Question

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

Answer 1

Hint: Substitute the values of highest and lowest principal quantum number possible in the Rydberg formula to find the wave number.
Formula: $\dfrac{1}{\lambda }={{Z}^{2}}{{R}_{\infty }}\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)$
where,
 $\dfrac{1}{\lambda }$ is the wavenumber,
 Z is the atomic number,
${{R}_{\infty }}$ is the Rydberg constant whose value is $1.09677*{{10}^{7}}{{m}^{-1}}$,
${{n}_{1}}$ is the principal quantum number of lower energy level,
${{n}_{2}}$ is the principal quantum number of upper energy level

Complete step by step answer:
We will now try to understand the emission spectrum of Hydrogen.
-When a hydrogen atom absorbs a photon, it causes the electron to experience a transition to a higher energy level, for example, n = 1 to n = 2.
-When a photon is emitted through a hydrogen atom, the electron undergoes a transition from a higher energy level to a lower, for example, n = 3 to n = 2.
-During this transition from a higher level to a lower level, there is the transmission of light.
The series of hydrogen emission spectrums have unique names based on the transition from higher energy to lower energy.
-Lyman Series: Transition of an electron from outer orbit having n>1 to n=1.
-Balmer Series: Transition of an electron from outer orbit having n>1 to n=2.
-Paschen Series: Transition of an electron from outer orbit having n>1 to n=3.
-Brackett Series: Transition of an electron from outer orbit having n>1 to n=4.
-Pfund Series: Transition of an electron from outer orbit having n>1 to n=5.
-Humphrey Series: Transition of an electron from outer orbit having n>1 to n=6.
It is important to remember that the wavelength emitted in the Balmer series alone lies in the visible range of the light spectrum.
The shortest wavelength transition in Balmer series is $n=\infty \to n=2$
Substituting the above values in the Rydberg formula,
$\dfrac{1}{\lambda }=1*1.09677*{{10}^{7}}*\left( \dfrac{1}{4}-0 \right)$
$\dfrac{1}{\lambda }=2741925{{m}^{-1}}$

Therefore, the wavenumber with the shortest wavelength transition is $2741925{{m}^{-1}}$.

Note: Do not get confused between the meaning of ${{n}_{1}}$and ${{n}_{2}}$.${{n}_{2}}$ must always have a larger value than ${{n}_{1}}$for the transition to exist and wavelength to be transmitted.