OCR A-Level Physics/Electrons, Waves and Photons NEW SPECIFICATION/Quantum physics

The photon model

Electromagnetic waves travel through space as waves, which is shown by their ability to diffract and interfere. However, when waves interact with matter, they do so as packets of discrete energy called quanta. A quantum of electromagnetic energy is called a photon.

The energy of a photon $E_{\gamma }$ is related to the frequency $f$ of the electromagnetic radiation it represents by the following equation;

$E_{\gamma }=hf$

Where $h$ is the Planck constant which is equal to $6.63\times 10^{-34}$ Js.

Frequency is related to wavelength $\lambda$ by the following equation;

$v=\lambda f$

Where $v$ is the speed of the wave. All photons travel at the speed of light in a vacuum $c$, therefore for a photon;

$c=\lambda f$

$f=\frac{c}{\lambda }$

$\therefore E_{\gamma }=\frac{hc}{\lambda }$

Where $\lambda$ is the wavelength of the photon and $c$ is the speed of light in a vacuum which is equal to $3\times 10^8$ ms^-1.

The electronvolt

The electronvolt (eV) is a unit of energy smaller that the Joule (J) which is more convenient for calculations at the quantum level. The electronvolt is defined as the energy gained by an electron accelerated through a potential difference of one Volt. Voltage $V$ is defined as the energy transferred $W$ per unit charge $Q$, which can be rearranged as;

$W=VQ$

For an electronvolt, the voltage is equal to one Volt, and the charge is equal to the charge of an electron $e$, therefore;

$W=1\times e$

$1$ eV $=1.6\times 10^{-19}$J

Calculating $h$ using LEDs

Light emitting diodes (LEDs) emit waves of light, and therefore photons. However, LEDs are diodes, so will only begin allowing current through them once a certain voltage $V_c$ called the threshold voltage is exceeded. Photons are emitted when the voltage of the circuit is equal to the threshold voltage and when electrons in the LED lose energy.

Therefore the energy lost by electrons is roughly equal to the energy gained by the photons. This is only an approximation because no LED is 100% efficient. We can use this to find an approximation for the Planck constant, $h$.

$E_e\approx E_{\gamma }$

$e{V}_c\approx \frac{hc}{\lambda }$

$V_c\approx (\frac{hc}{e})\frac{1}{\lambda }$

$V_c\propto \frac{1}{\lambda }$

Therefore, the greater the threshold frequency of the LED, the smaller the wavelength of the emitted photons of light. Plotting $V_c$ on the y-axis and $\frac{1}{\lambda }$ on the x-axis of a graph will give a straight line which passes roughly through the origin. The gradient $m$ of the line would be defined by the following relationship;

$m\approx \frac{hc}{e}$

So rearranging for $h$ we get;

$h\approx \frac{me}{c}$

The photoelectric effect

The photoelectric effect: incident photons (red) absorbed by the metal plate induce the emission of photoelectrons (blue)

The photoelectric effect is the phenomenon by which the absorption of EM radiation of sufficiently high frequency by a metal induces the emission of electrons from the surface. Electrons emitted by this phenomenon are known as photoelectrons.

The gold-leaf electroscope experiment

The photoelectric effect can be demonstrated using a gold-leaf electroscope. In this experiment, the metal plate, stem and leaf are charged negatively to begin with. The like charges repel such that the gold leaf rises. If the metal plate is discharged, the gold leaf will collapse back to a vertical position.

Diagram showing the parts of a gold-leaf electroscope

When visible light, no matter the intensity, is directed towards the metal plate, the gold leaf stays risen. When UV light, even at very low intensities, is directed towards the metal plate, the gold leaf collapses. Therefore the intensity of the EM radiation incident on the metal plate does not affect the emission of electrons, but the frequency of the light does. This proves that the incident light must arrive in discrete packets (photons) and that the plate will only emit electrons if the energy of the photons is above a certain threshold. The energy of a photon must be great enough to liberate an electron from the electrostatic influence of atomic protons.

