Question

A radiation of wavelength ${\lambda }_0$ is emitted, 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 the emitted radiation will be,

• A. $\frac{16}{25}{\lambda }_0$
• B. $\frac{20}{27}{\lambda }_0$
• C. $\frac{27}{20}{\lambda }_0$
• D. $\frac{25}{16}{\lambda }_0$

Answer 1

The correct option is B $\frac{20}{27}{\lambda }_0$

We know that,
$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\Rightarrow \frac{1}{{\lambda }_{3\to 2}}=R[\frac{1}{2^2}-\frac{1}{3^2}]$

$\Rightarrow \frac{1}{{\lambda }_0}=\frac{5R}{36}\Rightarrow {\lambda }_0=\frac{36}{5R}$

And, $\frac{1}{{\lambda }_{4\to 2}}=R[\frac{1}{2^2}-\frac{1}{4^2}]=\frac{3R}{16}$
$\Rightarrow {\lambda }_{4\to 2}=\frac{16}{3R}$

$\therefore \frac{{\lambda }_{4\to 2}}{{\lambda }_0}=\frac{\frac{16}{3R}}{\frac{36}{5R}}=\frac{20}{27}\Rightarrow {\lambda }_{4\to 2}=\frac{20}{27}{\lambda }_0$

Hence, $\left(B\right)$ is the correct answer.

Why this Question? This question is based on wavelength of emitted radiation.