Chapter 7, Problem 9PEB

To determine

The amount of energy of a photon of ultraviolet light greater than the energy of an average photon of visible light.

Answer to Problem 9PEB

The energy of a photon of ultraviolet light is $3.08\times {10}^{-19}\text{J}$ more than the energy of an average photon of visible light.

Explanation of Solution

Write the expression for the energy of a photon.

    $E=hf$                                                                                   (I)

Here, $E$ is the energy of a photon $h$ is the Planck's constant and $f$ is the frequency.

Write the expression for the speed of light.

    $\begin{array}{c}c=\lambda f\\ f=\frac{c}{\lambda }\end{array}$

Here, $c$ is the speed of light and $\lambda$ is the wavelength of light.

Substitute $\frac{c}{\lambda }$ for $f$ in equation (I) to find $E$.

    $E=\frac{hc}{\lambda }$                                                                               (II)

Write the expression for the difference in energy of ultraviolet light and the visible light.

    $\Delta E=E_{\text{UV}}-E_{\text{visible}}$                                                              (III)

Here, $E_{\text{UV}}$ is the energy of ultraviolet light and $E_{\text{visible}}$ is the energy of visible light.

Conclusion:

Substitute $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3.0\times {10}^8\text{m/s}$ for $c$ and $3.00\times {10}^{-7}\text{m}$ for $\lambda$ in equation (II) to find $E_{\text{UV}}$.

    $\begin{array}{c}{E}_{\text{UV}}=\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m/s}\right)}{3.00\times {10}^{-7}\text{m}}\\ =6.63\times {10}^{-19}\text{J}\end{array}$

Substitute $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3.0\times {10}^8\text{m/s}$ for $c$ and $5.60\times {10}^{-7}\text{m}$ for $\lambda$ in equation (II) to find $E_{\text{visible}}$.

    $\begin{array}{c}{E}_{\text{visible}}=\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m/s}\right)}{5.60\times {10}^{-7}\text{m}}\\ =3.55\times {10}^{-19}\text{J}\end{array}$

Substitute $6.63\times {10}^{-19}\text{J}$ for $E_{\text{UV}}$ and $3.55\times {10}^{-19}\text{J}$ for $E_{\text{visible}}$ in equation (III) to find $\Delta E$.

    $\begin{array}{c}\Delta E=6.63\times {10}^{-19}\text{J}-3.55\times {10}^{-19}\text{J}\\ =3.08\times {10}^{-19}\text{J}\end{array}$

Therefore, the energy of a photon of ultraviolet light is $3.08\times {10}^{-19}\text{J}$ more than the energy of an average photon of visible light.