Question

Calculate the shortest wavelength in the Balmer series of hydrogen atoms. In which region (infra-red, visible, ultraviolet) of the hydrogen spectrum does this wavelength lie?

Answer 1

Hint: You can start by describing Rutherford’s atomic model. Then explain the various series in the spectrum of Hydrogen atoms. Then use the equation$\dfrac{1}{\lambda } = 1.097 \times {10^7} \times \left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$and put ${n_1} = 2$and ${n_2} = \infty $ to reach the solution.

Complete answer:
The Balmer series is a part of the six series that are created after Rutherford’s model of the atom.
According to Rutherford’s model of an atom:
Most of the mass of an atom is due to the positively charged protons and is concentrated in a nucleus that is relatively smaller than the atom.
Electrons (the negatively charged particles) surround the nucleus and revolve around it at a very high speed in circular paths.
The atom is held together due to the electrostatic force of attraction.
Every atom has a unique emission spectrum, hydrogen also has a unique spectrum, which is divided into multiple spectral series. These spectral lines are the result of electrons jumping from one orbit to the other. Rydberg formula was used to divide these lines into 6 named series:
Lyman series – produced when electron from an outer orbit($n < 1$) to the first orbit
Balmer series - produced when electron from an outer orbit($n < 2$) to the second orbit
Paschen series – produced when electron from an outer orbit($n < 3$) to the third orbit
Brackett series – produced when electron from an outer orbit($n < 4$) to the fourth orbit
Pfund series – produced when electron from an outer orbit($n < 5$) to the fifth orbit
Humphreys series - produced when electron from an outer orbit($n < 6$) to the six orbit
So wavelength of light emitted in the Balmer series of a H-atom:
$\dfrac{1}{\lambda } = 1.097 \times {10^7} \times \left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
For shortest wavelength in the Balmer series, we know that ${n_1} = 2$and ${n_2} = \infty $
$\dfrac{1}{{{\lambda _{\min }}}} = 1.097 \times {10^7} \times \left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{\infty }} \right)$
${\lambda _{\min }} = \dfrac{4}{{1.097 \times {{10}^7}}} = 3.46 \times {10^{ - 7}}m$
The electromagnetic wave of the wavelength $3.46 \times {10^{ - 7}}m$ will lie in the ultraviolet region.

Note: In the solution while discussing that the emission spectrum of hydrogen atom is divided into multiple spectrum series, it is important to note that we mention the 6 named series. It must be made clear that there are more than 6 series in reality but we discussed only 6 because further series have not been properly named yet.