Q. The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561\ \mathring{\mathrm{A}}$. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is
A) $1215\ \mathring{\mathrm{A}}$
B) $1640\ \mathring{\mathrm{A}}$
C) $2430\ \mathring{\mathrm{A}}$
D) $4687\ \mathring{\mathrm{A}}$
Explanation
(a) $\frac{1}{\lambda_{H_{2}}} = RZ_{H}^{2}\left\lbrack \frac{1}{4} - \frac{1}{9} \right\rbrack = R(1)^{2}\left\lbrack \frac{5}{36} \right\rbrack$
$\frac{1}{\lambda_{He}} = RZ_{He}^{2}\left\lbrack \frac{1}{4} - \frac{1}{16} \right\rbrack = R(4)\left\lbrack \frac{3}{16} \right\rbrack$
$\frac{\lambda_{He}}{\lambda_{H_{2}}} = \frac{1}{4}\left\lbrack \frac{16}{3} \times \frac{5}{36} \right\rbrack = \frac{5}{27}$
$\lambda_{He} = \frac{5}{27} \times 6561 = 1215\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.