Question

The energy of the electron of hydrogen atom in its nth orbit is given by $E_n=-\frac{13.6}{n^2}$ electron volt(eV). Based on this formula.
(i) Draw different energy levels corresponding to $n=1,2,3,4,5,6$ and $\infty$.
(ii) Show Lyman and Baler series of emission spectrum of hydrogen atom by drawing various electronic transitions.
(iii) Find the ionisation energy of hydrogen atom.

Answer 1

Energy of electron in nth orbit of Hydrogen atom is given by $E_n=\frac{-13.6}{n^2}$
So energy of electron in $n=1$ to $n=6$ is calculated as shown in the above figure.
$E_1=\frac{-13.6}{1^2}=-13.6eV$
$E_2=\frac{-13.6}{2^2}=-3.4eV$
and so on.
Lyman series is obtained in the hydrogen atom when an electronic transition takes place from $n_2=2,3,4,5,6,...$ to $n_1=1$. Whereas, Balmer series is obtained in the hydrogen atom when an electronic transition takes place from $n_2=3,4,5,6,...$ to $n_1=2$. These transition is also shown in the above diagram.
(iii)
Ionisation energy of hydrogen atom $E_{ionisation}=E_{\infty }-E_1$
$⟹$ $E_{ionisation}=0-(-13.6)=13.6eV$