Question

The ionization potential of the hydrogen atom is $13.6eV$. What is the energy required to remove an electron from the $n=2$ state of the hydrogen atom.

Answer 1

Given:

The ionization potential of hydrogen atom $=13.6eV$

We know that,

$E\left(n\right)=\frac{-13.6}{n^2}eV/atom$

Here, the electron is to be removed from $n=2$. Therefore,

$\Delta E=E\left(\infty \right)-E\left(2\right)$

$\Delta E=0-\left(\frac{-13.6}{2^2}\right)$

$\Delta E=\frac{13.6}{4}$

$\Delta E=3.4eV$

Hence, this is the required result.