AP Physics 2: Quantum Physics - Your Last-Minute Guide 🚀

Hey there, future physics pro! Let's get you prepped and confident for your AP Physics 2 exam. This guide is designed to be your go-to resource, especially the night before the big day. We'll break down the key concepts, highlight must-know info, and tackle some practice problems together. Let's do this! 💪

7.5 Wave-Particle Duality and Quantum Phenomena

Wave-Particle Duality: The Basics 🤯

Remember how light can act like both a wave and a particle? Well, this duality isn't just for light; it applies to all fundamental particles! This idea is called wave-particle duality.

• Particles as Waves: On a tiny scale, particles (like electrons) can show wave properties. Think of the double-slit experiment where particles diffract, just like waves. 🌊
• Waves as Particles: Waves (like photons) can also show particle properties. Photons have momentum and energy related to their frequency and wavelength. 💡

Compton Effect: Proof of Photon Momentum

In the 1920s, Arthur Compton showed that when an X-ray photon hits an electron, momentum is conserved. This is known as the Compton effect. The scattered photon has a lower frequency than the original photon.

Caption: The Compton effect demonstrates the particle-like behavior of photons, showing that they have momentum.

De Broglie Wavelength: Particles as Waves

Louis de Broglie proposed that if a photon's momentum is $p = \frac{h}{\lambda}$, then the wavelength is $\lambda = \frac{h}{p}$. He suggested that particles could also have a wavelength. For a particle with mass m and velocity v, the de Broglie wavelength is:

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

• Key Point: For a noticeable wavelength, the mass has to be extremely small (like atomic scale). This is why we see wave-like behavior more with electrons than with everyday objects. ⚛️

Example Problem

Let's calculate the de Broglie wavelength of electrons accelerated through a 175 V potential difference:

1. Find the momentum: $p = \sqrt{2mK}$, where K is kinetic energy.
2. Kinetic Energy: $K = 175 \text{ eV} = 175 \times 1.6 \times 10^{-19} \text{ J}$
3. Plug into de Broglie: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$
4. Solve: $\lambda \approx 0.093 \text{ nm}$

Practice Question

Multiple Choice:

1. An atomic particle of mass m moving at speed v has a wavelength λ. What is the wavelength of a second particle with a speed 3v and the same mass? A) (1/9) λ B) (1/3) λ C) λ D) 3 λ E) 9 λ

2. A very slow proton has its kinetic energy doubled. What happens to the proton's corresponding de Broglie wavelength? A) The wavelength is decreased by a factor of √2 B) The wavelength is halved C) There is no change in the wavelength D) The wavelength is increased by a factor of √2 E) The wavelength is doubled

Free Response:

Electrons are accelerated from rest through a potential difference of 300 V. Calculate:

(a) The kinetic energy of the electrons in Joules. (b) The momentum of the electrons. (c) The de Broglie wavelength of the electrons.

Scoring Guide

(a) 1 point for using $KE = qV$ and correct conversion to Joules: $KE = (1.6 \times 10^{-19} C)(300 V) = 4.8 \times 10^{-17} J$

(b) 2 points for using $p = \sqrt{2mKE}$ and correctly calculating the momentum: $p = \sqrt{2(9.11 \times 10^{-31} kg)(4.8 \times 10^{-17} J)} = 9.35 \times 10^{-24} kg \cdot m/s$

(c) 2 points for using $\lambda = \frac{h}{p}$ and correctly calculating the wavelength: $\lambda = \frac{6.626 \times 10^{-34} J \cdot s}{9.35 \times 10^{-24} kg \cdot m/s} = 7.09 \times 10^{-11} m$

Relativistic Mass-Energy Equivalence

The Famous Equation: E=mc²

Relativistic mass-energy equivalence means that mass and energy are interchangeable. This is described by Einstein's famous equation:

$$E=mc^2$$

• Key Idea: As an object's velocity increases, its mass also increases. This is a consequence of the theory of relativity. 🚀
• Important Note: The speed of light (c) is a constant, so a small amount of mass can be converted into a huge amount of energy. 💥

Wave Phenomena: Interference and Diffraction

Why Only Waves Show Interference and Diffraction?

Interference and diffraction are wave phenomena. They happen when waves combine or bend around obstacles. You won't see these with particles because they don't have a wave nature.

• Interference: When two or more waves overlap, they create a new wave pattern. 🌊
• Diffraction: Waves bend around obstacles or through small openings. 🚧

Final Exam Focus 🎯

Alright, here’s the lowdown on what to focus on for the exam:

• High-Value Topics:

  • Wave-particle duality and its implications
  • De Broglie wavelength calculations
  • Relativistic mass-energy equivalence
  • Understanding interference and diffraction

• Common Question Types:

  • Multiple-choice questions testing your understanding of wave-particle duality and the Compton effect.
  • Free-response questions involving calculations of de Broglie wavelengths and energy-mass conversions.
  • Conceptual questions about why waves show interference and diffraction but particles don't.

Last-Minute Tips ⏰

• Time Management: Don't spend too long on a single question. If you get stuck, move on and come back to it later.
• Common Pitfalls: Watch out for unit conversions and make sure you understand the difference between wave and particle properties.
• Strategy: Start with the questions you know well to build confidence. Then tackle the harder ones. Show all your work, even if you're not sure – you might get partial credit! ✅

Wave-particle duality and de Broglie wavelength are high-yield topics. Make sure you understand the concepts and can apply the equations.

Practice Problems: 🧩

1. Which of the following graphs best represents the de Broglie wavelength λ of a particle as a function of the linear momentum p of the particle?

   

Answers:

1. B: de Broglie wavelength is given by, $p = h / \lambda$ … $mv = h / \lambda$ … $\lambda = h / mv$ … 3x $m = 1/3 \lambda$
2. A: From above $\lambda = h / mv$ … Since $K = \frac{1}{2}mv^2$ , 2x K means $\sqrt{2}$x v. So when we plug in the new velocity of $\sqrt{2}v$, the wavelength decreases by this factor since we divide.
3. D: From $p=h/ \lambda$, they are inverses.

Practice Question

Multiple Choice:

1. Which of the following graphs best represents the de Broglie wavelength λ of a particle as a function of the linear momentum p of the particle?

   

Free Response Question:

A beam of electrons is accelerated from rest through a potential difference of 100 V. These electrons then pass through a single slit of width 0.2 nm.

(a) Calculate the kinetic energy of the electrons in Joules. (b) Calculate the de Broglie wavelength of the electrons. (c) If the electrons are diffracted, what is the angle of the first minimum in the diffraction pattern?

Scoring Guide

(a) 1 point for using $KE = qV$ and correct conversion to Joules: $KE = (1.6 \times 10^{-19} C)(100 V) = 1.6 \times 10^{-17} J$

(b) 2 points for using $p = \sqrt{2mKE}$ and correctly calculating the momentum: $p = \sqrt{2(9.11 \times 10^{-31} kg)(1.6 \times 10^{-17} J)} = 5.4 \times 10^{-24} kg \cdot m/s$ 1 point for using $\lambda = \frac{h}{p}$ and correctly calculating the wavelength: $\lambda = \frac{6.626 \times 10^{-34} J \cdot s}{5.4 \times 10^{-24} kg \cdot m/s} = 1.23 \times 10^{-10} m$

(c) 2 points for using $d \sin{\theta} = m\lambda$ and correctly calculating the angle: 0.2 \times 10^{-9} \sin{\theta} = 1 \times 1.23 \times 10^{-10} $\theta = \sin^{-1}(0.615) = 37.9^\circ$

You've got this! Go ace that exam! 🌟