Q. Which transition in the hydrogen atomic spectrum will have the same wavelength as the transition, $n$=4 to $n$=2 of $He^{+}$ spectrum?
A) $n = 4\ to\ n = 3$
B) $n = 3\ to\ n = 2$
C) $n = 4\ to\ n = 2$
D) $n = 2\ to\ n = 1$
Explanation
(d)
$\frac{1}{\lambda} = Z^{2},\ R_{H}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$For\ He^{+},\frac{1}{\lambda} = 2^{2}.R_{H}\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack = 4 \times \frac{3}{16} = \frac{3}{4}$
$For\ H,\frac{1}{\lambda} = 1^{2}.R_{H}\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack = \frac{3}{4}$
Hence, for hydrogen $n = 2\ to\ n = 1$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.