Chapter 28, Problem 50A

(a)

To determine

To show:A hydrogen atom in level $2$ will be ionized by a photon of wavelength $332\text{ nm}$ .

Answer to Problem 50A

The atom will be ionized as energy of the photon ( $3.74\text{ eV}$ ) is greater than the ionization energy ( $-3.4\text{ eV}$ ) of the level.

Explanation of Solution

Given:

Wavelength of photon, $\lambda =332\text{ nm}$

Atomic level, $\text{n=2}$

Formula used:

Energy of an emitted photon

  $\begin{array}{l}E=hf\\ E=\frac{hc}{\lambda }\text{ [}f=\frac{c}{\lambda }]\end{array}$

where $E$ is the energy of the photon, $h$ is the Planck’s constant, $f$ is the photon’s frequency, $c$ is the speed of light and $\lambda$ is he wavelength of the photon.

The energy of $n$ th energy level of hydrogen is determined by

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

Calculation:

As per problem

Calculate the energy of the photon

  $\begin{array}{l}\lambda =332\text{ nm}\\ c=3\times {10}^8\text{ m/s}\\ h=6.63\times {10}^{-34}\text{ Js}\end{array}$

Substitute the values,

  $\begin{array}{l}E=\frac{hc}{\lambda }\\ E=\frac{(6.63\times {10}^{-34}\text{ Js})(3\times {10}^8\text{ m/s})}{332\times {10}^{-9}\text{ m}}\\ E=5.99\times {10}^{-19}\text{ J}\\ E=3.74\text{ eV}\end{array}$

Therefore the energy of photon is $3.74\text{ eV}$ .

Calculate the value of ionization energy of $\text{n=2}$ level

  $\begin{array}{l}{E}_n=\frac{-13.6\text{ eV}}{{\text{n}}^2}\\ E_2=\frac{-13.6\text{ eV}}{{(2)}^2}\\ E_2=-3.4\text{ eV}\end{array}$

Thus the ionization energy of the level is $-3.4\text{ eV}$ .

The energy of the photon is greater than the ionization energy of level $\text{n=2}$

Hence, the atom will be ionized.

Conclusion:

The atom will be ionized as energy of the photon is greater than the ionization energy of the level.

(b)

To determine

To calculate:The kinetic energy of the electron in joules.

Answer to Problem 50A

The kinetic energy of the atom is $K.E\text{.=}5.44\times {10}^{-20}\text{ J}$ .

Explanation of Solution

Given:

As calculated in part (a.)

The energy of the photon, $E=3.74\text{ eV}$

The ionization energy of $\text{n=2}$ level, $E_2=-3.4\text{ eV}$

Calculation:

The energy of the photon, $E=3.74\text{ eV}$

The ionization energy of $\text{n=2}$ level, $E_2=-3.4\text{ eV}$

The atom absorbs the photon’s energy to get ionized

The kinetic energy of the atom

  $\begin{array}{l}K.E=3.74\text{ eV+(-}3.4\text{ eV)}\\ K\text{.}E\text{.=0}\text{.34 eV}\\ K\text{.}E\text{.=(}0.34\text{ eV})(1.6\times {10}^{-19}\text{ J/eV)}\\ K.E\text{.=}5.44\times {10}^{-20}\text{ J}\end{array}$

Conclusion:

The kinetic energy of the atom is $K.E\text{.=}5.44\times {10}^{-20}\text{ J}$ .