AP Physics 2- 15.1 Quantum Theory and Wave-Particle Duality- Study Notes- New Syllabus
AP Physics 2- 15.1 Quantum Theory and Wave-Particle Duality – Study Notes
AP Physics 2- 15.1 Quantum Theory and Wave-Particle Duality – Study Notes – per latest Syllabus.
Key Concepts:
- Introduction to Quantum Theory
- Light as Photons and Wave-Particle Duality
- Wave Properties of Particles
Introduction to Quantum Theory
Why Quantum Theory Was Developed:
Classical mechanics could not explain several experimental observations of matter and energy at small scales.
- Key phenomena that required a new framework include:
- Atomic spectra – discrete emission and absorption lines.
- Blackbody radiation – classical theory predicted the “ultraviolet catastrophe,” which was resolved by Planck’s quantization of energy.
- Photoelectric effect – explained by Einstein using photons (light quanta), not waves alone.
Scope of Quantum Theory:
- Necessary to describe and predict the behavior of matter at atomic and subatomic scales.
- Replaces or extends classical ideas when they fail at microscopic levels.
Wave–Particle Duality:
- Quantum theory shows that fundamental particles (such as electrons, photons) can exhibit both wave-like and particle-like properties.
- Examples:
- Electrons form diffraction patterns like waves.
- Photons in the photoelectric effect behave as particles carrying discrete energy packets (\(\mathrm{E = hf}\)).
Example :
A blackbody is maintained at a temperature of \(\mathrm{T = 5800 \, K}\) (approximately the surface of the Sun). Use Wien’s displacement law to find the wavelength at which the blackbody emits the maximum intensity of radiation.
▶️ Answer/Explanation
Step 1: Relevant formula: \(\mathrm{\lambda_{\text{max}} T = b}\), where \(\mathrm{b = 2.9 \times 10^{-3} \, m \, K}\).
Step 2: Substitution: \(\mathrm{\lambda_{\text{max}} = \dfrac{2.9 \times 10^{-3}}{5800}}\).
\(\mathrm{\lambda_{\text{max}} \approx 5.0 \times 10^{-7} \, m = 500 \, nm}\).
Answer: Maximum emission occurs at about \(\mathrm{500 \, nm}\), in the visible spectrum (green light).
Example :
Light of wavelength \(\mathrm{400 \, nm}\) shines on a metal surface with a work function \(\mathrm{\phi = 2.0 \, eV}\). Find the maximum kinetic energy of the emitted electrons.
▶️ Answer/Explanation
Step 1: Photon energy: \(\mathrm{E = \dfrac{hc}{\lambda}}\).
\(\mathrm{E = \dfrac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{400 \times 10^{-9}}}\).
\(\mathrm{E = 4.97 \times 10^{-19} \, J = 3.1 \, eV}\).
Step 2: Apply Einstein’s photoelectric equation: \(\mathrm{K_{\text{max}} = E – \phi}\).
\(\mathrm{K_{\text{max}} = 3.1 – 2.0 = 1.1 \, eV}\).
Answer: The maximum kinetic energy of the electrons is \(\mathrm{1.1 \, eV}\).
Light as Photons and Wave-Particle Duality
Light can be modeled both as a wave and as discrete particles, called photons. This dual nature is essential to explain phenomena such as interference and diffraction (wave model) and the photoelectric effect (particle model).
Photon Properties:
- A photon is a massless, electrically neutral particle.
- The energy of a photon is directly proportional to its frequency.
Relevant Equation:
$\mathrm{E = h f}$ where \(\mathrm{E}\) is the photon energy, \(\mathrm{h = 6.63 \times 10^{-34} \, J \cdot s}\) is Planck’s constant, and \(\mathrm{f}\) is the frequency of light.
Photon Motion:
- Photons travel in straight lines unless they interact with matter (e.g., scattering, absorption, refraction).
- The speed of photons depends on the medium through which they travel.
- In free space, all photons move at the speed of light: $\mathrm{c = 3.00 \times 10^8 \, m/s}$
- In a medium with refractive index \(\mathrm{n}\), the photon speed is: $\mathrm{v = \dfrac{c}{n}}$ showing that photon speed is inversely proportional to the index of refraction.
