AS Physics - Light And All That


Photoelectric Effect

A negatively charged zinc plate holds a large amount of electrons, however, if light with a high enough frequency (ultraviolet light) hits the surface, the zinc loses charge quickly
This is because the high frequency gives the electrons enough energy to be released from the metal
This is shown using either a coulombmeter or a gold-leaf electroscope
The gold-leaf repelling from the electroscope shows that the electroscope is negatively charged. When ultraviolet light hits the zinc plate, the gold-leaf falls, showing a loss of charge
If the experiment is repeated with a sheet of glass between the light and the zinc plate, the gold-leaf will fall slower than without, because glass absorbs UV light
This process forms ‘photoelectrons,’ which are electrons which are given their energy by light (photo)

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Photoelectric Effect - Proof Against a Wave

Waves with bigger amplitudes carry larger amounts of energy and in light, bigger amplitudes mean a bright light

A dim blue light will make a lithium surface give off a few electrons,so there will be a mall, measurable current.
A brighter blue light will give a greater photoelectric current, as there are more photoelectrons.
This is as expected, but a red light of the same brightness gives off zero current, which is impossible to explain if light is a wave. This is because the strong red light would be a wave of greater amplitude, and so have enough energy to free electrons, however this doesn’t happen, proving light isn’t acting as a wave

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Light as a Particle

Planck suggested that electromagnetic energy comes in ‘lumps’ called quanta (sing = quantum). Einstein used this idea to explain the photoelectric effect.
He suggested that each quantum an provide the energy for one electron to escape the metal. If the quantum is too small, the electron cannot escape.
The name for a particle of light is ‘photon’
The photons of the dim blue light (on card 2) contain enough energy to eject some electrons from the lithium surface. A dim blue light has few photons, so few electrons get released.
The brighter blue light has more of these high-energy photons and so causes the lithium to release a larger amount of photoelectrons, giving a higher current
The red light does not contain any photons with high enough energy, but does release a large amount of them. This explains why the red light does not free any photoelectrons

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Planck’s Equation

Energy of a photon, E, in Joules = the Planck Constant, h, in Joules-second X frequency, f, in Hertz


This uses both the ideas for a particle (E) and for a wave (f)
Going through the spectrum of visible light, from red on the left to purple on the right, red light has the lowest energy, while purple has the highest energy.

To convert joules into electron-volts:
Energy in eV = Energy in joules / 1.6x10^-19 (J eV^-1)

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Photons on the Electromagnetic Spectrum

Radio waves - Lowest amount of energy in photons
Microwaves - 2nd lowest amount of energy in photons
Infrared waves - 3rd lowest amount of energy (less than red in visible light)
Visible light
Ultraviolet - Photons are more energetic, which explains why this light is hazardous
X-rays and Gamma rays - Both of these have even higher frequencies, and so their photons have even more energy (making them more dangerous, too)

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Work Function

In a graph, of ‘Kinetic energy of photoelectrons, eV’ on the y-axis, and the ‘Frequency of light, 10^14 Hz’ on the x-axis, then the x-intercept is the threshold frequency.
The threshold frequency is the minimum amount of energy required for electrons to be released from the surface of a metal. The gradient of the line is Planck’s constant, for every metal.
Where the kinetic energy is zero, this means that all of the kinetic energy is used to get the electron out of the metal (and has no energy to continue moving). The amount of energy needed to remove the electron is called the work function

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Photoelectric Equation

Kinetic energy of the electron = energy of photon - energy needed to be released from the metal


Maximum kinetic energy of electron, Ek = (the Planck constant, h, x frequency, f) - work function

Ek = hf - O OR 0.5 x mv^2 = hf - O

The second form of the equation can be used to work out the velocity of the electrons

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de Broglie

In an electron-tube experiment, most of the electrons pass straight through the very thin graphite, but some electrons pass through at certain angles only, showing rings on the fluorescent screen. This rings are like the interference maxima you see when Ishtar waves pass through a diffraction grating.
This is proof that electrons can act as a wave, as well as a particle (as waves can be diffracted by a gap and they can also superpose to give interference patterns)
The rings on the fluorescent screen are where the waves are in phase

The slits in a diffraction grating must be as small as an atom to have an effect, as de Broglie’s theory suggests that the wavelength of electron waves are about the size of an atom

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de Broglie’s Equation

Momentum of the particle = the Planck constant / wavelength of a wave

Or m v = h / λ So λ = h / m v

For photons, always use Planck’s equation, and for the wave behaviour of subatomic ‘particles’ like electrons, always use de Broglie’s equation

As the equation above shows, the wavelength is inversely proportional to the momentum, which means that the way in which kinetic energy, momentum, and wavelength change with p.d. V is: larger V => larger Ek => larger mv => smaller λ

To calculate a wavelength from a given p.d., use the equation eV = 0.5 x mv^2

v ^2 = 2eV / m multiply by m^2, m^2 v^2 = m^2(2eV / m)

m^2 v^2 = 2m eV

λ = h / ((2m eV)^0.5)

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