In a hydrogen atom, the ratio of the kinetic energies of an...

The kinetic energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is inversely proportional to the square of the principal quantum number $n$, i.e., $K \propto \dfrac{1}{n^2}$.

Given the ratio of kinetic energies in states A and B is $K_A : K_B = 9:1$.

Therefore, $\dfrac{K_A}{K_B} = \dfrac{n_B^2}{n_A^2} = 9$

State A is the first excited state, which corresponds to $n_A = 2$.

Substituting $n_A = 2$ into the equation:
$\dfrac{n_B^2}{2^2} = 9$
$n_B^2 = 9 \times 4 = 36$
$n_B = 6$

The principal quantum number $n=6$ corresponds to the $5^{\text{th}}$ excited state.

Assuming the first excited state is $n=1$ leads to the incorrect answer of the $2^{\text{nd}}$ excited state. Forgetting to square the quantum numbers leads to the $17^{\text{th}}$ excited state.

Answer: $5^{\text{th}}$ excited state