To introduce quantum physics and the photon model

What is quantum physics? Quantum physics explores how, at the very small scale, we use different models to describe electromagnetic radiation and subatomic particles like electrons It is amongst the most modern areas of physics having been established at the end of the nineteenth century and the beginning of the twentieth century.

What is an electromagnetic wave? An electromagnetic wave is a transverse wave. It transfers energy through space at the speed of light c=3x108 ms-1 There are different types, classified by wavelength. Longest wavelength is radio waves Shortest wavelength is gamma rays Visible light is a good example of an electromagnetic wave

Photons German physicist Max Planck discovered in 1900 that electromagnetic energy appeared in “small packets” known as quanta. His model proposed that electromagnetic radiation has a particulate nature – that is it is tiny packets of energy rather than a continuous wave. Einstein coined a new term for these packets, photons.

Key definitions A quantum (plural quanta) is a small discrete unit of energy A photon is a quantum associated with electromagnetic radiation The Planck Constant, h, has a value 6.626 x 10-34 and photon energies are always emitted in multiples of this

Photon energy The energy of a single photon is directly proportional to its frequency: E = hf E is the energy of the photon in Joules, J f is the frequency of the electromagnetic radiation in Hz h is the Planck constant. It has an experimental value of 6.626x10-34 Js

In terms of wavelength The wave equation c = f𝛌 c = speed of light (3x108ms-1), f = frequency and 𝛌 = wavelength Rearranging E = hf or E= ℎ𝑐 𝛌

Example A laser emits red light with a wavelength of 633nm. Calculate the energy if each red photon emitted E=hc/𝛌, = (6.63x10-34 x 3.00 x 108) / 633 x 10-9 = 3.14 x 10-19 J

Example 2 A monochromatic source of red light has a power of 12W and a wavelength of 720nm. Calculate The frequency of the light at this wavelength The energy of the light at this wavelength The number of photons emitted per second a) c= f𝛌. b) E = hf c) Energy of one photon is 2.8x10-19. 12W light source provides 12J of energy per sec. So divide total energy by energy of one photon to find number of photons per sec. 12÷2.8𝑥10−19

You try Q2-5 from first block on sheet

Questions of scale At the subatomic scale of the quantum level, the SI unit of energy (the Joule) is huge. Even the most energetic gamma photons only have energy of the order of tens of millijoules. So we use the electronvolt eV to measure energies at the quantum scale

The electronvolt The energy of 1 electronvolt (1eV) is defined as the kinetic energy gained by an electron when it moves through a potential difference of 1V Energy = Charge x potential difference = QV Energy = charge on the electron x 1V = 1.60 x 10-19C x 1V = 1.60 x 10-19J 1eV = 1.60 x 10-19J So if an electron is accelerated through a potential difference of 100V, it will gain 100eV of kinetic energy

Example To convert from eV to J → ×𝟏.𝟔𝟎× 𝟏𝟎 −𝟏𝟗 To convert from J to eV → ÷𝟏.𝟔𝟎× 𝟏𝟎 −𝟏𝟗 Convert 6eV into Joules Convert 50nJ into eV 6 x 1.6x10-19 = 50x10-9 ÷1.6𝑥10−19

You try Questions from the second block on the sheet NB charge on an alpha particle is +2e

Homework Read up in the text book page 244 about the experiment to determine the Planck constant This will be our experiment next week Plan your experiment, and know the graphs you will draw and how to calculate the Planck constant