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# If the wavelength of ${H_{alpha}}$line of Balmer series is ${\text{X}}\,{{\text{A}}^{\text{0}}}$, then the wavelength of ${H_{beta}}$line of Balmer series is: A. ${\text{X}}\dfrac{{{\text{108}}}}{{{\text{80}}}}{{\text{A}}^{\text{0}}}$B. ${\text{X}}\dfrac{{{\text{80}}}}{{{\text{108}}}}{{\text{A}}^{\text{0}}}$C. $\dfrac{{\text{1}}}{{\text{X}}}\dfrac{{{\text{80}}}}{{{\text{108}}}}{{\text{A}}^{\text{0}}}$D. $\dfrac{{\text{1}}}{{\text{X}}}\dfrac{{{\text{108}}}}{{{\text{80}}}}{{\text{A}}^{\text{0}}}$

Last updated date: 20th Jun 2024
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Hint: The hydrogen spectrum series is classified into five types. These are Lyman series, Balmer series, Paschen series, Bracket series and Pfund series.

We know that for Balmer series,
$\dfrac{1}{{{\lambda }}} = {\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{{\text{1}}}{{{\text{n}}_{\text{2}}^{\text{2}}}}} \right]$ …..(1)
For the alpha line, n = 3. Using this value in the above expression (1) we get,
$\dfrac{1}{{{{{\lambda }}_a}}} = {\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{{\text{1}}}{{{3^2}}}} \right] = \dfrac{1}{{\text{X}}}$
$\Rightarrow \dfrac{1}{{{{{\lambda }}_a}}} = \,{{R \times }}\,\dfrac{{\text{5}}}{{{\text{36}}}} = \,\dfrac{1}{{\text{X}}} \\ \Rightarrow {\text{R}} = \,\dfrac{{{\text{36}}}}{{\text{5}}}{{ \times }}\dfrac{{\text{1}}}{{\text{X}}} \\$ ……(2)
For the beta line, n = 4. Using this value in the above expression (1) we get,
$\Rightarrow \dfrac{1}{{{{{\lambda }}_{{\beta }}}}} = \,{\text{R}}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right] = \,\dfrac{3}{{16}}{\text{R}}\,$
$\Rightarrow \dfrac{1}{{{{{\lambda }}_{{\beta }}}}} = \,{\text{X}}\dfrac{{80}}{{108}}{{\text{A}}^{\text{0}}}\,$……(3)

So, the correct answer is Option C.

The general formula for the hydrogen emission spectrum is given by Johannes Rydbergis :
$\mathop \nu \limits^ - = \,109677\,\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Here,
${n_1}$= 1,2,3,4 …
${n_2}$ = n1 +1
$\mathop \nu \limits^ -$= wave number of the electromagnetic radiation.
The series can be classified according to the transition in the following way:
Transition from the first shell, i.e. (n=1) to any other shell gives rise to the Lyman series
Transition from the second shell i.e. (n=2) to any other shell gives rise to the Balmer series
Transition from the third shell i.e. (n=3) to any other shell gives rise to the Paschen series
Transition from the fourth i.e. (n=4) shell to any other shell gives rise to the Bracket series
Transition from the fifth shell i.e. (n=5) to any other shell gives rise to the Pfund series

Note:
The value of Rydberg constant for hydrogen is $109,677\,{\text{c}}{{\text{m}}^{ - 1}}$. The Rydberg constant is used for the calculation of the wavelengths in the hydrogen spectrum.