Question

The limit of the Balmer series is $3646{A^ \circ }$. The wavelength of the first member of this series will be
(A) $6563{A^ \circ }$
(B) $3646{A^ \circ }$
(C) $7200{A^ \circ }$
(D) $1000{A^ \circ }$

Answer 1

Hint The hydrogen spectrum is the simplest spectrum. We can observe some kind of regularity in these spectral lines. We can observe a set of lines that are called the spectral series. The first series was observed by Balmer and hence it is called the Balmer series. We have to find the empirical formula developed by Balmer to find the wavelength.
Formula used:
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{n^2}}}} \right)$
Where $\lambda $stands for the wavelength, $R$is a constant called the Rydberg constant, $n = 3,4,5...$

Complete Step by step solution
We know that, For Balmer series,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{n_2^2}}} \right)$
Where, ${n_2} = 3,4,5....\infty $
For the limit of the series, ${n_2} = \infty $
The wavelength can be written as,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)$
Since$\dfrac{1}{{{\infty ^2}}} \approx 0$, we can write
$\dfrac{1}{\lambda } = R\dfrac{1}{{{2^2}}}$
From this we can write,
$R = \dfrac{4}{\lambda }$
From the question, we know
$\lambda = 3646{A^ \circ }$
Substituting, we get
$R = \dfrac{4}{{3646}}$
For the first line,
${n_2} = 3$
We can write the wavelength as,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)$
This will be,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right)$
$ \Rightarrow \dfrac{1}{\lambda } = R\dfrac{5}{{36}}$
Substituting the value of $R$from the above equation, we get
$\dfrac{1}{\lambda } = \dfrac{4}{{3646}} \times \dfrac{5}{{36}}$
Solving the above equation, we get
$\lambda = 6562.8 = 6563{A^ \circ }$

The answer is: Option (A): $6563{A^ \circ }$

Additional Information
The set of lines in the hydrogen spectral line series are Lyman, Balmer, Paschen, Brackett, and Pfund. The Lyman series lies in the ultraviolet region. Paschen and Brackett series lie in the infrared region. Balmer series are found to lie in the visible region. The Pfund series lies in the far-infrared region. Balmer series was the first series to be discovered.

Note
This question can be easily done if we know the value of the Rydberg constant. The value of Rydberg constant is $R = 1.097 \times {10^7}{m^{ - 1}}$
We know that, the wavelength of Balmer series can be written as,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{n_2}^2}}} \right)$
For the first member of the series,${n_2} = 3$
Now, we can write the wavelength as,
$\dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)$
Solving the above equation, we get
$\dfrac{1}{\lambda } = 1.5236 \times {10^6}{m^{ - 1}}$
From this,
$\lambda = \dfrac{1}{{1.5236 \times {{10}^6}}} = 6563{A^ \circ }$