17 - Particle nature of light

Question 1

Intensity of light

Answer 1

• radiation flux density = power/area
• if a sphere is a distance of r from a source of light, the light spreads over an area of 4(pi)r^2, so radiation flux from the sun is power/4(pi)r^2

Question 2

The Photoelectric effect

Answer 2

• Photons are particles of light that carry a set amount of energy depending on their frequency (E = plancks constant*frequency)
• Work function of a metal = minimum energy a single photon has to have to remove an electron from the metals surface
• If light was a wave, increasing intensity would increase energy transferred, but as it acts like a particle unless each photon has sufficient energy to overcome the work function no electrons will move. This is why small amounts of UV work, but no IR will

Question 3

using the photoelectric effect

Answer 3

• Einstein’s equation = hf = work function+1/2mv^2
• any spare energy from light is given as the kinetic energy of the particle
• by shining light on a cathode to release electrons onto a negative anode, the repulsion of the cathode will try to force the electrons back. increasing the voltage until a current shows means the electrons have overcome the force of the anode, and completed the circuit. this can show us the work function of the metal, or the frequency of the light

Question 4

Atomic spectra

Answer 4

• the wavelengths of light given out by atoms of specific elements when their electrons are excited
• light is given off when electrons with energy jump to higher energy levels, and then fall back down, releasing the energy as light. as electrons only go up and down in whole steps, specific atoms only release specific frequencies of light
• as well as this, these elements only absorb the same frequencies of light, so an absorption spectra shows the whole spectrum minus those colours, while an emission spectra only shows those colours

Question 5

The electronvolt

Answer 5

• small energies are better expressed in electronvolts, or the energy an electron has when it moves through a p.d. of 1 volt
• this is the same as the charge on an electron, 1.6*10^-19, but in Joules not Coulombs
• converting between eV and J is just multiplying Joules by 1.610^-19 and dividing eV by 1.610^-19