Question

A 12.9 eV beam of electronic is used to bombard gaseous hydrogen at room temperature. Upto 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.

Answer 1

Energy of the electron in the n^​th state of an atom = \[- \frac{13 . 6 z^2}{n^2}\] eV
Here, z is the atomic number of the atom.
For hydrogen atom, z = 1

Energy required to excite an atom from initial state (n_i) to final state (n_f)  = \[- \frac{13 . 6}{{n_f}^2} + \frac{13 . 6}{{n_i}^2}\]eV
This energy must be equal to or less than the energy of the incident electron beam.

∴ \[- \frac{13 . 6}{{n_f}^2} + \frac{13 . 6}{{n_i}^2}\] 12.9
Energy of the electron in the ground state  = ∴  \[- \frac{13 . 6}{1^2} = - 13 . 6\]  eV

\[\therefore - \frac{13 . 6}{{n_f}^2} + 13 . 6 = 12 . 9\]

\[13 . 6 - 12 . 9 = \frac{13 . 6}{{n_f}^2}\]

\[{n_f}^2 = \frac{13 . 6}{0 . 7} = 19 . 43\]

\[ \Rightarrow n_f = 4 . 4\]

State cannot be a fraction number.
∴ \[n_f\] =4

Hence, the hydrogen atom would be excited up to 

\[4^{th}\] energy level.
Rydberg's formula for the spectrum of the hydrogen atom is given by:

\[\frac{1}{\lambda} = R\left[ \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right]\]

Here,

\[\lambda\]  is the wavelength. 
Rydberg's constant, R = \[1 . 097 \times {10}^7\] m^-1
For the first member of the Paschen series:

\[n_1 = 3 \]

\[ n_2 = 4\]

\[\frac{1}{\lambda} = 1 . 097 \times {10}^7 \left[ \frac{1}{3^2} - \frac{1}{4^2} \right]\]

\[\lambda = 18761\]Å

For the first member of Balmer series:

\[n_1 = 2 \]

\[ n_2 = 3\]

\[\frac{1}{\lambda} = 1 . 097 \times {10}^7 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]\]

\[\lambda = 6563\]Å