# Quantum Chemistry/Example 20

Write an example showing the determination of the initial electronic state of the hydrogen atom given the wavelength of its transition to the ground state.

## An electronic transition of H atom from an energy level to the ground state is observed with a corresponding wavelength of 102.5 nm. Determine the initial state of the electron from which transition has occurred.

The Rydberg's phenomenological equation will be used to solve the energy transition problem

 Rydberg's phenomenological equation for hydrogen ${\displaystyle \lambda =\mathbb {R} _{H}\left({\frac {1}{n_{1}^{2}}}\right)-\left({\frac {1}{n_{2}^{2}}}\right)}$

where the ${\displaystyle R_{H}}$  is Rydberg constant for hydrogen and is equal to 109737 ${\displaystyle cm^{-1}}$ . ${\displaystyle n_{1}}$  is the final energy level, ${\displaystyle n_{2}}$  is the initial energy level for hydrogen transition. Both ${\displaystyle n_{1}}$  and ${\displaystyle n_{2}}$  are integers and ${\displaystyle n_{2}>n_{1}}$ .

The wavelength is given as 103nm

Concert to SI units:

${\displaystyle \lambda =103nm*1^{-7}{\frac {cm}{nm}}=1.03^{-5}cm}$

The H atom's transition is from unknown energy level to the ground state.

So we can know the final state is the ground state which means ${\displaystyle n_{1}=1}$

The ${\displaystyle n_{2}}$  now can be calculated:

${\displaystyle {\frac {1}{n_{2}^{2}}}={\frac {1}{n_{1}^{2}}}-{\frac {1}{\lambda *R_{H}}}}$

${\displaystyle {\frac {1}{n_{2}^{2}}}={\frac {1}{1}}-{\frac {1}{0.00103cm*109737cm^{-1}}}}$

${\displaystyle n_{2}^{2}=8.67\approx {9}}$

Because the energy level always ${\displaystyle >0}$

${\displaystyle n_{2}=3}$

Therefore, the initial state of the electron from which transition has occurred is energy level 3.