Chapter 28, Problem 47A

To determine

The maximum energy a photon can have when an electron in an excited hydrogen atom drops through energy levels.

To determine what happens if the same amount of energy is given to the atom in the ground state.

Answer to Problem 47A

The maximum energy of photon emitted is $13.6\text{ eV}$ .

If $13.6\text{ eV}$ energy is supplied to the atom in the ground state, the electron will have enough energy to leave the nucleus.

Explanation of Solution

Given:

An electron in an excited hydrogen atom drops through energy levels.

Formula used:

In the line spectrum of hydrogen, the lowest energy level has an energy $-13.6\text{ eV}$ whereas the energy of $n$ th energy level is determined by the formula $E_n=-13.6{\text{ eV/n}}^2$ .

Calculation:

For the photon to have maximum energy, the electron makes a transition from the infinity level ( $n=\infty$ ) to the lowest level.

The energy of the infinity level is

  $\begin{array}{l}{E}_n=-13.6{\text{ eV/n}}^2\\ E_{n=\infty }=-13.6\text{ eV/}{\infty }^2\\ E_{n=\infty }=0\text{ eV}\end{array}$

So, the maximum energy of the photon is

  $\begin{array}{l}\Delta E=E_{\infty }-E_1\\ \Delta E=0\text{ eV-(-13}\text{.6 eV)}\\ \Delta E=13.6\text{ eV}\end{array}$

The ionization energy for hydrogen happens to be $13.6\text{ eV}$ .

Therefore, if $13.6\text{ eV}$ energy is supplied to the atom in the ground state, the electron will have enough energy to leave the nucleus.

Conclusion:

The maximum energy of photon emitted is $13.6\text{ eV}$ .

If $13.6\text{ eV}$ energy is supplied to the atom in the ground state, the electron will have enough energy to leave the nucleus.