Question

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer 1

Hint: The unit of wave number is the reciprocal meters. Wave number is the number of wavelengths in a unit distance and it is similar to frequency. The spectral lines in the Balmer series are caused by the transition of electrons from higher energy levels than $n = 3$ to the lower $n = 2$ level.
Formulas used: $\dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{4} - \dfrac{1}{{{n^2}}}} \right]$, here $\lambda$ is the wavelength, ${R_H}$ is the Rydberg constant, and $n$ is the level of upper orbit.

Complete step by step answer:
First, we look into what the spectral series is and why are they grouped.
The atomic emission spectrum of hydrogen has lines called spectral lines and they are divided and grouped into several series according to their wavelength.
The Balmer series contains the spectral lines produced when the electrons move from higher levels and then come back to the energy level with principal quantum number $2$. The Balmer series wavelengths can be calculated by the formula $\dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{4} - \dfrac{1}{{{n^2}}}} \right]$, here $\lambda$ is the wavelength, ${R_H}$ is the Rydberg constant, and $n$ is the level of upper orbit.
Also $\nu = \dfrac{1}{\lambda }$, where $\nu$ is the wave number.
Therefore, the equation becomes
$\Rightarrow \nu = {R_H}(\dfrac{1}{4} - \dfrac{1}{{{n^2}}})$
We also know that the wave number is inversely proportional to transition wavelength. Therefore, the higher the transition wavelength the lower the wave number and vice versa.
For a lower wavenumber the $n$ should be minimum and the smallest higher energy level in the Balmer series is $3$.
Therefore, substituting these values in the wave number equation,
 $\Rightarrow \nu = \left( {1.097 \times {{10}^7}} \right)\left[ {\dfrac{1}{4} - \dfrac{1}{{{3^2}}}} \right]$,
here we have substituted $n = 3$ and ${R_H} = 1.097 \times {10^7}{m^{ - 1}}$
$\Rightarrow \nu = \left( {1.097 \times {{10}^7}} \right)\left[ {\dfrac{1}{4} - \dfrac{1}{9}} \right]$
$\Rightarrow \nu = \left( {1.097 \times {{10}^7}} \right)\left[ {\dfrac{{9 - 4}}{{36}}} \right]$
On solving this, we get:
$\Rightarrow \nu = 1.5236 \times {10^6}{m^{ - 1}}$
Hence the wave number for the wavelength transition in Balmer series of atomic hydrogen is:
$\nu = 1.5236 \times {10^6}{m^{ - 1}}$

Note:The lowest value of the upper energy levels should be used as the value of $n$ in the spectrum series equations to get the wave number for the longest transition wavelength in the different series of spectrum of atomic hydrogen. Note that the $4$ in our equation arises due to the fact that in the Balmer series, the lower energy level is always $n = 2$. There are other series in the Hydrogen spectrum, like Lyman, Paschen, Brackett etc. Also note that waves emitted from the Balmer series are in the visible region of the electromagnetic spectrum.