Q. If the electron in hydrogen atom jumps from the third to second orbit, the wavelength of the emitted radiation in terms of Rydberg constant R is given by
A) $\lambda\ = \frac{36}{5R}$
B) $\lambda\ = \frac{5R}{36}$
C) $\lambda\ = \frac{5}{R}$
D) $\lambda\ = \frac{R}{6}$
Explanation
(a) From Bohr’s model of atom, the wave number is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ = $R$ $\left( \frac{1}{n_{1}^{\ \ \ 2}} - \frac{1}{n_{2}^{\ \ 2}} \right)$
where$R$ is Rydberg’s constant and $n_{1}$ and $n_{2}$ the energy levels.
Given, $n_{1}$=2, $n_{2}$=3
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda} = R\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda} = R\left\lbrack \frac{5}{36} \right\rbrack$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{36}{5R}$
This gives corresponding wavelength of Balmer series.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.