# The Photoelectric Effect

## The Photoelectric Effect Revision

**The Photoelectric Effect**

When light of sufficient **energy** shines upon a metal surface,** electrons** are emitted from the surface of the metal. This phenomenon is known as the **photoelectric effect**. As well as providing evidence for the** particle **nature of light, it also is the phenomenon behind solar cells.

**Photoelectrons**

**Photoelectrons** are the particles emitted by the metal when the **photoelectric effect **occurs. They are **electrons** but are given a different name based upon their origin.

**Threshold Frequency**

The **photoelectric effect** does not happen in all metals, all the time. For the **photoelectric effect** to occur, the photons in the incident light ray must have **sufficient energy **such that it is enough to liberate surface electrons from the metal.

To calculate the **energy** of each **photon** we can use the equation:

E = hf

- E= the
**energy****of a****photon**in joules \text{(J)} - h=
**Planck’s constant**(6.63\times 10^{-34} \text{ Js}) - f= the
**frequency**of light in hertz \text{(Hz)}

**Example Question:**

Calculate the energy of a photon of light of frequency 450 \text{ THz}.

**[2 marks]**

\begin{aligned} \bold{E} &= \bold{hf} \\ E &= 6.63 \times 10^{-34} \times \textcolor{10a6f3}{450 \times 10^{12}}\\ &= \bold{2.98 \times 10^{19}} \textbf{ J} \end{aligned}

As we can see from the above equation, the **energy** of a **photon** and **frequency** are directly proportional.

Therefore, the minimum **energy** needed for the **photoelectric effect **to happen is directly proportional to a minimum **frequency** of light. This minimum frequency is known as the **threshold frequency**. The **threshold frequency** is different in each metal that we observe.

We also know from GCSE physics that v = f \lambda and as light is an **electromagnetic wave **which travels at the speed of the light (c). From this we get

c = f \lambda

which when rearranged for **frequency** gives:

f = \dfrac{c}{ \lambda}

Combining these equations gives us another equation we can use to calculate the **energy** of a **photon**:

\bold{E = \dfrac{hc}{ \lambda}}

**Example Question: **

A photon has of energy 4.2 \times 10^{-19} \text{ J} . What is its wavelength?

**[2 marks]**

E = \dfrac{hc}{ \lambda} \\ \begin{aligned} \bold{ \lambda} &= \bold{\dfrac{hc}{E}} \\ &= \dfrac{hc}{E} \\ &= \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{\textcolor{f95d27}{4.2 \times 10^{-19}}} \\ &= \bold{4.74 \times 10^{-7}} \textbf{ m or } \bold{474} \textbf{ nm} \end{aligned}

As previously discussed, the **threshold frequency** of a metal is the minimum frequency of light needed to cause** photoelectric emission**. If a ray of light has a** frequency** lower than the **threshold frequency**, then it will not have enough **energy** to liberate an **electron** from the surface of the metal. However, if the light has **frequency** equal to the** threshold frequency** or greater, then **photoelectric emission** will occur.

**Light as a Photon**

As previously mentioned, the **photoelectric effect **provides evidence for the particle nature of light. As only a single **electron** in the metal can absorb a single **photon** of light, this brings the idea that **photons** must be tiny “packets” of quantised energy.

**The Electronvolt**

When using the equation listed previously for calculating the **energy** of a **photon**, our answer will be given in **Joules** \left(\text{J}\right).

However, cause each individual **photon** of light has such small amounts of **energy**, another unit called the **electronvolt** is used which is more suited for expressing very small amounts of **energy**.

\bold{1} \textbf{ electronvolt} is defined as the energy gained by an **electron** travelling through a **potential difference** of 1\text{ V} and is equal to 1.6 \times 10^{-19} \textbf{ J}.

To convert from \text{eV} to \text{J}, multiply by 1.6 \times 10^{-19} \text{ J}.

To convert from \text{J} to \text{eV}, divide by 1.6 \times 10^{-19} \text{ J}.

**Work Function**

Another key term often used when discussing the **photoelectric effect **is the **work function** (\Phi). The **work function **is defined as the minimum **energy** needed for a **photoelectron **to be released from the surface of the metal.

As the** energy** of a **photon** can be calculated using the equation, E = hf, we can see that the **work function **and** threshold frequency** are related. As** threshold frequency** is the minimum **frequency **needed for** photoelectric emission** to occur, we can say that:

\Phi = h \times f_{\text{threshold}}

- \Phi = the
**work function**in joules \text{(J)} - h=
**Planck’s constant**(6.63\times 10^{-34} \text{ Js}) - f_{\text{threshold}}= the
**threshold frequency**in hertz \text{(Hz)}

As only one **photon** can be absorbed by each **electron**, for the **photoelectric effect** to occur the **photon** must have equal to, or more **energy** than the **work function **of the metal. Any extra **energy** above the **work function** contributes to the **kinetic energy** of the **photoelectron**.

From this we can derive a new equation:

E_{\text{kmax}} = E_{\text{photon}} – \Phi

- E_{\text{kmax}} is the maximum
**kinetic energy****of the****photoelectron**emitted by the**photoelectric effect**in joules \text{(J)} - E_{\text{photon}}= the
**energy of the photon**in joules \text{(J)} - \Phi = the
**work function**in joules \text{(J)}

**Example Question:**

The work function of a metal surface is 1.55 \text{ eV}. UV light of wavelength 430 \text{ nm} is incident on the metal.

Calculate the maximum kinetic energy of the photoelectrons.

**[3 marks]**

E_{\text{kmax}} = E_{\text{photon}} – \Phi

\\ \begin{aligned} E &= \dfrac{hc}{\lambda} \\ &= \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{\textcolor{10a6f3}{430 \times 10^{-9}}} \\ &= 4.626 \times 10^{-19} \text{ J} \end{aligned}

\\ E_{\text{kmax}} = 4.626 \times 10^{-19} \text{ J} – (\textcolor{f21cc2}{1.55 \text{ eV}} \times 1.6 \times 10^{-19}) = \bold{2.15 \times 10^{-19}} \textbf{ J}

**Stopping Potential**

We can prevent the **photoelectric effect **happening in a piece of metal by applying a **potential difference** to the metal plate. This **potential difference** is known as the **stopping potential** Vs and is the **potential difference **required to stop the **photoelectric effect **from occurring.

As the **voltage** of the power source in the diagram above is increased, the metal plate becomes more positively charged and has a greater attraction holding the **electrons** in place, therefore increasing the amount of **energy** needed by the **photon** to cause **photoelectric emission**.

## The Photoelectric Effect Example Questions

**Question 1:** Define the threshold frequency of a metal.

**[2 marks]**

Threshold frequency is the **minimum frequency** of electromagnetic waves incident on a metal **needed to remove an electron from the surface**.

**Question 2:** Sodium metal has a work function of 2.28 \text{ eV}. What is this in Joules?

**[2 marks]**

**Question 3:** The work function of a metal surface is 3.70 \text{ eV}. UV light of wavelength 233 \text{ nm} is incident on the metal.

Calculate the maximum kinetic energy of the photoelectrons in \text{eV}.

**[4 marks]**

In \text{eV}:

\dfrac{2.62 \times 10^{-19}}{1.6 \times 10^{-19}} = \bold{1.64} \textbf{ eV}## You May Also Like...

### MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.