5.5 Graph between $K{E_{max}}$ and frequency

Question 4. What is the energy (in $eV$) of a photon of wavelength $12400{\text{ }}\mathop {\text{A}}\limits^0 $?

Solution: Using Planck's formula, we have $$\begin{equation} \begin{aligned} E = hf = \frac{{hc}}{\lambda } \\ {\text{where }}h = 6.63 \times {10^{ - 34}}{\text{ }}J{\mkern 1mu} s;c = 3 \times {10^8}{\text{ }}m{\mkern 1mu} {s^{ - 1}};\lambda = 12400 \times {10^{ - 10}}{\text{ }}m \\ E = \frac{{\left( {6.63 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{12400 \times {{10}^{ - 10}}}} = 1{\text{ }}eV{\text{ (approx)}} \\\end{aligned} \end{equation} $$

Note: In general, a photon of wavelength $\lambda $ ( in $\mathop {\text{A}}\limits^0 $ ) will have energy ( in $eV$ ) as given by $E = 12400/\lambda $.

Question 5. The stopping potential for photoelectrons emitted from a surface illuminated by light wavelength of $5893 {\text{ }}\mathop {\text{A}}\limits^0 $ in $0.36 V$. Calculate the maximum kinetic energy of photoelectrons, the work function of the surface, and the threshold frequency.

Solution: We know that $$\begin{equation} \begin{aligned} \quad \quad KE{_{max}} = hf - \phi = \left( {\frac{{hc}}{\lambda }} \right) - \phi \\ {\text{or }}\phi = \frac{{hc}}{\lambda } - KE{_{max}} \\ {\text{Also, }}KE{_{max}} = e{V_s} = 0.36{\text{ }}eV \\ \Rightarrow \quad \quad \phi = \frac{{\left( {6.62 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{5893 \times {{10}^{ - 10}}}} - 0.36 \times {10^{ - 19}} \\ \quad \quad \;\,\,\quad = 1.746{\text{ }}eV \\\end{aligned} \end{equation} $$

The threshold frequency is given by $$\begin{equation} \begin{aligned} {f_0} = \frac{\phi }{h} = \frac{{2.794 \times {{10}^{ - 19}}}}{{6.62 \times {{10}^{ - 34}}}} = 4.22 \times {10^{14}}{\text{ }}Hz \\ \\\end{aligned} \end{equation} $$

Question 6. A beam of light has three wavelengths $4144{\text{ }}\mathop {\text{A}}\limits^0 $, $4972{\text{ }}\mathop {\text{A}}\limits^0 $ and $6216{\text{ }}\mathop {\text{A}}\limits^0 $ with a total intensity of $3.6 \times {10^{ - 3}}\ W{m^{ - 2}}$ equally distributed among the three wavelengths. The beams fall normally on an area $1.0\ c{m^2}$ of a clean metallic surface of work function $2.3$ $eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2$ seconds.

Solution: We know that threshold wavelength (${{\lambda _0}}$) = $hc$/$\phi $

$$\begin{equation} \begin{aligned} {\lambda _0} = \frac{{\left( {6.63 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{2.3 \times \left( {1.6 \times {{10}^{ - 19}}} \right)}} = 5404{\text{ }}\mathop {\text{A}}\limits^0 \\ \\\end{aligned} \end{equation} $$

Thus, wavelength $4144{\text{ }}\mathop {\text{A}}\limits^0 $ and $4972{\text{ }}\mathop {\text{A}}\limits^0 $ will emit electrons from the metal surface.

Energy incident on the surface per unit time for each wavelength

$$\begin{equation} \begin{aligned} = {\text{Intensity of each wavelength }} \times {\text{ Area of the surface}} \\ = \frac{{3.6 \times {{10}^{ - 3}}}}{3} \times \left( {1.0{\text{ c}}{{\text{m}}^2}} \right) = 1.2 \times {10^{ - 7}}\ W \\\end{aligned} \end{equation} $$

Number of photons $n_1$ due to wavelength $4144{\text{ }}\mathop {\text{A}}\limits^0 $,

$$\begin{equation} \begin{aligned} {n_1} = \frac{{\left( {2.4 \times {{10}^{ - 7}}} \right)\left( {4144 \times {{10}^{ - 10}}} \right)}}{{\left( {6.63 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}} = 0.5 \times {10^{12}} \\ \\\end{aligned} \end{equation} $$

Number of photons $n_1$ due to wavelength $4144{\text{ }}\mathop {\text{A}}\limits^0 $,

$$\begin{equation} \begin{aligned} {n_2} = \frac{{\left( {2.4 \times {{10}^{ - 7}}} \right)\left( {4972 \times {{10}^{ - 10}}} \right)}}{{\left( {6.63 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}} = 0.575 \times {10^{12}} \\ {\text{So, }}N = {n_1} + \,\,{n_2} = 0.5 \times {10^{12}} + 0.575 \times {10^{12}} \\ \quad \quad \quad \quad \;\;\,\, = 1.075 \times {10^{12}} \\\end{aligned} \end{equation} $$