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

Open in App

Solution

Potential energy of a electron in Bohr's modal is given by (assuming Coulombic force) $U = \frac{- K z e^{2}}{γ}$
where, $γ =$ radious
& radius for Bohr's orbital (for mono-electronic system) = $0.529 \frac{n^{2}}{z}$
so, $\frac{U_{i}}{U_{r}} = \frac{γ_{2}}{γ_{1}} = \frac{n_{2}^{2}}{n_{1}^{2}}$ ...(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 $n_{2} = 5.$

Suggest Corrections

Similar questions

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