Question

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

• A. $\frac{3}{16}\times {10}^5$
• B. $\frac{3}{16}\times {10}^{15}$
• C. $\frac{9}{16}\times {10}^{15}$
• D. $\frac{3}{4}\times {10}^{15}$

Answer 1

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

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{16}\Rightarrow \lambda =\frac{16}{3R}=\frac{16}{3}\times {10}^{-5}cm$
Frequency n = $\frac{c}{\lambda }=\frac{3\times {10}^{10}}{\frac{16}{3}\times {10}^{-5}}=\frac{9}{16}\times {10}^{15}Hz$