JEE Main 2025 Apr 7 Shift 1, Physics Q16: The Line Spectra of the Hydrogen Atom

The largest wavelength in a series corresponds to the smallest photon energy, i.e. the transition between the two lowest energy levels of that series.

Lyman series (final level $n=1$). Largest wavelength is the $n=2\to 1$ transition:

$\frac{1}{{\lambda }_1}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3R}{4}\Rightarrow {\lambda }_1=\frac{4}{3R}$

Balmer series (final level $n=2$). Largest wavelength is the $n=3\to 2$ transition:

$\frac{1}{{\lambda }_2}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5R}{36}\Rightarrow {\lambda }_2=\frac{36}{5R}$

Take the ratio:

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{4/(3R)}{36/(5R)}=\frac{4\cdot 5}{3\cdot 36}=\frac{20}{108}=\frac{5}{27}$

$\therefore$ Option (2).