2

`1/lambda=1.097xx10^7[1/1^2-1/2^2]`

λ = 1215 Å

For the first member of Balmer series:

n_{1} = 2

n_{2} = 3

`1/lambda=1.097xx10^7[1/2^2-1/3^2]`

⇒λ = 6563 Å