Find Smallest Final Quantum Number from Potential Energy Ratio in Hydrogen Atom

According to Bohr's model for the hydrogen atom, the radius of the $n$-th orbit is proportional to $n^2$, i.e., $r_n \propto n^2$.

The potential energy $V_n$ of the electron in the $n$-th orbit is inversely proportional to the radius $r_n$: $V_n = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r_n} \propto -\frac{1}{n^2}$

The ratio of the initial potential energy $V_i$ to the final potential energy $V_f$ is given by: $\frac{V_i}{V_f} = \frac{-\frac{1}{n_i^2}}{-\frac{1}{n_f^2}} = \left(\frac{n_f}{n_i}\right)^2$

Given that $\frac{V_i}{V_f} = 6.25$, we can write: $\left(\frac{n_f}{n_i}\right)^2 = 6.25 = \frac{25}{4}$

Taking the positive square root on both sides (since principal quantum numbers $n_i$ and $n_f$ are positive integers): $\frac{n_f}{n_i} = \frac{5}{2} \implies n_f = \frac{5}{2} n_i$

For $n_f$ to be an integer, $n_i$ must be a multiple of $2$.

To find the smallest possible value of $n_f$, we take the smallest possible integer value for $n_i$, which is $n_i = 2$: $n_f = \frac{5}{2} \times 2 = 5$

Thus, the smallest possible value of $n_f$ is $5$.