Chapter 28, Problem 49A

To determine

To calculate:The kinetic energy of the electron ejected from the atom.

Answer to Problem 49A

The electron is ejected with kinetic energy $\text{0}\text{.4 eV}$ from the atom.

Explanation of Solution

Given info:

Energy of the photon = $14.0\text{ eV}$

Formula used:

For hydrogen atom, ionization energy is $-13.6\text{ eV}$ .

Kinetic energy = excited energy $\text{+}$ ionization energy.

Calculation:

As per problem,

Energy of the photon = $14.0\text{ eV}$

The energy of the photon excites the electron. If it exceeds the ionization energy the electron is ejected from the nucleus with kinetic energy given as

Kinetic energy = excited energy $\text{+}$ ionization energy.

For hydrogen atom,

Ionization energy = $-13.6\text{ eV}$ .

Therefore,

Kinetic energy of the electron ejected is

  $\begin{array}{l}K.E.=14.0\text{ eV+(-}13.6\text{ eV)}\\ K.E\text{.=0}\text{.4 eV}\end{array}$

Conclusion:

Thus, the electron is ejected with kinetic energy $\text{0}\text{.4 eV}$ from the atom.