Question

The maximum wavelength of the Lyman series is,
A. $\dfrac{4}{{3R}}$
B. $\dfrac{1}{{{R^2}}}$
C. $\dfrac{C}{R}$
D. $\dfrac{1}{{RC}}$

Answer 1

Hint: The electrons, when coming from the higher energy level to the lower energy level of the principal quantum number some lines are produced, these lines are called the Lyman series. The longest wavelength corresponds to the lowest energy transition. Consider the highest orbits for the principal quantum number.
Formula used:
$ \Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n^2}_1}} - \dfrac{1}{{{n^2}_2}}} \right]$
Where,
$\lambda $ is the wavelength,$R$ is the Rydberg constant,$n$ is the principal quantum number where the electrons come.

Complete step by step answer:
The Lyman series arises when the electrons of the higher energy levels jump to the principal quantum number one.
The formula for calculating the maximum wavelength is given as,
$ \Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n^2}_1}} - \dfrac{1}{{{n^2}_2}}} \right]$
Where,
$\lambda $ is the wavelength, ${R_H}$ is the Rydberg constant. The value of the Rydberg constant is $109678c{m^{ - 1}}$. Have the value of the Rydberg constant as $R$itself. $n$ is the principal quantum number where the electrons come.
The values are given. For the Lyman series, the value of the quantum number is 1. Substitute the values in the formula.
$ \Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n^2}_1}} - \dfrac{1}{{{n^2}_2}}} \right]$
$ \Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{1}{1} - \dfrac{1}{0}} \right]$
From the given equation we easily see that the electron with maximum wavelength for the Lyman series is ejected when the transition happens from the next highest orbit.
The next highest orbit is the 2nd of the principal quantum number.
$ \Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n^2}_1}} - \dfrac{1}{{{n^2}_2}}} \right]$
Substitute $n = 2$in the given formula.
$ \Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{1}{1} - \dfrac{1}{4}} \right]$
Simplifying the given equation,
$ \Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{{4 - 1}}{4}} \right]$
$ \Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{3}{4}} \right]$
$ \Rightarrow \lambda = \dfrac{4}{{3R}}$
Therefore, the maximum wavelength of the Lyman series is $\dfrac{4}{{3R}}$.

So, the correct answer is “Option A”.

Note:
In hydrogen atoms, the lowest energy level has the principal quantum number is one. The value of the highest energy level can be infinite. In the electromagnetic spectrum, the Lyman series lies in the ultraviolet region. The electromagnetic spectrum has the other series also namely Balmer, Paschen, Bracket, and Pfund series along with the Lyman series.