Question

A particular hydrogen like ion emits radiation of frequency $2.92\times {10}^{15}\text{Hz}$ when it makes transition from $n=3$ to $n=1$. The frequency (in $\text{Hz}$) of radiation emitted, in transition from $n=2$ to $n=1$, will be:

• A. $2.46\times {10}^{15}$
• B. $4.38\times {10}^{15}$
• C. $0.44\times {10}^{15}$
• D. $6.57\times {10}^{15}$

Answer 1

The correct option is A $2.46\times {10}^{15}$

The wavelength of the radiation emitted during the transition of an electron, is given by,

$\frac{1}{\lambda }=R(\frac{1}{p^2}-\frac{1}{n^2})$

Here, $R$ is Rydberg's constant.

As, $c=\nu \lambda \Rightarrow \lambda =\frac{c}{\nu }$

$\therefore \frac{\nu }{c}=R(\frac{1}{p^2}-\frac{1}{n^2})$

For the transition from $3\to 1$

$\frac{{\nu }_1}{c}=R(\frac{1}{1^2}-\frac{1}{3^2})$

$\therefore {\nu }_1=\frac{8cR}{9}......\left(1\right)$

For the transition from $2\to 1$

$\frac{{\nu }_2}{c}=R(\frac{1}{1^2}-\frac{1}{2^2})$

$\therefore {\nu }_2=\frac{3cR}{4}......\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$, we get

$\frac{{\nu }_1}{{\nu }_2}=\frac{32}{27}$

$\therefore {\nu }_2={\nu }_1\times \frac{27}{32}=2.92\times {10}^{15}\times \frac{27}{32}$


${\nu }_2=2.46\times {10}^{15}\text{Hz}$

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