Example :
A photon of light has a frequency of \(\mathrm{6.0 \times 10^{14} \, Hz}\). Calculate the energy of this photon in joules and in electron volts (1 eV = \(\mathrm{1.6 \times 10^{-19} \, J}\)).
▶️ Answer/Explanation
Step 1: Use photon energy equation: $\mathrm{E = h f}$ $\mathrm{E = (6.63 \times 10^{-34})(6.0 \times 10^{14})}$
$\mathrm{E = 3.98 \times 10^{-19} \, J}$
Step 2: Convert to electron volts: $\mathrm{E = \dfrac{3.98 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.49 \, eV}$
Final Answer: $\mathrm{3.98 \times 10^{-19} \, J \, \, (2.49 \, eV)}$
Example:
Light of wavelength \(\mathrm{500 \, nm}\) passes through a glass medium of refractive index \(\mathrm{n = 1.5}\). Calculate the speed of photons in the medium and their frequency.
▶️ Answer/Explanation
Step 1: Speed of photons in medium:
$\mathrm{v = \dfrac{c}{n} = \dfrac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8 \, m/s}$
Step 2: Frequency of photons (unchanged by medium):
$\mathrm{f = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8}{500 \times 10^{-9}} = 6.0 \times 10^{14} \, Hz}$
Final Answer: Photon speed in glass = $\mathrm{2.0 \times 10^8 \, m/s}$ Frequency = $\mathrm{6.0 \times 10^{14} \, Hz}$
Wave Properties of Particles
Quantum theory extends the concept of wave-particle duality to matter. Particles such as electrons, neutrons, and even atoms can demonstrate wave-like behavior under appropriate experimental conditions.
Evidence: Variations of Young’s double-slit experiment performed with electrons (or other particles) show interference patterns, confirming that particles can behave like waves.
De Broglie Hypothesis:
Louis de Broglie proposed that every moving particle has an associated wavelength, known as the de Broglie wavelength:
$\mathrm{\lambda = \dfrac{h}{p}}$ where \(\mathrm{h}\) is Planck’s constant and \(\mathrm{p}\) is the momentum of the particle.
Dependence on Momentum:
- The wavelength increases when momentum decreases.
- Low-momentum particles (e.g., slow electrons) have longer de Broglie wavelengths, making their wave nature more observable.
- High-momentum (fast-moving) particles have very short wavelengths, so their wave properties are less noticeable.
Quantum Significance:
Quantum theory is necessary when the de Broglie wavelength is comparable to the size of the system under study (e.g., atoms, electrons in crystals).
Quantization of Energy and Momentum:
In bound systems (like electrons in an atom), only certain discrete values of energy and momentum are allowed. This explains atomic spectra and the stability of matter.
Example :
An electron is accelerated from rest through a potential difference of \(\mathrm{150 \, V}\). Calculate its de Broglie wavelength.
▶️ Answer/Explanation
Step 1: Energy gained by the electron: $\mathrm{E = q \Delta V = (1.6 \times 10^{-19})(150) = 2.4 \times 10^{-17} \, J}$
Step 2: Momentum of the electron: $\mathrm{p = \sqrt{2 m E} = \sqrt{2 (9.11 \times 10^{-31})(2.4 \times 10^{-17})}}$ $\mathrm{p \approx 6.6 \times 10^{-24} \, kg \, m/s}$
Step 3: de Broglie wavelength: $\mathrm{\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34}}{6.6 \times 10^{-24}} \approx 1.0 \times 10^{-10} \, m}$
So the electron’s wavelength is about \(\mathrm{0.1 \, nm}\), which is comparable to atomic spacing — making wave behavior significant.
Example :
A baseball of mass \(\mathrm{0.15 \, kg}\) is thrown at \(\mathrm{40 \, m/s}\). Find its de Broglie wavelength and comment on whether wave properties are observable.
▶️ Answer/Explanation
Step 1: Momentum of baseball: $\mathrm{p = mv = (0.15)(40) = 6.0 \, kg \, m/s}$
Step 2: de Broglie wavelength: $\mathrm{\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34}}{6.0} \approx 1.1 \times 10^{-34} \, m}$
Step 3: Interpretation: This wavelength is unimaginably small compared to atomic scales (\(\sim 10^{-10} \, m\)). Therefore, the wave properties of macroscopic objects like a baseball are not observable.