Question

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 wavelength of the first member of Lyman and first member of Balmer series.

Answer 1

Energy of the electron in the $nth$ state of an atom $=\frac{-13.6z^2}{n^2}eV$

Here, $z$ is the atomic number of the atom.

For the hydrogen atom, $z$ is equal to $1$.

Energy required to excite an atom from the initial state (ni) to the final state (nf) $=-13.6/n_f^2+13.6/n_i^{2}eV$

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

∴ $-13.6/n_f^2+13.6/n_i^{2}eV=12.5$ (Energy of the electron in the ground state $=-13.612=-13.6eV$)

$=-13.6/n_f^2+13.6eV=12.5$

$\Rightarrow 13.6-12.5=13.6/n_f^2$

$\Rightarrow n_f=\sqrt{\frac{13.6}{1.1}}=3.51$

Since $n_f$ cann't be fraction
$\therefore n_f=3$

Hence hydrogen atom would be excited upto 3rd energy level

First member of lyman:

$\frac{1}{{\lambda }_H}=R_H{\left(1\right)}^2[\frac{1}{1^2}-\frac{1}{2^2}]$

$R=1.097\times {10}^{-7}{m}^{-1}$

$\lambda =1215\dot{A}$

First member of Balmer:

$\frac{1}{{\lambda }_{23}}=R_H{\left(1\right)}^2[\frac{1}{2^2}-\frac{1}{3^2}]$

$\lambda =6563\dot{A}$