Question

An electron jumps from the $4^{th}$ orbit to the $2^{nd}$ orbit of hydrogen atom. Given Rydberg's constant $R={10}^5{\text{cm}}^{-1}.$
The frequency of the emitted radiation will be (in $\text{Hz}$)

• A. $\frac{9}{16}\times {10}^{15}$
• B. $\frac{6}{16}\times {10}^{15}$
• C. $\frac{3}{6}\times {10}^{15}$
• D. $\frac{3}{16}\times {10}^{15}$

Answer 1

The correct option is A $\frac{9}{16}\times {10}^{15}$

The wavelength of the emitted radiation will be,

$\frac{1}{\lambda }=R[\frac{1}{n_2^2}-\frac{1}{n_4^2}]$

$\frac{1}{\lambda }={10}^5[\frac{1}{2^2}-\frac{1}{4^2}]={10}^5[\frac{1}{4}-\frac{1}{16}]$

$\frac{1}{\lambda }=\frac{3}{16}\times {10}^5{\text{cm}}^{-1}=\frac{3}{16}\times {10}^7{\text{m}}^{-1}$

$\nu =\frac{c}{\lambda }=3\times {10}^8\times \frac{3}{16}\times {10}^7$

$\nu =\frac{9}{16}\times {10}^{15}\text{Hz}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.