Which energy level transition among the following will have the least wavelength?
Which energy level transition among the following will have the least wavelength?
$n_4 \longrightarrow n_3$
$n_4 \longrightarrow n_2$
$n_4 \longrightarrow n_1$
$n_2 \longrightarrow n_1$
Solution
According to energy, $E=\frac{h}{m v}$ or $E=\frac{h c}{\lambda}$
i.e $E \propto \frac{1}{\lambda}$
Here, energy of electrons is inversely proportional to wavelength i.e. greater the difference between the transition, lesser will be the wavelength.
$
\begin{aligned}
& \Delta n \Rightarrow n_4-n_3=n_1 \\
& n_4 \rightarrow n_2=n_2 \\
& n_4 \rightarrow n_1=n_3 \text { i.e. highest energy level transition } \\
& n_2 \rightarrow n_1=n_1 \\
& \therefore n_4 \rightarrow n_1 \text { having higher energy level transition. }
\end{aligned}
$
$\therefore n_4 \rightarrow n_1$ having higher energy level transition. So, it has least wavelength
So, it has least wavelength
$
\frac{1}{\lambda}=R_{\mathrm{H}}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]
$
$\therefore R_{\mathrm{H}}=$ Rydberg constant,
$
\begin{aligned}
& \frac{1}{\lambda}=R_{\mathrm{H}}\left[\frac{1}{1^2}-\frac{1}{4^2}\right] \\
& \frac{1}{\lambda}=R_{\mathrm{H}} \times \frac{15}{16} \\
& \frac{1}{\lambda}=109737 \times \frac{15}{16} \\
& \lambda=\frac{1}{102878.43} \\
& \lambda=9.72 \mathrm{~nm}
\end{aligned}
$
hence, $n_4 \rightarrow n_1$ having least wavelength