In hydrogen atoms, transition from the state $n=6$ to $n=1$ result in ultraviolet radiation. Infrared…
In hydrogen atoms, transition from the state $n=6$ to $n=1$ result in ultraviolet radiation.
Infrared radiation will be obtained in the transition
$\mathrm{n}=6$ to $\mathrm{n}=2$
$\mathrm{n}=5$ to $\mathrm{n}=3$
$\mathrm{n}=3$ to $\mathrm{n}=5$
$\mathrm{n}=4$ to $\mathrm{n}=2$
Solution
$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
For $\mathrm{UV}_1=1$ and $\mathrm{n}_2=6, \lambda \propto \frac{1}{\mathrm{R}} \frac{36}{35}$
For infrared the wavelength must be greater the UV wavelength.
For $\mathrm{n}_1=3$ and $\mathrm{n}_2=5, \lambda \propto \frac{1}{\mathrm{R}} \frac{225}{16}$
This is the largest wavelength we can get. So, it is the answer.