The wavelength of radiation emitted is ' $\lambda_0$ ' when an electron jumps from the second excited state…
The wavelength of radiation emitted is ' $\lambda_0$ ' when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be $\frac{20}{x} \lambda_0$. The value of $x$ is
$3$
$9$
$13$
$27$
Solution
According to Rydberg's formula, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
When electron jumps from $2^{\text {nd }}$ exited state to first exited state, $\mathrm{n}_2=3, \mathrm{n}_1=2, \lambda=\lambda_0$, we get
$\frac{1}{\lambda_0}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$
When electron jumps from $3^{\text {rd }}$ exited state to $2^{\text {nd }}$ orbit,
$\mathrm{n}_2=4, \mathrm{n}_1=2$, we get
$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^2}-\frac{1}{2^2}\right)$
$\begin{aligned}
\therefore \quad \frac{\lambda}{\lambda_0} & =\frac{\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)} \\
& =\frac{5}{36} \times \frac{16}{3}=\frac{20}{27}
\end{aligned}$
$\begin{aligned}
& \therefore \quad \lambda=\frac{20}{27} \lambda_0 \\
& \Rightarrow \mathrm{x}=27
\end{aligned}$