Grade 11

Advanced

10 min

Lesson 10 of 12

Einstein's Photon Theory

Not Started

Applying the conservation of energy to the photoelectric effect using Einstein's famous equation.

Why can a dim beam of ultraviolet light eject electrons from a metal surface, while a blindingly bright red light does absolutely nothing? The answer changed physics forever.

Learning Objectives

1.Calculate photon energy using the Planck constant and frequency
2.Apply Einstein's photoelectric equation to find the maximum kinetic energy of ejected electrons
3.Determine the stopping potential required to halt the flow of photoelectrons

The Quantum Packet

In the early 1900s, classical physics failed to explain why the color of light mattered more than its brightness when knocking electrons off metal. Albert Einstein solved this by proposing that light isn't just a continuous wave, but a stream of discrete 'packets' called photons. The energy of a single photon ( $E$ ) is directly proportional to its frequency ( $f$ ). This relationship is defined by the formula $E = hf$ , where $h$ is Planck’s constant ( $6.63 \times 10^{-34} \text{ J}\cdot\text{s}$ ). Because blue light has a higher frequency than red light, each individual blue photon carries more 'punch' than a red one, regardless of how many photons are in the beam.

Calculate the energy of a single photon of green light with a frequency of $5.5 \times 10^{14} \text{ Hz}$ .

1. Identify the knowns: $f = 5.5 \times 10^{14} \text{ Hz}$ and $h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s}$ . 2. Use the formula: $E = hf$ . 3. Substitute: $E = (6.63 \times 10^{-34}) \times (5.5 \times 10^{14})$ . 4. Calculate: $E = 3.65 \times 10^{-19} \text{ J}$ .

Quick Check

If you double the frequency of a light source, what happens to the energy of each individual photon?

Look at the relationship between E and f in the formula.

Answer

The energy of each photon doubles, as energy is directly proportional to frequency ( $E = hf$ ).

The Work Function and Kinetic Energy

When a photon hits a metal, it transfers all its energy to a single electron. However, electrons are held in the metal by an 'entry fee' called the work function (

\Phi

). If the photon's energy is less than

\Phi

, nothing happens. If it is greater, the electron is ejected, and the leftover energy becomes the electron's maximum kinetic energy (

K_{max}

). This is a perfect example of the Conservation of Energy. Einstein’s photoelectric equation is written as:

hf = \Phi + K_{max}

This explains why increasing light intensity (more photons) only increases the number of electrons, while increasing frequency increases their speed.

A metal has a work function of $\Phi = 2.3 \text{ eV}$ . If a photon with $4.0 \text{ eV}$ of energy strikes the metal, what is the maximum kinetic energy of the ejected electron?

1. Use Einstein's equation: $E_{photon} = \Phi + K_{max}$ . 2. Rearrange for $K_{max}$ : $K_{max} = E_{photon} - \Phi$ . 3. Substitute values: $K_{max} = 4.0 \text{ eV} - 2.3 \text{ eV}$ . 4. Result: $K_{max} = 1.7 \text{ eV}$ .

Quick Check

What happens if a photon strikes a metal with energy exactly equal to the work function?

Think about the 'leftover' energy.

Answer

The electron is liberated from the metal but has zero kinetic energy ( $K_{max} = 0$ ).

Stopping Potential

How do we actually measure the speed of these tiny electrons? We use an electric field to oppose their motion. The stopping potential ( $V_s$ ) is the voltage required to stop even the fastest electron from reaching a collector plate. At this point, the maximum kinetic energy is equal to the work done by the electric field: $K_{max} = eV_s$ , where $e$ is the elementary charge ( $1.6 \times 10^{-19} \text{ C}$ ). This allows us to convert kinetic energy from Joules to electron-volts (eV), a much more convenient unit for atomic physics.

Light with a wavelength of $300 \text{ nm}$ ( $f = 1.0 \times 10^{15} \text{ Hz}$ ) shines on a metal with a work function of $2.5 \text{ eV}$ . Find the stopping potential $V_s$ .

1. Calculate photon energy in Joules: $E = hf = (6.63 \times 10^{-34})(1.0 \times 10^{15}) = 6.63 \times 10^{-19} \text{ J}$ . 2. Convert $E$ to eV: $E_{eV} = \frac{6.63 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.14 \text{ eV}$ . 3. Find $K_{max}$ : $K_{max} = E - \Phi = 4.14 \text{ eV} - 2.5 \text{ eV} = 1.64 \text{ eV}$ . 4. Determine $V_s$ : Since $K_{max} = eV_s$ , the stopping potential is simply $1.64 \text{ V}$ .

Key Takeaways

1Light is quantized into photons with energy $E = hf$ .
2The photoelectric effect proves the particle nature of light through the conservation of energy: $hf = \Phi + K_{max}$ .
3The stopping potential ( $V_s$ ) is a direct measure of the maximum kinetic energy of ejected electrons ( $K_{max} = eV_s$ ).

Quick Check

Multiple Choice

Which of the following would increase the maximum kinetic energy of photoelectrons?

Multiple Choice

If the stopping potential for a certain experiment is $3.0 \text{ V}$ , what is the maximum kinetic energy of the electrons?

True/False

A very bright red light (low frequency) is more likely to eject electrons than a dim violet light (high frequency).

What's Next?

Review Tomorrow

In 24 hours, try to write down Einstein's photoelectric equation from memory and explain what each of the three terms represents.

Practice Activity

Look up the work functions for Gold, Sodium, and Aluminum. Calculate which of these would eject electrons when hit by visible light ( $f \approx 6 \times 10^{14} \text{ Hz}$ ).

Browse

Browse

Einstein's Photon Theory

Learning Objectives

The Quantum Packet

The Work Function and Kinetic Energy

Stopping Potential

Key Takeaways

Quick Check

What's Next?

Einstein's Photon Theory

Learning Objectives

The Quantum Packet

The Work Function and Kinetic Energy

Stopping Potential

Key Takeaways

Quick Check

What's Next?