8. Topic 8(COLON) Atomic, Nuclear and Quantum Physics

Lesson 8.4: Quantum Phenomena And Wave-particle Duality

#### Lesson focus #### Learning outcomes Students should be able to:.

Lesson 8.4: Quantum Phenomena and Wave–Particle Duality

Introduction

Welcome to Lesson 8.4 of Foundation Physics! In this lesson, we will explore fascinating concepts in quantum physics that change the way we understand the universe. Our journey will take us through the duality of light and matter, as well as some intriguing phenomena that arise at the atomic level. By the end of this lesson, you should be able to:

  • Understand the photon model and calculate photon energy using the equations $E = hf$ and $E = \frac{hc}{\lambda}$.
  • Explain the photoelectric effect, including the work function and threshold frequency using the photoelectric equation.
  • Describe electron diffraction and the de Broglie wavelength.
  • Discuss discrete energy levels and the resulting emission and absorption spectra, and provide a qualitative understanding of the uncertainty principle.
  • Apply the photoelectric equation and understand the limitations of the wave model of light.

Photon Model and Photon Energy

Before diving into quantum phenomena, let’s establish a foundational concept in quantum physics: the photon model.

What is a Photon?

A photon is a basic unit of light and electromagnetic radiation. Unlike classical particles, photons have no mass and travel at the speed of light ($c$), approximately $3 \times 10^8$ m/s. The energy $E$ of a photon is directly related to its frequency $f$ and wavelength $\lambda$ by the equations:

$$

$E = hf $

$$

where $h$ is Planck's constant, approximately equal to $6.626 \times 10^{-34}$ J·s. Alternatively, we can express the energy of a photon in terms of its wavelength:

$$

$E = \frac{hc}{\lambda}$

$$

Example:

Let’s calculate the energy of a photon with a wavelength of $500$ nm (nanometers, or $500 \times 10^{-9}$ m).

Using the second equation, we have:

$$

E = $\frac{hc}{\lambda}$ = $\frac{(6.626 \times 10^{-34} \text{ J·s})(3 \times 10^8 \text{ m/s})}{500 \times 10^{-9} \text{ m}}$ $\approx 3$.$97 \times 10^{-19}$ $\text{ J}$

$$

The Photoelectric Effect

The photoelectric effect is a phenomenon in which electrons are emitted from the surface of a material (usually metals) when it is exposed to light of sufficient frequency.

Work Function and Threshold Frequency

The work function ($\phi$) is the minimum energy required to remove an electron from the surface of a material. The threshold frequency ($f_0$) is the minimum frequency required for this effect to occur. It is related to the work function as follows:

$$

$\phi = hf_0$

$$

Photoelectric Equation

The kinetic energy ($KE$) of the emitted electron can be given by the photoelectric equation:

$$

$KE = hf - \phi$

$$

Example:

Suppose we have a metal with a work function of $2.0$ eV. To find the cutoff frequency where photoelectrons just begin to get emitted:

Convert the work function from electron volts to joules ($1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$):

$$

$\phi$ = $2.0 \text{ eV}$ = $2.0 \times 1$.$602 \times 10^{-19}$ $\text{ J}$ $\approx 3$.$204 \times 10^{-19}$ $\text{ J}$

$$

Then using the relation:

$$

f_0 = $\frac{\phi}{h}$ $\approx$ $\frac{3.204 \times 10^{-19}}{6.626 \times 10^{-34}}$ $\approx 4$.$83 \times 10^{14}$ $\text{ Hz}$

$$

Electron Diffraction and De Broglie Wavelength

Moving on, let’s discuss how matter, specifically electrons, displays wave-like behavior. This concept is encapsulated in the de Broglie wavelength equation:

$$

$\lambda = \frac{h}{p}$

$$

where $p$ is the momentum of the particle. For an electron, this indicates that every particle has a wavelength associated with it, leading to phenomena like electron diffraction. This wave-particle duality is pivotal in understanding quantum mechanics.

Example:

For an electron, we can use the relation for momentum, $p = mv$, with mass $m = 9.11 \times 10^{-31}$ kg and velocity $v$. If we consider an electron moving with a speed of $10^6$ m/s:

$$

$\lambda$ = $\frac{h}{mv}$ = $\frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(10^6)}$ $\approx 7$.$27 \times 10^{-10}$ $\text{ m}$

$$

Discrete Energy Levels and Spectra

Atoms have quantized energy levels. When electrons transition between these levels, photons are emitted or absorbed, resulting in distinctive spectral lines.

Emission and Absorption Spectra

  • Emission Spectrum: Occurs when electrons fall to a lower energy level, releasing energy in the form of light.
  • Absorption Spectrum: Occurs when electrons move to a higher energy level by absorbing energy.

Each element emits light at certain wavelengths, leading to unique spectral fingerprints that can be used for identification.

The Uncertainty Principle

Finally, we touch upon the uncertainty principle, formulated by Werner Heisenberg. It states that the position ($x$) and momentum ($p$) of a particle cannot both be precisely known at the same time:

$$\Delta x \Delta p \geq \frac{\hbar}{2}

$$

where $\hbar$ is the reduced Planck’s constant. This principle underscores the fundamental limits of measurement in quantum mechanics.

Conclusion

In this lesson, we have explored iconic concepts in quantum phenomena and wave-particle duality. We learned how photons behave, the significance of the photoelectric effect, and how particles exhibit both wave and particle characteristics. By understanding these principles, we lay the groundwork for further study in modern physics, such as quantum mechanics and relativity.

Study Notes

  • Photons: Light particles, no mass, energy $E = hf$.
  • Photoelectric Effect: Emission of electrons when light hits a material.
  • Work Function ($\phi$): Minimum energy to remove an electron.
  • Kinetic Energy Equation: $KE = hf - \phi$.
  • De Broglie Wavelength: $\lambda = \frac{h}{p}$, showing wave properties of matter.
  • Spectroscopy: Study of emission and absorption spectra for element identification.
  • Uncertainty Principle: Limits of measuring position and momentum together.

Practice Quiz

5 questions to test your understanding