The ionization energy of the electron in the hydrogen atom in its grounds state is $13.6 \mathrm{eV}$. The…
The ionization energy of the electron in the hydrogen atom in its grounds state is $13.6 \mathrm{eV}$. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between :
$n=4$ to $n=3$ states
$n=3$ to $n=2$ states
$n=3$ to $n=1$ states
$n=2$ to $n=1$ states
Solution
$\begin{aligned} & \frac{\mathrm{n}(\mathrm{n}-1)}{2}=6 \\ \Rightarrow & \mathrm{n}=4\end{aligned}$
For maximum wavelength energy difference between states should be minimum because
$\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}$
So, transition state in $\mathrm{n}=4$ to $\mathrm{n}=3$