Chapter 5, Problem 41Q

To determine

The photon energy and photon wavelength and the transition which absorb or emit the visible light.

Answer to Problem 41Q

The wavelength of transition from $0\text{ eV}$ to $1\text{ eV}$ is $1240\text{ nm}$ and this wavelength corresponds to infrared region. The wavelength of transition from $0\text{ eV}$ to $3\text{ eV}$ is $413.5\text{ nm}$ and this wavelength corresponds to the visible region, and the wavelength of transition from $1\text{ eV}$ to $3\text{ eV}$ is $620\text{ nm}$ and this wavelength corresponds to the visible region.

Explanation of Solution

Given:

The first energy level is $0\text{ eV}$.

The second energy level is $1\text{ eV}$.

The third energy level is $3\text{ eV}$.

Concept used:

Draw the energy level diagram for a photon.

The possible transitions of an atom are shown in the above figure, the possible transitions are, $0\text{ eV}$ to $1\text{ eV}$, $1\text{ eV}$ to $3\text{ eV}$ and $0\text{ eV}$ to $3\text{ eV}$.

Write the expression for the Energy of the photon.

$E=\frac{hc}{\lambda }$   ............ (1)

Here, $E$ is the energy, $h$ is the plank’s constant, $c$ is the speed of light and $\lambda$ is the wavelength of the photon.

Rearrange Equation (1) for $\lambda$.

$\lambda =\frac{hc}{E}$   ............ (2)

Calculation:

For transition between $0\text{eV}$ to $1\text{ eV}$.

Substitute $1\text{ eV}$ for $E$, $4.135\times {10}^{-15}\text{}\text{eV}\cdot \text{s}$ for $h$, ${\lambda }_1$ for $\lambda$ and $3\times {10}^8\text{ m/s}$ for $c$ in equation (2).

$\begin{array}{c}{\lambda }_1=\frac{\left(4.135\times {10}^{-15}\text{ eV}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{1\text{ eV}}\\ =1240\times {10}^{-9}\text{m}\left(\frac{1\text{ nm}}{{10}^{-9}\text{}\text{m}}\right)\\ =1240\text{ nm}\end{array}$

The wavelength of transition from $0\text{ eV}$ to $1\text{ eV}$ is $1240\text{ nm}$ and this wavelength corresponds to the infrared region.

For transition between $0\text{eV}$ to $3\text{ eV}$.

Substitute $3\text{ eV}$ for $E,$

$4.135\times {10}^{-15}\text{}\text{eV}\cdot \text{s}$ for $h$, ${\lambda }_2$ for $\lambda$ and $3\times {10}^8\text{ m/s}$ for $c$ in equation (2).

$\begin{array}{c}{\lambda }_2=\frac{\left(4.135\times {10}^{-15}\text{ eV}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{3\text{ eV}}\\ =413.5\times {10}^{-9}\text{m}\left(\frac{1\text{ nm}}{{10}^{-9}\text{}\text{m}}\right)\\ =413.5\text{ nm}\end{array}$

The wavelength of transition from $0\text{ eV}$ to $3\text{ eV}$ is $413.5\text{ nm}$ and this wavelength corresponds to the visible region.

For transition between $1\text{eV}$ to $3\text{ eV}$.

Substitute $2\text{ eV}$ for $E,$

$4.135\times {10}^{-15}\text{}\text{eV}\cdot \text{s}$ for $h$, ${\lambda }_3$ for $\lambda$ and $3\times {10}^8\text{ m/s}$ for $c$ in equation (2).

$\begin{array}{c}{\lambda }_3=\frac{\left(4.135\times {10}^{-15}\text{ eV}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\text{2 eV}}\\ =620\times {10}^{-9}\text{m}\left(\frac{1\text{ nm}}{{10}^{-9}\text{}\text{m}}\right)\\ =620\text{ nm}\end{array}$

The wavelength of transition from $1\text{ eV}$ to $3\text{ eV}$ is $620\text{ nm}$ and this wavelength corresponds to the visible region.

Conclusion:

Thus, the wavelength of transition from $0\text{ eV}$ to $1\text{ eV}$ is $1240\text{ nm}$ and this wavelength corresponds to infrared region. The wavelength of transition from $0\text{ eV}$ to $3\text{ eV}$ is $413.5\text{ nm}$ and this wavelength corresponds to the visible region, and the wavelength of transition from $1\text{ eV}$ to $3\text{ eV}$ is $620\text{ nm}$ and this wavelength corresponds to the visible region.