Q. If the shortest wavelength of H-atom in Lyman series is $x$, the longest wavelength in Balmer series of $He^{+}$ is

A) $\frac{36x}{5}$
B) $\frac{5x}{9}$
C) $\frac{x}{4}$
D) $\frac{9x}{5}$
Explanation

(d)
$\overline{v} = \frac{1}{\lambda} = R^{'}Z^{2}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
For shortest wavelength (maximum energy) in Lyman series of hydrogen $Z = 1,n_{1} = 1,n_{2} \longrightarrow \infty$ and
$\lambda\mathbf{=}x$
$\frac{1}{x} = R'$
For longest wavelength (minimum energy) in Balmer series of $He^{+},Z = 2$ and $n_{1} = 2,n_{2} = 3$
$\frac{1}{\lambda} = R^{'}2^{2}\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack$
$\frac{1}{\lambda} = \frac{4}{x}\left\lbrack \frac{1}{4} - \frac{1}{9} \right\rbrack$
$\frac{1}{\lambda} = \frac{4}{x}\frac{5}{36}$
$\lambda = \frac{9x}{5}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.