The energy of the electron of the hydrogen atom in the $n t h$ orbit is $E_{n} = - \frac{13.6}{n^{2}} e V$ . Draw the energy level diagram for the hydrogen atom and show transitions corresponding to lives of Lyman and Balmer series.

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Solution

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.
$E_{1} = \frac{- 13.6}{1^{2}} = - 13.6 e V$
$E_{2} = \frac{- 13.6}{2^{2}} = - 3.4 e V$
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.

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The energy of the electron of the hydrogen atom in the nth orbit is En=−13.6n2eV. Draw the energy level diagram for the hydrogen atom and show transitions corresponding to lives of Lyman and Balmer series.

The energy of the electron of the hydrogen atom in the $n t h$ orbit is $E_{n} = - \frac{13.6}{n^{2}} e V$ . Draw the energy level diagram for the hydrogen atom and show transitions corresponding to lives of Lyman and Balmer series.