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Question: If the wavelength of \[{H_{alpha}}\]line of Balmer series is \[{\text{X}}\,{{\text{A}}^{\text{0}}}\]...

If the wavelength of Halpha{H_{alpha}}line of Balmer series is XA0{\text{X}}\,{{\text{A}}^{\text{0}}}, then the wavelength of Hbeta{H_{beta}}line of Balmer series is:
A. X10880A0{\text{X}}\dfrac{{{\text{108}}}}{{{\text{80}}}}{{\text{A}}^{\text{0}}}
B. X80108A0{\text{X}}\dfrac{{{\text{80}}}}{{{\text{108}}}}{{\text{A}}^{\text{0}}}
C. 1X80108A0\dfrac{{\text{1}}}{{\text{X}}}\dfrac{{{\text{80}}}}{{{\text{108}}}}{{\text{A}}^{\text{0}}}
D. 1X10880A0\dfrac{{\text{1}}}{{\text{X}}}\dfrac{{{\text{108}}}}{{{\text{80}}}}{{\text{A}}^{\text{0}}}

Explanation

Solution

The hydrogen spectrum series is classified into five types. These are Lyman series, Balmer series, Paschen series, Bracket series and Pfund series.

Complete step by step answer:
We know that for Balmer series,
1λ=R[1221n22]\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,
1λa=R[122132]=1X\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}}}\,$$……(3) **So, the correct answer is Option C.** **Additional Information:** The general formula for the hydrogen emission spectrum is given by Johannes Rydbergis : $$\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.