First slide

Bohr Model of the Hydrogen atom.

Question

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

Moderate

A

$\frac{27}{5}$
B

$\frac{5}{27}$
C

$\frac{4}{9}$
D

$\frac{9}{4}$

Solution

The wavelength of a spectral line in the Lyman series is
$\frac{1}{λ_{L}} = R (\frac{1}{1^{2}} - \frac{1}{n^{2}}), n = 2, 3, 4, \dots \dots$
and that in the Balmer series is
$\frac{1}{λ_{B}} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}}), n = 3, 4, 5$
$For the longest wavelength in the Lyman series, n = 2$
$∴ \frac{1}{λ_{L}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = R (\frac{1}{1} - \frac{1}{4}) = R (\frac{4 - 1}{4}) = \frac{3 R}{4}$
$or λ_{L} = \frac{4}{3 R}$
For the longest wavelength in the Balmer series, n = 3
$∴ \frac{1}{λ_{B}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{1}{4} - \frac{1}{9}) = R (\frac{9 - 4}{36}) = \frac{5 R}{36}$
$or λ_{B} = \frac{36}{5 R}$
$Thus, \frac{λ_{L}}{λ_{B}} = \frac{\frac{4}{3 R}}{\frac{36}{5 R}} = \frac{4}{3 R} \times \frac{5 R}{36} = \frac{5}{27}$

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