Question

How much energy does it take to ionize a hydrogen atom that is in its first excited state?

Short answer

The excited hydrogen atom's ionization energy will be $3.4$ eV or less.

Step-by-step solution

Given information

The given energy of the first excited state of a hydrogen atom

Finding ionization energy of hydrogen

The energy of an electron in a quantum state is calculated as follows:

$E_n=-\frac{13.6\mathrm{eV}}{n^2}$

For the first excited state, the quantum number n is given by $n=2$. Hence the energy of the electron in the first excited state is

$\begin{array}{rcl}{E}_2& =& -\frac{13.6\mathrm{eV}}{2^2}\\ & =& -3.4\mathrm{eV}\end{array}$

The energy of the electron after ionization is

$E_{\text{ion}}=0$

As a result, the energy we must provide is

$\begin{array}{rcl}\Delta E& =& E_{\text{ion}}-E_2\\ & =& 0-(-3.4\mathrm{eV})\\ & =& 3.4\mathrm{eV}\end{array}$

As a result, ionizing a hydrogen atom from its first excited state will require $3.4$ eV of energy.