Question

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Answer 1

Hint: Transition from n = 4 to n =2 means the electrons jumped from higher energy level orbit to lower energy level orbit by losing energy. There is a relationship between wavenumber and number of orbits where the electron is present.
As per Balmer formula
\[\overset{-}{\mathop{\text{V}}}\,={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]\]
Where ${{R}_{H}}$ = Rydberg constant
${{n}_{1}}$ = lowest energy level orbital in the atom
${{n}_{2}}$ = highest energy level orbital in the atom.

Complete step by step answer:
- In the question it is given that an electron jumps from an energy level of 2 to an energy level of 4.
- Means here ${{n}_{1}}$ = 2 and ${{n}_{2}}$ = 4.
- We know that ${{R}_{H}}$ = Rydberg constant = 109678
- By substituting all the known values in the below equation we will get a wave number.
\[\begin{align}
& \overset{-}{\mathop{\text{V}}}\,={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right] \\
& =109678\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right] \\
& =109678\times \dfrac{3}{16} \\
\end{align}\]
- We know that there is a relation between wavenumber and wavelength and it is as follows.
\[\overset{-}{\mathop{\text{V}}}\,=\dfrac{1}{\lambda }\]
Where $\lambda $ = Wavelength of the light.
- Therefore
\[\begin{align}
& \lambda =\dfrac{16}{109678\times 3} \\
& \lambda =486nm \\
\end{align}\]

- So, the electrons release 486 nm of light when it jumps from higher energy level to lower energy level in the atom.

Note: If an electron present in an atom absorbs energy then it jumps from lowest energy level to highest energy level and when an electron jumps from higher to lower energy level then it loses energy in the form of light called emission of energy.