Question

The Rydberg constant for hydrogen atom is $1.097 \times {10^7}{m^{ - 1}}$. Calculate the long wavelength limit of the Lyman series.

Answer 1

Hint: In order to solve this question, firstly we will apply the concept of Lyman series through the version of Rydberg formula i.e. $\dfrac{1}{\lambda } = {R_H}\left( {1 - \dfrac{1}{{{n^2}}}} \right)$. Then we will substitute the value of Rydberg constant for hydrogen atom to get the required result.

Formula used-
1. $\dfrac{1}{\lambda } = {R_H}\left( {1 - \dfrac{1}{{{n^2}}}} \right)$
2. $\dfrac{1}{{{\lambda _L}}} = R\left( {\dfrac{1}{1} - \dfrac{1}{{{2^2}}}} \right) = \dfrac{1}{R}$

Complete Step-by-Step solution:
Firstly, the concept of Lyman series through the version of Rydberg formula-
$\dfrac{1}{\lambda } = {R_H}\left( {1 - \dfrac{1}{{{n^2}}}} \right)$
$\left( {{R_H} \approx 1.0967 \times {{10}^7}{m^{ - 1}} \approx \dfrac{{13.6eV}}{{hc}}} \right)$
Where n is a natural number greater than or equal to 2 (i.e. n = 2, 3, 4, …)
Wavelengths of the lines emitted lie in the Lyman series if the transition of the electron takes place from higher energy levels n>1 to ground state energy level n = 1.
So, when transition of electron take place either from n = 3 to n = 1 or n = 4 to n = 1, then wavelengths of lines so emitted lie in Lyman series whereas transitions either from n = 3 to n = 2 or n = 4 to n = 2 correspond to Balmer series.
Now, for calculating longest wavelength-
$n = 2$
$\dfrac{1}{{{\lambda _L}}} = R\left( {\dfrac{1}{1} - \dfrac{1}{{{2^2}}}} \right) = \dfrac{1}{R}$
Or $R\left( {1 - \dfrac{1}{{{2^2}}}} \right)$
Or$R \times \left( {\dfrac{3}{4}} \right)$
$ \Rightarrow \lambda = \dfrac{4}{{3R}}$
Here, we are given that-
$R = 1.097 \times {10^7}{m^{ - 1}}$
So, substituting the value of R in calculating longest wavelength-
We get-
${X_L} = \dfrac{4}{{3 \times 1.097 \times {{10}^7}}}$
Or ${X_L} = 12.15 \times {10^{ - 8}}m$
Or ${X_L} = 1215A$
Hence, the long wavelength limit of Lyman series is 1215A.

Note- While solving this question, we should not forget to write the unit of measurement of length i.e. Angstrom with the calculated value of the long wavelength limit. Angstrom is a unit of length used to measure small distances where one angstrom is equal to one ten-billionth of a meter.