Using the Photoelectric Effect Equation

Using the photoelectric effect equation

It is useful to observe the photoelectric effect equation represented graphically.

It can be seen from the graph that $E_{k}$ is plotted on the $y-$axis and $f$ is plotted on the $x-$axis. Using the straight line equation, $y=mx+c$, we can identify

\begin{align*} {E}_{k\max}& = hf-W_{0} \\ \underbrace{{E}_{k\max}}_{y}& = h\underbrace{f}_x-hf_{0} \end{align*}This allows us to conclude that the slope of the graph $m$ is Planck's constant $h$. Also, the $x$ intercept is the cut-off frequency $f_{0}$.

Example: The Photoelectric Effect Using Silver

Question

Ultraviolet radiation with a wavelength of $\text{250}$ $\text{nm}$ is incident on a silver foil (work function $W_0$ = $\text{6.9} \times \text{10}^{-\text{19}}$ $\text{J}$). What is the maximum kinetic energy of the emitted electrons?

Step 1: Determine what is required and how to approach the problem

We need to determine the maximum kinetic energy of an electron ejected from a silver foil by ultraviolet radiation.

The photoelectric effect tells us that:

\begin{align*} {E}_{k\max}& = {E}_{\mathrm{photon}}-W_0 \\ {E}_{k\max}& = h\frac{c}{\lambda }-W_0 \end{align*}

We also have:

Work function of silver: $W_{0~\text{silver}} =\text{6.9} \times \text{10}^{-\text{19}}\text{ J}$
UV radiation wavelength = $\text{250}$ $\text{nm}$ = $\text{250} \times \text{10}^{-\text{9}}$ $\text{m}$ = $\text{2.50} \times \text{10}^{-\text{7}}$ $\text{m}$
Planck's constant: $h=\text{6.63} \times \text{10}^{-\text{34}}\text{ m$^{2}$·kg·s$^{-1}$}$
Speed of light: $c=\text{3} \times \text{10}^{\text{8}}\text{ m·s$^{-1}$}$

Step 2: Solve the problem

\begin{align*} {E}_{k}& = \frac{hc}{\lambda }-W_{0~\text{silver}} \\ & = \left[ \text{6.63} \times \text{10}^{-\text{34}} \times \frac{\text{3} \times \text{10}^{\text{8}}}{\text{2.5} \times \text{10}^{-\text{7}}}\right]-\text{6.9} \times \text{10}^{-\text{19}}\\ & = \text{1.06} \times \text{10}^{-\text{19}}\text{ J} \end{align*}

The maximum kinetic energy of the emitted electron will be $\text{1.06} \times \text{10}^{-\text{19}}$ $\text{J}$.

Example: The Photoelectric Effect Using Gold

Question

If we were to shine the same ultraviolet radiation ($f=\text{1.2} \times \text{10}^{\text{15}}\text{ Hz}$) on a gold foil (work function = $\text{8.2} \times \text{10}^{-\text{19}}$ $\text{J}$ ) would any electrons be emitted from the surface of the gold foil?

Step 1: Calculate the energy of the incident photons

For the electrons to be emitted from the surface, the energy of each photon needs to be greater than the work function of the material.

\begin{align*} {E}_{\mathrm{photon}}& = hf\\ & = \text{6.63} \times \text{10}^{-\text{34}}\times \text{1.2} \times \text{10}^{\text{15}}\\ & = \text{7.96} \times \text{10}^{-\text{19}}\text{ J} \end{align*}

Therefore each photon of ultraviolet light has an energy of $\text{7.96} \times \text{10}^{-\text{19}}$ $\text{J}$.

Step 2: Write down the work function for gold.

\begin{align*} W_{0~\mathrm{gold}}& = \text{8.92} \times \text{10}^{-\text{19}}\text{ J} \end{align*}

Step 3: Is the energy of the photons greater or smaller than the work function?

\begin{align*} \text{7.96} \times \text{10}^{-\text{19}}\text{ J}& < \text{8.92} \times \text{10}^{-\text{19}}\text{ J}\\ {E}_{\mathrm{photons}}& < W_{0~\mathrm{gold}} \end{align*}

Since the energy of each photon is less than the work function of gold, the photons do not have enough energy to knock electrons out of the gold. No electrons would be emitted from the gold foil.

Example: Exam Past Question

Question

A metal surface is illuminated with ultraviolet light of wavelength $\text{330}$ $\text{nm}$. Electrons are emitted from the metal surface.

The minimum amount of energy required to emit an electron from the surface of this metal is $\text{3.5} \times \text{10}^{-\text{19}}$ $\text{J}$.

Name the phenomenon illustrated above.

(1 mark)
Give ONE word or term for the underlined sentence in the above paragraph.

(1 mark)
Calculate the frequency of the ultraviolet light.

(4 marks)
Calculate the kinetic energy of a photoelectron emitted from the surface of the metal when the ultraviolet light shines on it.

(4 marks)
The intensity of the ultraviolet light illuminating the metal is now increased. What effect will this change have on the following:
1. Kinetic energy of the emitted photoelectrons (Write down only INCREASES, DECREASES or REMAINS THE SAME.)
  
  (1 mark)
2. Number of photoelectrons emitted per second (Write down only INCREASES, DECREASES or REMAINS THE SAME.)
  
  (1 mark)
Overexposure to sunlight causes damage to skin cells.
1. Which type of radiation in sunlight is said to be primarily responsible for this damage?
  
  (1 mark)
2. Name the property of this radiation responsible for the damage.
  
  (1 mark)

[TOTAL: 14 marks]

Question 1

Photo-electric effect

(1 mark)

Question 2

Work function

(1 mark)

Question 3

\begin{align*} c & = f \lambda \\ \text{3} \times \text{10}^{\text{8}} & = f(\text{330} \times \text{10}^{-\text{9}})\\ \therefore f & = \text{9.09} \times \text{10}^{\text{14}}\text{ Hz} \end{align*}

\begin{align*} E & = \frac{hc}{\lambda} \\ & = \frac{(\text{6.63} \times \text{10}^{-\text{34}})(\text{3} \times \text{10}^{\text{8}})}{(\text{330} \times \text{10}^{-\text{9}})} \\ & = \text{6.03} \times \text{10}^{-\text{19}}\text{ J} \end{align*}\begin{align*} E & = hf \\ \text{6.03} \times \text{10}^{-\text{19}}\text{ J} & = (\text{6.63} \times \text{10}^{-\text{34}})f\\ \therefore f & = \text{9.09} \times \text{10}^{\text{14}}\text{ Hz} \end{align*}

(4 marks)

Question 4

Option 1:

\begin{align*} E & = W_{o} + K \\ \frac{hc}{\lambda} & = W_{o} + K \\ \therefore \frac{(\text{6.63}\times10^{-34})(3\times10^8)}{330\times10^{-9}} & = \text{3.5}\times10^{-19} + K\\ \therefore K & = \text{2.53}\times10^{-19}\text{ J} \end{align*}

Option 2:

\begin{align*} E & = W_{o} + K \\ hf & = W_{o} + K \\ \therefore (\text{6.63}\times10^{-34})(\text{9.09}\times10^{14} & = \text{3.5}\times10^{-19} + K\\ \therefore K & = \text{2.53}\times10^{-19}\text{ J} \end{align*}

(4 marks)

Question 5.1

Remains the same.

(1 mark)

Question 5.2

Increases

(1 mark)

Question 6.1

Ultraviolet radiation

(1 mark)

Question 6.2

High energy or high frequency

(1 mark)

[TOTAL: 12 marks]

Using the Photoelectric Effect Equation