Q. The wavelength of radiation emitted is $\lambda_{0}$ when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of hydrogen atom,the wavelength of radiation emitted will be
A) $\frac{16}{25}\lambda_{0}$
B) $\frac{20}{27}\lambda_{0}$
C) $\frac{27}{20}\lambda_{0}$
D) $\frac{25}{16}\lambda_{0}$
Explanation
(b) Wavelength ($\lambda)$ during transition from $n_{2}$to $n_{1}$ is given by
$\frac{1}{\lambda} =$ $R\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$\Rightarrow \frac{1}{\lambda_{3 \rightarrow 2}}$ $= \ R$ $\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack =$ $\frac{5R}{36}$
and$\frac{1}{\lambda_{4 \rightarrow 2}}$ = $r\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack =$ $\frac{3R}{16}$
$\therefore\ \ \ \frac{\lambda_{4 \rightarrow 2}}{\lambda_{3 \rightarrow 2}} = \ \frac{20}{27}$
$\Rightarrow \ \lambda_{4 \rightarrow 2} = \frac{20}{27}\lambda_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.