Chapter 4, Problem 4.30P

Interpretation Introduction

Interpretation:

Threshold frequency of metal has to be calculated if metal is irradiated with light having wavelength of $\text{390}\text{nm}$ and electrons are ejected with a speed of $5.00\times {10}^5\text{m}\cdot {\text{s}}^{-1}$.

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.

    $E=h\nu -h{\nu }_0$

This excess energy is the kinetic energy of the electron that is being ejected.  Hence, the equation can be written as shown below.

    $E_k=h\nu -h{\nu }_0$

Explanation of Solution

Wavelength of the incident light is given as $\text{390}\text{nm}$.  Electrons are ejected at a speed of $5.00\times {10}^5\text{m}\cdot {\text{s}}^{-1}$.  Kinetic energy of an electron and the threshold frequency of metal is related by the equation given below.

    $E_k=h\nu -h{\nu }_0$        (1)

Kinetic energy can be expressed considering the velocity and mass of the electron as shown below.

    $E_k=\frac{1}{2}m{v}^2$        (2)

Frequency of the incident light can be given by the equation shown below.

    $v=\frac{c}{\lambda }$        (3)

The binding energy of the metal can be calculated as shown below.

    $\begin{array}{c}E=h{\nu }_0+E_k\\ h\nu =E_0+E_k\\ E_0=h\nu -E_k\\ E_0=\frac{hc}{\lambda }-\frac{1}{2}m{v}^2\end{array}$

Speed of ejected electron $(v_0)$ is given as $5.00\times {10}^5\text{m}\cdot {\text{s}}^{-1}$.  Mass of electron $(m)$ is $9.10\times {10}^{-31}\text{kg}$.  Substituting the values in above equation.

    $\begin{array}{c}{E}_0=\frac{hc}{\lambda }-\frac{1}{2}m{v}^2\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^8\text{m}\cdot {\text{s}}^{-1}}{390\times {10}^{-9}\text{m}}-\frac{(9.10\times {10}^{-31}\text{kg)(}5.00\times {10}^5\text{m}\cdot {\text{s}}^{-1}{)}^2}{2}\\ =\frac{\text{19}\text{.8640854}\times {10}^{-26}\text{J}}{390\times {10}^{-9}}-\frac{(9.10\times {10}^{-31}\text{kg)(}5.00\times {10}^5\text{m}\cdot {\text{s}}^{-1}{)}^2}{2}\\ =\text{509}\text{.335523076923}\times {\text{10}}^{\text{-21}}\text{J}-\text{113}\text{.75}\times {10}^{-21}\text{J}\\ =\text{395}\text{.25}\times {\text{10}}^{\text{-21}}\text{J}\end{array}$

Threshold frequency of metal can be calculated as shown below.

    $\begin{array}{c}E=h{\nu }_0+E_k\\ h\nu =E_0+0(\because \text{KE}=0)\\ h\nu =\text{395}\text{.25}\times {\text{10}}^{\text{-21}}\text{J}\\ \nu =\frac{\text{395}\text{.25}\times {\text{10}}^{\text{-21}}\text{J}}{6.626\times {10}^{-34}\text{J}\cdot \text{s}}\\ =59.6\times {10}^{13}{\text{s}}^{-1}\\ =5.96\times {10}^{14}{\text{s}}^{-1}\end{array}$

Therefore, the threshold frequency of metal is calculated as $5.96\times {10}^{14}{\text{s}}^{-1}$.