The ratio of ionization energy of a hydrogen atom in fifth excited...

The ionization energy of a hydrogen atom in the $n$-th state is given by $IE_n = \dfrac{13.6}{n^2}$ eV.

For the fifth excited state, the principal quantum number is $n_1 = 6$.

For the second excited state, the principal quantum number is $n_2 = 3$.

The ratio of their ionization energies is:

$\dfrac{IE_6}{IE_3} = \dfrac{\dfrac{13.6}{6^2}}{\dfrac{13.6}{3^2}}$

$\dfrac{IE_6}{IE_3} = \dfrac{3^2}{6^2} = \dfrac{9}{36} = \dfrac{1}{4}$

The ratio is $1:4$.

Answer: $1:4$