First slide

Bohr Model of the Hydrogen atom.

Question

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

Easy

A

1
B

4
C

0.5
D

2

Solution

The wavelength of last line of Balmer series
$\frac{1}{λ_{B}} = R c (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R c}{4}$
The wavelength of last line of Lyman series
$\frac{1}{λ_{B}} = R c (\frac{1}{1^{2}} - \frac{1}{\infty^{2}}) = R c$
$∴ \frac{λ_{B}}{λ_{L}} = \frac{4}{1} = 4$

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