Q. If the wavelength of the first line of the balmer series of hydrogen is $6561\mathring{\mathrm{A}}$, the wavelength of the second line of the series should be
A) $13122\ \mathring{\mathrm{A}}$
B) $3280\ \mathring{\mathrm{A}}$
C) $4860\ \mathring{\mathrm{A}}$
D) $2187\ \mathring{\mathrm{A}}$
Explanation
(c) For Balmer series, $n_{1} - 2,\ n_{2} = 3$ for 1st line and $n_{2} = 4$ for second line
$\frac{\lambda_{1}}{\lambda_{2}} = \left( \frac{\frac{1}{2^{2}} - \frac{1}{4^{2}}}{\frac{1}{2^{2}} - \frac{1}{3^{2}}} \right) = \frac{3/16}{5/16} = \frac{3}{16} \times \frac{36}{5} = \frac{27}{20}$
$\lambda_{2} = \frac{20}{27}\lambda_{1} = \frac{20}{27} \times 6561 = 4860\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.