DHS 19 Quantum Physics (Lecture Slides)

2025 DHS Y6 Physics H2: Quantum Physics 16/4/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 1 H2 Quantum Physics 1 Content  Energy of a photon  The photoelectric effect  Wave-particle duality  Energy levels in atoms  Line spectra  X-ray spectra  The uncertainty principle 2 Learning Outcomes: 3 Intro: Modern Physics and Quantum Mechanics • Developed from 1900s to explain atomic-scale phenomena • Energy exists in discrete packets (quanta) • Wave-particle duality • Uncertainty principle • Changed our understanding of nature: • Particles are described by quantum wavefunctions (probabilistic in nature) • Light and matter have dual wave-particle nature • Led to modern technologies: • Electronics and computers • Medical imaging • Lasers 4 1 2 3 4

2025 DHS Y6 Physics H2: Quantum Physics 16/4/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 2 H2 Quantum Physics • Energy of a photon • The photoelectric effect 5 Photons •Concept introduced by Max Planck and Albert Einstein •Photon: A discrete packet (quantum) of energy of electromagnetic radiation •Behaves like a particle •Energy of a photon: E = hf where: E = energy of photon h = Planck's constant (6.63 × 10−³⁴ J s) f = frequency of radiation 6 Question 1 Determine the energy in joule of a high-energy gamma photon of frequency 1026 Hz. Solution:    34 26 1 8 6.63 10 J s 10 s 6.63 10 J E hf         7 (have frequencies above 3 x 1019 Hz) Electron-volts • Q1 shows that the energy for a high energy photon is far less than 1.0 J • Joule is not a convenient unit for measuring photon energies • The electronvolt (eV) is often used instead 1 eV is the energy gained by an electron when it is accelerated through a potential difference of 1 V • Energy gained = QΔV = (1 e) (1 V) = (1.60 × 10–19 C) (1 V) = 1.60 × 10–19 J • Therefore, 1 eV = 1.60 × 10–19 J 8 5 6 7 8

2025 DHS Y6 Physics H2: Quantum Physics 16/4/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 3 Question 2 Visible light has wavelength spanning from 400 nm (violet) to 700 nm (red). Find, in eV, the maximum energy of a photon of visible light. Solution: Emax = hfmax = hc/λmin = (6.63 × 10–34) (3.00 ×108) / (400 ×10–9) = 4.97 × 10–19 J = 3.11 eV 9 Question 3 Determine the number of photons emitted per second by a 60 W violet light source. Wavelength of the violet light source is 400 nm. Solution: Power of the light source = Total energy of light source per second = number of photons per second× energy of a violet photon 60 = n/t × 4.97 ×10–19 (from Question 2) n/t = 1.21×1020 photons per second 10 Photoelectric effect The emission of electrons from a cold metal surface when electromagnetic radiation of a sufficiently high frequencyfalls on it. 11 (conduction electrons in a metal which are not attached to a particular atom) Photoelectric effect (mechanism) • A photon of energy hf is incident on the metal surface • All the energy of the photon is being absorbed by the electron Note: not all photons get to interact with electrons • Some energy is used to overcome the forces from positive ions holding the electron in the metal, the rest is KE of the emitted electron. 12 9 10 11 12

2025 DHS Y6 Physics H2: Quantum Physics 16/4/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 4 Potential well and Photoelectric Effect photon energy (hf) max KE of electron Einstein’s Photoelectric Equation Using the idea of a photon, by conservation of energy: The work function energy Φ of a metal is the minimum energy of photon to cause emission of electron from surface of a metal. •It depends on the nature of the material of the metal [e.g., Ca: 2.9 eV, Al: 4.08 eV, Fe: 4.9 eV] and its surface conditions (like contamination, which can reduce Φ). 14 21 max2hf mv  Einstein’s Photoelectric Equation • The threshold frequency fo is the lowest frequency of electromagnetic radiation that gives rise to the ejection of electrons from the metal surface. • The threshold wavelength λo is the highest wavelength of electromagnetic radiation that gives rise to the ejection of electrons from the metal surface. • At this frequency/wavelength, the amount of energy supplied by each photon is just able to overcome the work function energy, hence KE of emitted electrons is 0. 15 21 max2hf mv  o o hchf   Characteristics of the photoelectric effect •A single photon can only interact, and hence exchange its energy with a single electron (one-to-one interaction). It is extremely unlikely for any electron to simultaneously receive energy from multiple photons. •Not all photons (of sufficient energy) get to interact with electrons. Most are reflected from metal surface. •The photoelectrons are emitted in all random directions with varying speeds. 16 13 14 15 16

2025 DHS Y6 Physics H2: Quantum Physics 16/4/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 5 Characteristics of the photoelectric effect When Incident Radiation Frequency < Threshold Frequency: •Photon still transfers energy to an electron •Electron cannot escape the metal surface •Energy absorbed appears as electron kinetic energy •Electrons collide with metal ions, losing kinetic energy •Result: Metal heats up Example: Metal plate near a table lamp gets hot 17 Question 4 The maximum KE of the electrons emitted from a metallic surface is 1.0 eV when the frequency of the incident radiation is 7.5 ×1014 Hz. Calculate (a) the work function energy of the metal in joules, (b) the threshold wavelength for emission of electrons from the metal. (a) (6.63 × 10−34)(7.5 × 1014) = Φ + (1.0) (1.60 × 10−19) Φ = 3.37 × 10−19 J (b) 3.37  10−19 = (6.63  10−34) (3.0  108 / o) o = 5.90  10−7 m 18 21 max2hf mv  o hc   Question 5 Suggest why the emitted electrons are likely to have a range of kinetic energy (instead of just the maximum KE) for any one frequency of the electromagnetic radiation. 19 • Maximum KE corresponds to the electrons emitted from the metal surface • Photon may interact with other electrons not at the surface; electrons emitted from deeper in the metal lose energy escaping • Some electrons may collide with other particles whilst escaping; collisions reduce final kinetic energy Photoelectric Effect Experimental Setup 20 Components: • Evacuated glass chamber • Emitter (E): Metal electrode (negative terminal) • Collector (C): Metal electrode (positive terminal) • Voltage source • Ammeter • Monochromatic light source 17 18 19 20

2025 DHS Y6 Physics H2: Quantum Physics 16/4/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 6 Photoelectric Effect Experimental Setup 21 Observations: 1.In darkness: No current 2.Light on (f > threshold): • Photoelectrons emitted from E • Current detected Key Points: • V: Potential of collector relative to emitter • Photoelectrons emitted with various kinetic energies • Electric field accelerates electrons towards C • All emitted electrons contribute to photocurrent i i V 21

2025 DHS Y6 Physics H2: Quantum Physics 24/4/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 1 Photoelectric Effect Experimental Setup 1 • Reversing e.m.f. source polarity creates opposing electric field • Adjusting field strength prevents less energetic electrons from reaching collector • Stopping potential (Vs): Minimum reverse voltage to stop current flow • When V = −Vs, current i stops • Vs determines maximum kinetic energy of emitted electrons i V −Vs Using Stopping Potential to Find Maximum Kinetic Energy 2 Electron movement: • From emitter to collector • Potential decreases by Vs • EPE of electron increases, KE decreases • Most energetic electron leaves E with KEmax has zero KE at C Energy conservation:      s max s 21 max max2 0 0 0 KE E e KE mv e KE V V PE            maximum kinetic energy with which electrons leave the emitter The minimum potential, Vs, that is applied to stop the most energetic electrons is the stopping potential. Classical wave theory: Light is a propaga