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Bohr Model of the Hydrogen atom.

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Question

The second line of Balmer series has wavelength 4861 $\overset{o}{A}$ The wavelength of the first line of Balmer series is

Moderate

A

1216 $\overset{o}{A}$
B

6563 $\overset{o}{A}$
C

4340 $\overset{o}{A}$
D

4101 $\overset{o}{A}$

Solution

$\begin{array}{l} For first member, \\ \frac{1}{λ_{1}} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}] \\ or \frac{1}{λ_{1}} = R [\frac{1}{4} - \frac{1}{9}] \\ or \frac{1}{λ_{1}} = R [\frac{9 - 4}{36}] \\ or λ_{1} = \frac{36}{5 R} \\ For second member \\ \frac{1}{λ_{2}} = R [\frac{1}{2^{2}} - \frac{1}{4^{2}}] or \frac{1}{λ_{2}} = R [\frac{1}{4} - \frac{1}{16}] \\ or \frac{1}{λ_{2}} = R [\frac{4 - 1}{16}] or \frac{1}{λ_{2}} = \frac{3 R}{16} \\ or λ_{2} = \frac{16}{3 R} \\ Now, \frac{λ_{1}}{λ_{2}} = \frac{36}{5 R} \times \frac{3 R}{16} or \frac{λ_{1}}{λ_{2}} = \frac{27}{20} \\ or λ_{1} = \frac{27}{20} \times λ_{2} = \frac{27}{20} \times 4861 \overset{o}{A} = 6562 .4 \overset{o}{A} \end{array}$

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