Chapter 27, Problem 73A

To determine

To Plot: The stopping potential versus frequency.

To Find: The work function, the threshold wavelength, and the value of $\frac{h}{e}$

Answer to Problem 73A

Work function is ${\varphi }_o=2.055\text{eV}$ .

Threshold wave length is ${\nu }_o=0.4959\times {10}^{15}\text{Hz}$ .

Value of ${\left(\frac{h}{e}\right)}_{\exp }=4.17\times {10}^{-15}\text{J}\cdot \text{s/C}$

Explanation of Solution

Given data:

Wavelength$\lambda \left(\text{nm}\right)$	Stopping potential $V_o\left(\text{eV}\right)$
200	4.20
300	2.06
400	1.05
500	1.01
600	0.03

Formula used:

Frequency can be calculated as:

  $v=\frac{c}{\lambda }$

  $c=$ Speed of the light

  $\lambda =$ Wavelength

By Einstein’s photoelectric equation: -

  $K.E_{\max }=hv-\varphi$

  $K.E_{\max }=$ Kinetic energy of ejected electron

  $\varphi =$ Work function

  $h=$ Planck’s constant

  $v=$ Frequency of light $=\frac{c}{\lambda }$

Calculation:

Wavelength $\lambda \left(\text{nm}\right)$	Frequency $v=\frac{c}{\lambda }$
200	$\frac{3\times {10}^8}{200\times {10}^{-9}}=15\times {10}^{14}\text{Hz}$
300	$\frac{3\times {10}^8}{300\times {10}^{-9}}=10\times {10}^{14}\text{Hz}$
400	$\frac{3\times {10}^8}{400\times {10}^{-9}}=7.5\times {10}^{14}\text{Hz}$
500	$\frac{3\times {10}^8}{500\times {10}^{-9}}=6\times {10}^{14}\text{Hz}$
600	$\frac{3\times {10}^8}{600\times {10}^{-9}}=5\times {10}^{14}\text{Hz}$

From the above table graph can be drawn as:

  

The photoelectric equation can be written as:

  ${\left(KE\right)}_{\max }=h\nu -{\varphi }_o$

  ${\left(KE\right)}_{\max }$ = Stopping potential

Here, the stopping potential is the function of the frequency.

Taking stopping potential on y axis and frequency on x axis, equation can be written as:

  $y=mx-{\varphi }_o$

From the above graph, slope can be determined as:

  $\begin{array}{l}m=\frac{4.20-0.03}{\left(15-5\right)\times {10}^{14}}\\ m=4.17\times {10}^{-15}\text{ eV/Hz}\end{array}$

The equation can be written as:

  $y=\left(4.17\times {10}^{-15}\right)x-{\varphi }_o$

To find the work function ${\varphi }_o$ , substitute the point $\left(15\times {10}^{14},4.20\right)$ in the equation

  $\begin{array}{l}4.20=\left(15\times {10}^{14}\right)\left(4.17\times {10}^{-15}\right)-{\varphi }_o\\ {\varphi }_o=\left(15\times 4.17\right)\left(0.01\right)-4.20\\ {\varphi }_o=\left(6.255-4.20\right)\\ {\varphi }_o=2.055\text{eV}\end{array}$

Or

  ${\varphi }_o=3.288\times {10}^{-19}\text{ J}$

Now, the relation between the threshold, frequency and work function is:

  $\begin{array}{l}h{\nu }_o={\varphi }_o\\ {\nu }_o=\frac{{\varphi }_o}{h}\\ {\nu }_o=\frac{3.288\times {10}^{-19}}{6.63\times {10}^{-34}}\\ {\nu }_o=0.4959\times {10}^{15}\text{Hz}\end{array}$

Now,

  $\begin{array}{l}{\left(\frac{h}{e}\right)}_{\exp }=\frac{m}{e}=\frac{4.17\times {10}^{-15}\text{eV}}{e}\\ {\left(\frac{h}{e}\right)}_{\exp }=\frac{m}{e}=\frac{\left(4.17\times {10}^{-15}\times 1.6\times {10}^{-19}\right)\text{J}\cdot \text{s}}{\left(1.6\times {10}^{-19}\right)\text{C}}\\ {\left(\frac{h}{e}\right)}_{\exp }=4.17\times {10}^{-15}\text{J}\cdot \text{s/C}\end{array}$

The accepted value of $\left(\frac{h}{e}\right)$ is:

  $\begin{array}{l}{\left(\frac{h}{e}\right)}_{acc}=\frac{6.63\times {10}^{-34}}{1.6\times {10}^{-19}}\\ {\left(\frac{h}{e}\right)}_{acc}=4.143\times {10}^{-15}\text{J}\cdot \text{s/C}\end{array}$

Now,

  $\begin{array}{l}{\left(\frac{h}{e}\right)}_{\exp }-{\left(\frac{h}{e}\right)}_{acc}=\left(4.17-4.143\right)\times {10}^{-15}\\ {\left(\frac{h}{e}\right)}_{\exp }-{\left(\frac{h}{e}\right)}_{acc}=0.027\times {10}^{-15}\text{J}\cdot \text{s/C}\\ {\left(\frac{h}{e}\right)}_{\exp }=0.027\times {10}^{-15}\text{J}\cdot \text{s/C+}{\left(\frac{h}{e}\right)}_{acc}\\ {\left(\frac{h}{e}\right)}_{\exp }\approx {\left(\frac{h}{e}\right)}_{acc}\end{array}$

Experimental and expected values are approximately equal.

Conclusion:

  ${\left(\frac{h}{e}\right)}_{\exp }\approx {\left(\frac{h}{e}\right)}_{acc}$

Experimental and expected values are approximately equal.