Question

If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å, the wavelength of the second line of the series should be

• A. 13122 Å
• B. 3280 Å
• C. 4860 Å
• D. 2187 Å

Answer 1

Answer: (C)

4860 Å

Answer 2

For Balmer series $n_{1} = 2,n_{2} = 3$ for 1^st line and $n_{2} = 4$ for second line.

$\frac{\lambda_{1}}{\lambda_{2}} = \frac{\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right)}{\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)} = \frac{3/16}{5/36} = \frac{3}{16} \times \frac{36}{5} = \frac{27}{10}$

$\lambda_{2} = \frac{20}{27}\lambda_{1} = \frac{20}{27} \times 6561 = 4860Å$