Question

The ratio of minimum wavelength of Lyman and Balmer series will be:
A. 1.25
B. 0.25
C. 5
D. 10

Answer 1

In the hydrogen atom, the electrons de-excite themselves by continuously making transitions from the higher energy level to the lower energy level.These transitions are divided in a group of five series. While undergoing these transitions, an electron emits a photon or radiation of particular wavelength. This wavelength is given by Rydberg’s formula. Lyman series is formed when a transition takes place from the higher energy levels to the first energy level. Balmer series is formed when a transition takes place from the higher energy levels to the second energy level.

Formula Used:
The Rydberg’s formula is given by:
$\dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_0}^2}}} \right]$
where, ${n_i}$ is the principal quantum number of the lower energy level, ${n_0}$ is the principal quantum number of the upper energy level and ${R_H}$ is Rydberg's constant for hydrogen atoms.

Complete step by step answer:
The radiation spectrum of hydrogen is grouped into different series. Each series is named after its discoverer. These series are Lyman series, Balmer series, Paschen series, Brackett series and Pfund series. According to Bohr’s theory, the wavelength $\lambda$ of radiations emitted from hydrogen atom is given by;
$\dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_0}^2}}} \right]$
This equation is also known as Rydberg’s formula.

For the Lyman series, ${n_i} = 1$ and ${n_0} = 2,3,4....$. The photon makes a transition from the upper energy level, that is any of the 2nd, 3rd and 4th levels to the lower energy level, that is 1st atomic level. This photon emits wavelengths of the radiation given by $\dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{{{\left( 1 \right)}^2}}} - \dfrac{1}{{{n_0}^2}}} \right]$. The longest and the shortest wavelengths of the Lyman series are 121.6nm and 91.2nm respectively.
The Balmer series or radiation correspond to ${n_i} = 2$ and ${n_0} = 3,4,5....$.. The wavelength of the radiation emitted by Balmer transitions is given by: $\dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{{{\left( 2 \right)}^2}}} - \dfrac{1}{{{n_0}^2}}} \right]$. The longest and the shortest wavelengths of the Balmer series are 656.3nm and 364.8nm respectively.
Therefore, the ratio of minimum wavelength of Lyman and Balmer series will be
$\therefore\dfrac{{91.2nm}}{{364.8nm}} = \dfrac{1}{4} = 0.25$

Hence, option B is the correct answer.

Note: The five series, namely Lyman series, Balmer series, Paschen series, Brackett series and Pfund series are applicable only for Hydrogen atoms. An electron can revolve only in particular orbits and that is given by $n$.
The Rydberg’s formula is given more precisely by
$\dfrac{1}{\lambda } = {Z^2}{R_H}\left[ {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_0}^2}}} \right]$
where, $Z$ is the atomic number.