Q. Of the following transition in the hydrogen atom, the one which gives an emission line of the highest frequency is
A) $n$ =1 to $\ n$=2
B) $n$ =2 to $\ n$=1
C) $n$ =3 to $\ n$=10
D) $n$ =10 to $\ n$=3
Explanation
(b) E$_{1}$ = $- 13.6\ –( - 3.4)\ = \ - 10.$2 eV
$E_{2}\ $= $- \ 3.4\ –\ (\ - \ 13.6)\ = \ + \ 10.$2 eV
$E_{3}$ =$\ - \ 0.136\ –\ (\ - 1.51)\ = \ - \ 1.374$ eV
$E_{4}\ =$ $- \ 1.51\ –\ (\ - \ 0.136)\ = - \ 1.374\ $eV
When an electron makes transition from higher energy level having energy $E_{2}(n_{2})\ $to lower energy level having energy$E_{1}(n_{1})$, then a photon of frequency $v$ is emitted.
Here, for emission line $E_{1}\ $is maximum hence, it will have the highest frequency emission line.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.