Question

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

• A. $\frac{3}{16}\times {10}^{15}\text{Hz}$
  
• B. $\frac{8}{16}\times {10}^{15}\text{Hz}$
  
• C. $\frac{9}{16}\times {10}^{15}\text{Hz}$
  
• D. $\frac{3}{4}\times {10}^{15}\text{Hz}$
  

Answer 1

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


We know that,

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

Here, $n_1=2$, $n_2=4$

$\Rightarrow \frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{4^2})$

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

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

Using,
$\nu =\frac{c}{\lambda }$

$\Rightarrow \nu =\frac{3\times {10}^8}{\left(\frac{16}{3}\right)\times {10}^{-7}}$

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

Hence, option $\left(C\right)$ is correct.

Why this question? This question gives an idea to calculate the frequency of emitted radiation.