Q. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant =$1.097 \times 10^{7}m^{- 1}$)
A) 91 nm
B) 192 nm
C) 406 nm
D) $9.1 \times 10^{- 8}nm$
Explanation
(a)
$\frac{1}{\lambda} = {\overline{v}}_{H} = R_{H}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$= 1.097 \times 10^{7}\left\lbrack \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right\rbrack$
$\lambda = \frac{1}{1.097 \times 10^{7}}m$
$= 9.11 \times 10^{- 8}m$
$= 91.1 \times 10^{- 9}m$
$= 91.1nm$
$(1\ nm = 10^{- 9}\ m)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.