Question

Calculate the maximum wavelength of Balmer series in the hydrogen spectrum. Calculate the corresponding wavenumber.
$R = 1.097 \times {10^7}\;{m^{ - 1}}$

Answer 1

Hint: In this solution, we are going to use Rydberg's formula to calculate the wavelength of Balmer series. The maximum wavelength in the Balmer series is obtained corresponding to minimum energy transition, that is, when electron transition occurs between ${2^{nd}}$ and ${3^{rd}}$ shell.

Formula Used:
Rydberg’s Formula:
$\dfrac{1}{\lambda } = R \times {Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\;{m^{ - 1}}$
Where,
$R$is Rydberg’s Constant
$Z$ is the Atomic Number
${n_1}$ is lower energy state
${n_2}$ is higher energy state
$\lambda $ is Wavelength corresponding to electron’s transition between shells ${n_1}$and ${n_2}$

Complete step by step answer:
Given:
$Z = 1$ (As atomic number of Hydrogen is 1)
$R = 1.097 \times {10^7}{m^{ - 1}}$
${n_1} = 2$ (As ground state of Balmer series is 2)
Now, for the maximum wavelength to be emitted, the transition of the electron should take place in such a way that the corresponding energy released should be minimum.
And, for minimum energy, the transition of electrons should take place between two closest energy states.
As we know that, For Balmer series, the ground state ${n_1} = 2$ so, for minimum energy, the closest energy state would be the just next energy state, that is, ${n_2} = 3$
Substituting the values of $R,Z,{n_1}\;and\;{n_2}$in Rydberg’s Formula, we get,
\[
   \Rightarrow \dfrac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{m^{ - 1}}} \right) \times {\left( 1 \right)^2}\left( {\dfrac{1}{{{{\left( 2 \right)}^2}}} - \dfrac{1}{{{{\left( 3 \right)}^2}}}} \right) \\
   \Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7} \times 1\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right)\;{m^{ - 1}} \\
   \Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{5}{{36}}} \right)\;{m^{ - 1}} \\
   \Rightarrow \dfrac{1}{\lambda } = 1.524 \times {10^6}\;{m^{ - 1}} \\
\]
$
   \Rightarrow \lambda = 0.656 \times {10^{ - 6}}m \
   \Rightarrow \lambda = 656 \times {10^{ - 9}}\;m \
   \Rightarrow \lambda = 656\;nm \
$
Therefore, the maximum wavelength of the Balmer series in the hydrogen spectrum is $656\;nm$.
And, the wave number is given by $\dfrac{1}{\lambda }$ . So, the wave number corresponding to maximum wavelength of Balmer series in the hydrogen spectrum is given by:
Wave number $\dfrac{1}{\lambda } = 1.524 \times {10^6}\;{m^{ - 1}}$

Additional information:
The maximum wavelength of Balmer series in the hydrogen spectrum corresponds to a transition between ${2^{nd}}$ and ${3^{rd}}$ shell, similarly for minimum wavelength of Balmer series in the hydrogen spectrum will corresponds to a transition between ${2^{nd}}$ shell and $\infty $ .

Note:
While applying Rydberg's Formula, always remember that ${n_2}$ is a higher shell while ${n_1}$ is a lower shell. If both interchanged wave numbers or wavelengths will come out to be negative which is not possible.