Question

Calculate the Rydberg constant if $H{e^ + }$ ions are known to have the wavelength difference between the longest wavelength lines of the Balmer and Lyman series equal to $133.7nm$ .
A.$1.095 \times {10^5}c{m^{ - 1}}$
B.$9.11 \times {10^7}c{m^{ - 1}}$
C.$2.493 \times {10^4}c{m^{ - 1}}$
D.$1.095 \times {10^8}c{m^{ - 1}}$

Answer 1

Light is a form of electromagnetic radiation. The line spectra of hydrogen consists of several lines named after their discoverers. These are Lyman, Balmer, Paschen, Bracket and Pfund. There are two types of spectra observed when any substance interacts with any type of radiation - absorption and emission.

Complete step by step answer:
We know that light is a form of electromagnetic radiation. Light has both particle and wave nature. A ray of white light spreads out in the series of colored bands which is called a spectrum. Such a spectrum is a continuous spectrum. This is called a continuous spectrum because different colors merge into one another. Visible light is a part of such a spectrum. The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. The emission spectra of the atoms in gaseous phase do not show a continuous spread of wavelength from red to violet. Instead they emit light only at specific wavelengths with presence of dark spaces in between.
Such spectra is called line spectra since the emitted radiation is identified by the presence of bright lines in the spectra. So the emission spectra of $H{e^ + }$ ions will split into a series of lines having discrete frequencies. These lines are named after their discoverers. The visible lines in the spectra obey the formula-
$\dfrac{1}{\lambda } = {R_z}(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$
Where ${R_z}$ is Rydberg’s constant, ${n_1}$ and ${n_2}$ correspond to the number of series of lines. The first five series of lines that correspond to ${n_1} = 1,2,3,4,5$ are known as Lyman, Balmer, Paschen, Bracket and Pfund series respectively.
If ${n_1} = 1$ then ${n_2} = 2,3...$ then that series is called a Lyman series.
For Balmer series ${n_1} = 2$ and ${n_2} = 3$ then the wavelength of first line of Balmer series is-
$\dfrac{1}{{{\lambda _1}}} = {R_z}(\dfrac{1}{{{2^2}}} -\dfrac {1}{{{3^2}}}) = \dfrac{{5R}}{9}c{m^{ - 1}}$
For Lyman series ${n_1} = 1$ and ${n_2} = 2$ then the wavelength of first line of Lyman series is-
$\dfrac{1}{{{\lambda _2}}} = {R_z}(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}) = \dfrac{{3R}}{1}c{m^{ - 1}}$
Now the wavelength difference between the Balmer and Lyman series is given. $\Delta \lambda = 133.7nm = 133.7 \times {10^{ - 9}}m$
So,
$\Delta \lambda = {\lambda _1} - {\lambda _2} = \dfrac{1}{{3R}} - \dfrac{9}{{5R}} = \dfrac{{22}}{{15R}}$
Therefore, $\Delta \lambda = \dfrac{{22}}{{15R}}$ so
$R = \dfrac{{22}}{{15 \times \Delta \lambda }} = \dfrac{{22}}{{15 \times 133.7 \times {{10}^{ - 9}}}} = 1.095 \times {10^7}{m^{ - 1}} = 1.095 \times {10^5}c{m^{ - 1}}$
Therefore, the correct option is (A) .

Note: Spectral lines are expressed in wavenumbers $(c{m^{ - 1}})$ . Hydrogen atoms have the simplest line spectrum. Only the lines of the Balmer series appear in the visible region of the electromagnetic spectrum. The study of emission or absorption spectra is called spectroscopy.