Q. The electronic transitions from $n$=2 to $n$=1 will produce shortest wavelength in (where $n$=principle quantum state)
A) $Li^{2 +}$
B) $He^{+}$
C) $H$
D) $H^{+}$
Explanation
(a)
$\frac{1}{\lambda} = Z^{2}.R_{H}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$\Longrightarrow \frac{1}{\lambda} = (Z)^{2}.R_{H}\left\{ \frac{1}{1} - \frac{1}{4} \right\} = \frac{3}{4}R_{H}Z^{2}$
$\therefore\lambda \propto \frac{1}{Z^{2}}$
Hence for shortest $\lambda,\ Z$ must be maximum, which is for $Li^{2 +}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.