Question

An electron jumps from the $4^{th}$ orbit to the $2^{nd}$ orbit of hydrogen atom $(R={10}^{5}C{m}^{-1})$. Frequency is in $Hz$ of emitted radiation will be

• A. $\frac{3\times {10}^5}{16}$
• B. $\frac{3\times {10}^{15}}{16}$
• C. $\frac{9\times {10}^{15}}{16}$
• D. $\frac{9\times {10}^5}{16}$

Answer 1

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

$\begin{array}{c}\text{Electron jumps from}{4}^{\text{th}}\text{orbit to}\\ 2^{\text{nd}}\text{orbit.}\\ \text{Let wavelength be "x" then}\\ \frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]\end{array}$

$\text{Here,}\begin{array}{cc}& R={10}^5{cm}^{-1}\\ & z=1\\ & n_1=2,n_2=4\\ \frac{1}{\lambda }& ={10}^5\left[\frac{1}{2^2}\frac{1}{4^2}\right]\\ \frac{1}{\lambda }& ={10}^5[\frac{1}{4}-\frac{1}{16}]\\ \frac{1}{\lambda }& ={10}^5\left[\frac{3}{16}\right]\\ \frac{1}{\lambda }& =\frac{3\times {10}^5}{16}-\left(1\right)\end{array}$

$\begin{array}{c}\text{Frequency is}\frac{c}{\lambda }\\ c=3\times {10}^{8}m=3\times {10}^{10}cm\\ \text{From frequency,}v=3\times {10}^{10}\times \frac{3\times {10}^5}{16}\\ v=\frac{9\times {10}^{15}}{16}\end{array}$