Einstein's photoelectric equation

Albert Einstein derived an equation to describe the results of the gold-leaf electroscope experiment and the photoelectric effect. This is known as the photoelectric equation, and states that the energy $E_{\gamma }$ of an incident photon is equal to sum of: the minimum energy required to liberate an electron from the surface of the metal, $\varphi$ (also known as the work function); and the maximum kinetic energy that electron has on release, $K{E}_{max}$.

$E_{\gamma }=\varphi +K{E}_{max}$

$hf=\varphi +K{E}_{max}$

When the energy of the photon is less than the work function, no electrons are emitted. When the energy of the photon is equal to or greater than the work function, electrons will be emitted.

Threshold frequency

The threshold frequency $f_0$ is the minimum frequency a photon has to have to be able to liberate an electron from the surface of the metal. Thus, when $f=f_0$, $K{E}_{max}=0$.

$\therefore h{f}_0=\varphi$

$f_0=\frac{\varphi }{h}$

Threshold wavelength

The threshold frequency ${\lambda }_0$ is the minimum wavelength a photon has to have to be able to liberate an electron from the surface of the metal. Thus, when $\lambda ={\lambda }_0$, $K{E}_{max}=0$.

$\therefore \frac{hc}{{\lambda }_0}=\varphi$

$hc=\varphi {\lambda }_0$

${\lambda }_0=\frac{hc}{\varphi }$

Maximum velocity of a liberated photoelectron

The kinetic energy of a particle and the velocity of a particle are related by the equation;

$E_k=\frac{1}{2}m{v}^2$

A liberated electron has a maximum kinetic energy $K{E}_{max}$ which is therefore related to its maximum velocity $v_{max}$ by the equation;

$K{E}_{max}=\frac{1}{2}{m}_e(v_{max}{)}^2$

$\therefore hf-\varphi =\frac{1}{2}{m}_e(v_{max}{)}^2$

$v_{max}=\sqrt{\frac{2(hf-\varphi )}{m_e}}$

Where $m_e$ is the mass of an electron which is equal to $9.11\times 10^{-31}$kg.

Wave-particle duality

We already know that EM radiation can behave as both waves and particles. In 1923, Louis de Broglie proposed that all matter, including electrons, has a dual wave-particle nature.

Electron diffraction

The wave-particle duality of electrons can be proved using an electron diffraction tube. Electrons are accelerated through a thin sheet of graphite which acts as a diffraction grating, producing a pattern of rings on the screen. This pattern shows that electrons must interfere constructively and destructively to form maxima and minima respectively. This implies that the electrons must have a phase difference, implying that they behave as waves.

Because the electrons diffract, the wavelength of the electrons must be similar to the size of the gap between carbon atoms in the graphite.

The de Broglie equation

A particle travelling through space with momentum $p$ has an associated wavelength $\lambda$ given by the de Broglie equation;

$\lambda =\frac{h}{p}=\frac{h}{mv}$

Where $h$ is the Planck constant which is equal to $6.63\times 10^{-34}$ Js.

This can be validated by measuring the gap between carbon atoms in graphite, which has a similar order of magnitude to the predicted de Broglie wavelength of an electron when the electron is travelling with the correct velocity $v$ that causes diffraction.

'Derivation' the de Broglie equation

The de Broglie equation can be 'derived' using the energy of a photon and Einstein's mass-energy equation. However, this is not a 'true' derivation because in order to be applied to objects whose speed is not equal to the speed of light in a vacuum, we need to change our notation from $c$ to $v$ in the final step, which ignores the fact that Einstein's mass-energy equation uses $c$ as a constant.

$E=m{c}^2$

$E=\frac{hc}{\lambda }$

$\therefore \frac{hc}{\lambda }=m{c}^2$

$\frac{h}{\lambda }=mc$

$\frac{\lambda }{h}=\frac{1}{mc}$

$\lambda =\frac{h}{mc}$

$\lambda =\frac{h}{p}=\frac{h}{mv}$