The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the $$5^{th}$$ excited state of a hydrogen atom is :

For a hydrogen atom, the energies of an electron in the $$n$$-th state are related by the virial theorem:

$$\text{Kinetic Energy (KE)} = -E_n$$

$$\text{Potential Energy (PE)} = 2E_n$$

$$\text{Total Energy} = E_n$$

where $$E_n$$ is the total energy of the electron in the $$n$$-th state (which is negative).

The ratio of the magnitude of kinetic energy to the magnitude of potential energy is:

$$\frac{|KE|}{|PE|} = \frac{|-E_n|}{|2E_n|} = \frac{|E_n|}{2|E_n|} = \frac{1}{2}$$

This ratio is independent of the quantum number $$n$$ and holds for any state, including the 5th excited state ($$n = 6$$).

The correct answer is $$\frac{1}{2}$$.