Question

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $n_i$ to another with quantum number $n_f$. $V_i$ and $V_f$ are respectively the initial and final potential energies of the electron. If $\frac{V_f}{V_i}=6.25$, then the smallest possible $n_f$ is

Answer 1

Potential energy of a electron in Bohr's modal is given by (assuming Coulombic force) $U=\frac{-Kz{e}^2}{\gamma }$
where, $\gamma =$ radious
& radius for Bohr's orbital (for mono-electronic system) =$0.529\frac{n^2}{z}$
so, $\frac{U_i}{U_r}=\frac{{\gamma }_2}{{\gamma }_1}=\frac{n_2^2}{n_1^2}$ ...(1)

As, given $\frac{n_2}{n_1}=6.25$
by comparing above equation to equation (1).
$\frac{n_2}{n_1}=\sqrt{6.25}\frac{n_2}{n_1}=2.5$

So, $\frac{n_2}{n_1}=\frac{5}{2}$
it can be written as,
$n_2=5,n_1=2$
so the final value of $n_2=5.$