Question

If the wavelength of the first line of the Balmer series of hydrogen is $6561 \, ?$, the wavelength of the second line of the series should be

• A. $13122 ?$
• B. $3280 ?$
• C. $4860 ?$
• D. $2187 ?$

Answer 1

Answer: (C)

$4860 ?$

Answer 2

For Balmer series, $n_1 = 2, n_2 = 3$ for $1^{st}$ line and $n_2 = 4$ for second line. $\frac{\lambda_{1}}{\lambda_{2}} = \frac{\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)}{\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}$ $= \frac{3/16}{5/36} = \frac{3}{16} \times\frac{36}{5}$ $=\frac{27}{20}$ $\lambda_{2} = \frac{20}{27}\lambda_{1}$ $= \frac{20}{27}\times6561$ $= 4860 ?$