Chapter 4, Problem 4.61P

Interpretation Introduction

Interpretation:

The color of visible light that has enough energy to eject an electron from metal foil that has threshold frequency of $5.45\times {10}^{14}\text{Hz}$ has to be identified.

Concept Introduction:

The minimum frequency of radiation that is required to eject an electron from metal is known as threshold frequency $({\nu }_0)$ of the metal.  This value differs for each and every metal.  Minimum energy that is required to eject an electron is $h{\nu }_0$.

Energy that is required in excess to eject an electron is given by the equation shown below.

    $E=h\nu -h{\nu }_0$

Explanation of Solution

Threshold frequency for metal is given as $5.45\times {10}^{14}\text{Hz}$.  This is same as $5.45\times {10}^{14}{\text{s}}^{\text{-1}}$.  The minimum energy that is required to eject an electron the given metal foil is calculated as shown below.

    $\begin{array}{c}E=h{\nu }_0\\ =6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 5.45\times {10}^{14}{\text{s}}^{\text{-1}}\\ =\text{36}\text{.1117}\times {10}^{-20}\text{J}\\ =\text{36}\text{.11}\times {10}^{-20}\text{J}\\ \text{=3}\text{.611}\times {\text{10}}^{\text{-19}}\text{J}\end{array}$

Violet light is said to have wavelength of $\text{400}\text{nm}$.  Therefore, the energy of photon that has wavelength of $\text{400}\text{nm}$ is calculated as shown below.

    $\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{400\times {10}^{-9}\text{m}}\\ =\text{0}\text{.0496602135}\times {10}^{-17}\text{J}\\ =4.96\times {10}^{-19}\text{J}\end{array}$

Indigo light is said to have wavelength of $\text{445}\text{nm}$.  Therefore, the energy of photon that has wavelength of $\text{445}\text{nm}$ is calculated as shown below.

    $\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{445\times {10}^{-9}\text{m}}\\ =0.04463839416\times {10}^{-17}\text{J}\\ =4.45\times {10}^{-19}\text{J}\end{array}$

Blue light is said to have wavelength of $\text{475}\text{nm}$.  Therefore, the energy of photon that has wavelength of $\text{475}\text{nm}$ is calculated as shown below.

    $\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{475\times {10}^{-9}\text{m}}\\ =\text{0}\text{.04181912715}\times {10}^{-17}\text{J}\\ =4.18\times {10}^{-19}\text{J}\end{array}$

Green light is said to have wavelength of $\text{510}\text{nm}$.  Therefore, the energy of photon that has wavelength of $\text{510}\text{nm}$ is calculated as shown below.

    $\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{510\times {10}^{-9}\text{m}}\\ =\text{0}\text{.0389491870}\times {10}^{-17}\text{J}\\ =3.89\times {10}^{-19}\text{J}\end{array}$

Yellow light is said to have wavelength of $\text{570}\text{nm}$.  Therefore, the energy of photon that has wavelength of $\text{570}\text{nm}$ is calculated as shown below.

    $\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{570\times {10}^{-9}\text{m}}\\ =\text{0}\text{.034849272}\times {10}^{-17}\text{J}\\ =3.48\times {10}^{-19}\text{J}\end{array}$

Orange light is said to have wavelength of $\text{590}\text{nm}$.  Therefore, the energy of photon that has wavelength of $\text{590}\text{nm}$ is calculated as shown below.

    $\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{590\times {10}^{-9}\text{m}}\\ =\text{0}\text{.0336679413}\times {10}^{-17}\text{J}\\ =3.37\times {10}^{-19}\text{J}\end{array}$

Red light is said to have wavelength of $\text{650}\text{nm}$.  Therefore, the energy of photon that has wavelength of $\text{650}\text{nm}$ is calculated as shown below.

    $\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{650\times {10}^{-9}\text{m}}\\ =\text{0}\text{.0305601313}\times {10}^{-17}\text{J}\\ =3.06\times {10}^{-19}\text{J}\end{array}$

In order to eject an electron from the metal foil, the energy of photon has to be at least $\text{3}\text{.611}\times {\text{10}}^{\text{-19}}\text{J}$.  Therefore, electrons will be ejected by Green, Blue, Indigo, and Violet light.