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