Grade 12

Advanced

7 min

Lesson 5 of 12

The Photoelectric Effect and Quantization

Not Started

Examining the experiment that proved light behaves as a particle (photon) and challenged classical wave theory.

Imagine throwing thousands of ping-pong balls at a glass window; no matter how many you throw, the window never breaks. But then, you throw a single marble, and it shatters instantly. Why does light behave exactly like this marble when hitting a metal surface?

Learning Objectives

1.By the end of this lesson, you will be able to explain why the photoelectric effect proves light is quantized into photons.
2.You'll understand how to calculate photon energy and the maximum kinetic energy of ejected electrons.
3.You'll be able to determine the work function and stopping potential for different metallic surfaces.

The Classical Crisis

In the late 1800s, physicists believed light was purely a wave. According to classical theory, if you shined a very bright light (high intensity) on a metal, the electrons should eventually absorb enough energy to vibrate and break free, regardless of the light's color. However, experiments showed something shocking: if the light was below a certain threshold frequency ( $\nu_0$ ), no electrons were ejected, no matter how bright the light was. Conversely, even the dimmest light of a high frequency caused electrons to eject instantly. This suggested that light doesn't deliver energy continuously like a wave, but in discrete 'packets' called photons.

Quick Check

According to classical wave theory, what should happen if you increase the intensity (brightness) of light hitting a metal surface?

Think about how waves carry energy continuously.

Answer

Classical theory predicted that increasing intensity would eventually provide enough energy to eject electrons, regardless of the light's frequency.

Einstein’s Solution: Quantization

Albert Einstein solved the mystery by proposing that light is made of particles called photons. The energy (

E

) of a single photon is directly proportional to its frequency (

f

), defined by the equation:

E = hf

where

h

is Planck’s constant (

6.63 \times 10^{-34} \text{ J}\cdot\text{s}

). When a photon hits a metal, it transfers its entire energy to a single electron. If that energy is greater than the work function (

\Phi

), the minimum energy needed to 'unstick' an electron from the metal, the electron is ejected.

Calculate the energy of a single photon of violet light with a frequency of $7.5 \times 10^{14} \text{ Hz}$ .

1. Identify the formula: $E = hf$ 2. Plug in the constants: $E = (6.63 \times 10^{-34} \text{ J}\cdot\text{s}) \times (7.5 \times 10^{14} \text{ Hz})$ 3. Calculate: $E \approx 4.97 \times 10^{-19} \text{ J}$

The Photoelectric Equation

Einstein’s conservation of energy for this process is written as:

hf = \Phi + K_{max}

Here,

hf

is the incoming energy,

\Phi

is the 'entry fee' (work function) to leave the metal, and

K_{max}

is the maximum kinetic energy the electron has left over. If the photon energy is exactly equal to the work function (

hf = \Phi

), the frequency is called the threshold frequency (

f_0

). Any energy beyond

\Phi

goes into the electron's speed.

A metal has a work function of $\Phi = 2.0 \text{ eV}$ . If a photon with $3.5 \text{ eV}$ of energy hits it, what is the maximum kinetic energy of the ejected electron?

1. Use the equation: $K_{max} = hf - \Phi$ 2. Substitute the values: $K_{max} = 3.5 \text{ eV} - 2.0 \text{ eV}$ 3. Result: $K_{max} = 1.5 \text{ eV}$ 4. Note: To convert to Joules, multiply by $1.6 \times 10^{-19} \text{ J/eV}$ .

Quick Check

If you double the intensity of the light without changing its frequency, what happens to the maximum kinetic energy of the ejected electrons?

Does intensity change the energy of an individual photon?

Answer

The maximum kinetic energy stays the same; however, the number of electrons ejected per second increases.

Stopping Potential

How do we measure

K_{max}

? We use an opposing voltage called the stopping potential (

V_s

). By applying a negative voltage to the receiving plate, we can create an electric field that pushes the electrons back. When the voltage is high enough to stop even the fastest electron, the current drops to zero. The relationship is:

K_{max} = eV_s

where

e

is the elementary charge (

1.6 \times 10^{-19} \text{ C}

). This allows us to measure the energy of microscopic electrons using a simple voltmeter!

Light with a wavelength of $\lambda = 400 \text{ nm}$ shines on a metal with a work function of $2.2 \text{ eV}$ . Calculate the stopping potential $V_s$ .

1. Find photon energy in eV: $E = \frac{hc}{\lambda}$ . Using $hc \approx 1240 \text{ eV}\cdot\text{nm}$ , $E = \frac{1240}{400} = 3.1 \text{ eV}$ . 2. Calculate $K_{max}$ : $K_{max} = E - \Phi = 3.1 \text{ eV} - 2.2 \text{ eV} = 0.9 \text{ eV}$ . 3. Relate to stopping potential: Since $K_{max} = eV_s$ , a kinetic energy of $0.9 \text{ eV}$ requires a stopping potential of $0.9 \text{ V}$ .

Key Takeaways

1Light behaves as discrete packets of energy called photons, proving the particle nature of light.
2The energy of a photon depends only on its frequency ( $E=hf$ ), not its intensity.
3The Photoelectric Equation ( $hf = \Phi + K_{max}$ ) represents the conservation of energy during the interaction.
4Stopping potential ( $V_s$ ) is a direct measure of the maximum kinetic energy of emitted electrons ( $K_{max} = eV_s$ ).

Quick Check

Multiple Choice

Which of the following would increase the maximum kinetic energy of ejected electrons?

Multiple Choice

What is the 'threshold frequency'?

True/False

In the photoelectric effect, there is a significant time delay between the light hitting the metal and the electron being ejected if the light is very dim.

What's Next?

Review Tomorrow

In 24 hours, try to sketch the graph of Maximum Kinetic Energy ( $K_{max}$ ) vs. Frequency ( $f$ ). What does the x-intercept and the slope represent?

Practice Activity

Look up the work function for Gold and Cesium. Calculate which one would eject electrons when exposed to visible green light ( $550 \text{ nm}$ ).

Browse

Browse

The Photoelectric Effect and Quantization

Learning Objectives

The Classical Crisis

Einstein’s Solution: Quantization

The Photoelectric Equation

Stopping Potential

Key Takeaways

Quick Check

What's Next?

The Photoelectric Effect and Quantization

Learning Objectives

The Classical Crisis

Einstein’s Solution: Quantization

The Photoelectric Equation

Stopping Potential

Key Takeaways

Quick Check

What's Next?