Question

If ${\lambda _1}$ and ${\lambda _2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then ${\lambda _1}:{\lambda _2}$ is:
${\text{A}}{\text{. 1 : 3}} \\\ {\text{B}}{\text{. 1 : 30}} \\\ {\text{C}}{\text{. 7 : 50}} \\\ {\text{D}}{\text{. 7 : 108}} \\\$

Answer 1

We need to use the expression for wavelength obtained from the Bohr’s model for hydrogen atom. In the case of the Lyman series, the first member de-excites from n = 2 to n = 1 while in case of Paschen series, the first member de-excites from n = 4 to n = 3.

Formula used: According to Bohr’s model of hydrogen atom, we have the following relation for the wavelengths of radiation emitted by electrons jumping between various energy levels of the hydrogen atom.
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Here $\lambda$ is the wavelength of the radiation emitted by an electron when it jumps from higher energy level ${n_2}$ to lower energy level ${n_1}$, R is called the Rydberg’s constant. Its value is given as
$R = 10973731.6{m^{ - 1}}$

Complete step by step answer:
We are given that ${\lambda _1}$ and ${\lambda _2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively.
Lyman series is the name given to those wavelengths which are emitted when an electron de-excites to level n = 1 from higher levels. The first member of the Lyman series de-excites from n = 2 to n = 1. Therefore, we can write that
$\dfrac{1}{{{\lambda _1}}} = R\left( {\dfrac{1}{{{{\left( 1 \right)}^2}}} - \dfrac{1}{{{{\left( 2 \right)}^2}}}} \right) = R\left( {1 - \dfrac{1}{4}} \right) = \dfrac{3}{4}R{\text{ }}...\left( i \right)$
The Paschen series is the name given to those wavelengths which are emitted when an electron de-excites to level n = 3 from higher levels. The first member of the Lyman series de-excites from n = 4 to n = 3. Therefore, we can write that
$\dfrac{1}{{{\lambda _2}}} = R\left( {\dfrac{1}{{{{\left( 3 \right)}^2}}} - \dfrac{1}{{{{\left( 4 \right)}^2}}}} \right) = R\left( {\dfrac{1}{9} - \dfrac{1}{{16}}} \right) = R\left( {\dfrac{{16 - 9}}{{9 \times 16}}} \right){\text{ = }}\dfrac{7}{{9 \times 16}}R{\text{ }}...\left( {ii} \right)$
Now we will divide equation (ii) by equation (i). Doing so, we get the following result.
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{7R}}{{9 \times 16}} \times \dfrac{4}{{3R}} = \dfrac{7}{{108}} = 7:108$

So, the correct answer is “Option D”.

Note: The electrons revolving around the hydrogen atom exist in different energy levels. The ground state is the lowest energy level corresponding to n = 1. The electrons get excited to higher energy levels when they absorb energy. The higher energy states are unstable and readily de-excite to lower energy levels to emit various wavelengths depending on energy difference between levels.