Question

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

• A. $\frac{9}{16}\times {10}^{13}$
• B. $\frac{9}{16}\times {10}^{15}$
• C. $\frac{9}{16}\times {10}^{10}$
• D. $\frac{4}{3}\times {10}^{13}$

Answer 1

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

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

Here, $n_1=2,n_2=4$

So, $\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{16}$

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

Frequency, $f=\frac{c}{\lambda }=\frac{3\times {10}^8}{\frac{16}{3}\times {10}^{-7}}=\frac{9}{16}\times {10}^{15}\text{Hz}$

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

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