Question

Find the longest wavelength in the Balmer series of hydrogen.

Answer 1

Electron jumps from higher energy orbital to the second orbital in the Balmer series. The energy emitted for transitions of electrons can be expressed as, $E=\dfrac{hc}{\lambda }$. We will use the equation for energy released in Balmer transition to calculate the longest wavelength in the Balmer series. To find the longest wavelength, energy released should be minimum.

Formula used:
$\dfrac{1}{\lambda }={{R}_{H}}{{z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$

Complete step-by-step answer:
The Balmer series is a particular part of the electromagnetic spectrum of hydrogen with definite starting and final orbital through which the electronic transition occurs.
Now, for electronic transitions between two orbits, when an electron jumps from higher orbital to lower orbital, energy is released in the form of a spectrum.
Balmer series correspond to transitions from higher orbital to second innermost orbital, whatever be the orbital from which electron jumps.
Now, for such electronic transitions, we have a formula for calculating the wavelength of the spectrum when an electron jumps from higher to lower orbital, i.e.
$\dfrac{1}{\lambda }={{R}_{H}}{{z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$
Where,
${{n}_{1}}$ corresponds to the final orbital in which electron lands,
${{n}_{2}}$ corresponds to the initial orbital from which electron jumps,
${{R}_{H}}$ is the Rydberg constant,
And, $z$ is the atomic number of element in which transition occurs
Thus, for the hydrogen atom, $z=1$
For the Balmer series: ${{n}_{1}}=2$ and ${{n}_{2}}=3,4,5...$
Now, for computing maximum $\lambda$ by the above formula, for fixed ${{n}_{1}}$:
$\dfrac{1}{\lambda }={{R}_{H}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$
We have, ${{n}_{2}}$ should be minimum
So, for maximum $\lambda$ of the Balmer series: ${{n}_{1}}=2$ and ${{n}_{2}}=3$
Thus, for the Balmer series:
$\dfrac{1}{{{\lambda }_{\max }}}={{R}_{H}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)={{R}_{H}}\left( \dfrac{1}{4}-\dfrac{1}{9} \right)=\dfrac{5{{R}_{H}}}{36}$
${{\lambda }_{\max }}\left( B \right)=\dfrac{36}{5{{R}_{H}}}$
Putting,
${{R}_{H}}=1.097\times {{10}^{7}}{{m}^{-1}}$
We get,

& {{\lambda }_{\max }}\left( B \right)=\dfrac{36}{5\times 1.097\times {{10}^{7}}} \\\ & {{\lambda }_{\max }}\left( B \right)=656.3nm \\\ \end{aligned}$$ Hence, the longest wavelength in the Balmer series of hydrogen is $$\dfrac{36}{5{{R}_{H}}}$$ **Note:** Students should remember the definition of the Balmer series and should not get confused with orbital numbers related to the transition in the Balmer series. While considering maximum and minimum wavelengths in the Balmer series, students should estimate the final orbital with taking care of signs and relation in the formula used.