Physics

Dawn of Modern Physics

4,447 words, 10 sections

This chapter is syllabus unit Physics 14, Dawn of Modern Physics, and it carries 227 distinct questions across 76 official papers between 2000 and 2026, examined by Balochistan, DUHS, Federal, KMU, NUMS and UHS, with a mean item difficulty of 1.62 and 1239 active questions behind it in the practice bank. Physics is worth 36 marks of the 180-mark paper. The weight is concentrated in one place: quantum theory and the photoelectric effect account for 149 of the questions, special relativity for 44, and wave-particle duality with the de Broglie hypothesis for 34. A student who can use $E = hf$ , Einstein's photoelectric equation, the relativity factor and $\lambda = h/mv$ will answer most of the numericals, and the chapter rewards exact recall of a small set of constants.

Planck's quantum theory states that energy is emitted or absorbed not continuously but in discrete packets called quanta. A quantum of electromagnetic radiation is a photon, and the energy of one photon of frequency $f$ is $E = hf$ , or in terms of wavelength $E = \dfrac{hc}{\lambda}$ . The energy of a beam at the level of a single quantum is therefore proportional to frequency and inversely proportional to wavelength, so a violet photon carries more energy than a red one, an ultraviolet photon more than a visible one, and a gamma photon more than either. The Planck constant is $h = 6.63 \times 10^{-34}\text{ J s}$ , and it is the ratio of the energy difference between two quantum states to the frequency of the radiation emitted, $h = \dfrac{\Delta E}{f}$ .

A photon is a quantum of energy with zero rest mass and zero charge. Being massless, it has no rest energy, yet it carries momentum, $p = \dfrac{h}{\lambda} = \dfrac{E}{c} = \dfrac{hf}{c}$ . Einstein's particle model of light treats each photon as a self-contained packet that strikes one electron at a time, which is why the interaction is instantaneous and why the energy of each photon, not the brightness of the beam, decides what a single electron receives. An atom occupying a single quantum state has constant energy and neither emits nor absorbs radiation until it changes state, and when it does the radiation appears as photons.

Worked example

(a) Find the energy of a photon of frequency $1.2 \times 10^{14}\text{ Hz}$ . Then $E = hf = 6.63 \times 10^{-34}\text{ J s} \times 1.2 \times 10^{14}\text{ Hz} = 7.96 \times 10^{-19}\text{ J}$ . (b) A photon of wavelength $1240\text{ nm}$ has $E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{1240 \times 10^{-9}} = 1.6 \times 10^{-19}\text{ J} = 1.0\text{ eV}$ . The step students get wrong is forgetting that $\text{Hz} = \text{s}^{-1}$ in part (a), or leaving the answer in joules when the options are in electron volts in part (b); $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ is the conversion.

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Introduction

1. The quantum idea and the photonPlanck's quantum theory states that energy is emitted or absorbed not continuously but in discrete packets called quanta. A quantum of electromagnetic radiation