First slide

Bohr Model of the Hydrogen atom.

Question

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

Moderate

A

$\frac{7}{29}$
B

$\frac{9}{31}$
C

$\frac{5}{27}$
D

$\frac{3}{23}$

Solution

The wavelength of different spectral lines of Lyman series is given by
$\frac{1}{λ_{L}} = R [\frac{1}{l^{2}} - \frac{1}{n^{2}}] w h e r e n = 2, 3, 4, ....$
where subscript L refers to Lyman. For longest wavelength, n = 2
$∴ \frac{1}{λ_{L_{l o n g e s t}}} = R [\frac{1}{1^{2}} - \frac{1}{2^{2}}] = \frac{3}{4} R - - - (i)$
The wavelength of different spectral series of Balmer series is given by
$\frac{1}{λ_{B}} = R [\frac{1}{2^{2}} - \frac{1}{n^{2}}] w h e r e n = 3, 4, 5, ....$
where subscript B refers to Balmer. For longest wavelength, n= 3
$∴ \frac{1}{λ_{B_{l o n g e s t}}} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = R [\frac{1}{4} - \frac{1}{9}] = \frac{5 R}{36} - - - (i i)$
Divide (ii) by (i), we get
$\frac{λ_{L_{l o n g e s t}}}{λ_{B_{l o n g e s t}}} = \frac{5 R}{36} \times \frac{4}{3 R} = \frac{5}{27}$

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