THE PHOTON
The Particle Nature of Light
- In classical wave theory, electromagnetic (EM) radiation is assumed to behave as a wave
- This is demonstrated by the fact EM radiation exhibits phenomena such as diffraction and interference
- However, experiments from the last century, such as the photoelectric effect and atomic line spectra, can only be explained if EM radiation is assumed to behave as particles
- These experiments have formed the basis of quantum theory, which will be explored in detail in this section
- Photons are fundamental particles which make up all forms of electromagnetic radiation
- A photon is a massless “packet” or a “quantum” of electromagnetic energy
- What this means is that the energy is not transferred continuously, but as discrete packets of energy
- In other words, each photon carries a specific amount of energy, and transfers this energy all in one go, rather than supplying a consistent amount of energy
Exam Tip
Make sure you learn the definition for a photon: discrete quantity / packet / quantum of electromagnetic energy are all acceptable definitions
Calculating Photon Energy
The energy of a photon can be calculated using the formula:
$
E=h f
$
Using the wave equation, energy can also be equal to:
$
E=h \frac{c}{\lambda}
$
- Where:
- $E=$ energy of the photon $(J)$
- $\mathrm{h}=$ Planck’s constant $(\mathrm{J} \mathrm{s})$
- $\mathrm{c}=$ the speed of light $\left(\mathrm{m} \mathrm{s}^{-1}\right)$
- $f=$ frequency in Hertz $(\mathrm{Hz})$
- $\lambda=$ wavelength $(\mathrm{m})$
- This equation tells us:
- The higher the frequency of EM radiation, the higher the energy of the photon
- The energy of a photon is inversely proportional to the wavelength
- A long-wavelength photon of light has a higher energy than a shorter-wavelength photon
Worked example: Calculating photon energy
Light of wavelength $490 \mathrm{~nm}$ is incident normally on a surface, as shown in the diagram.
The power of the light is $3.6 \mathrm{~mW}$. The light is completely absorbed by the surface.
Calculate the number of photons incident on the surface in $2.0 \mathrm{~s}$.
Answer/Explanation
Step 1:
Write down the known quantities
Wavelength, $\lambda=490 \mathrm{~nm}=490 \times 10^{-9} \mathrm{~m}$
Power, $P=3.6 \mathrm{~mW}=3.6 \times 10^{-3} \mathrm{~W}$
Time, $t=2.0 \mathrm{~s}$
Step 2:
Write the equations for wave speed and photon energy
wave speed: $\quad \mathrm{c}=\mathrm{f} \lambda \rightarrow \mathrm{f}=\frac{c}{\lambda}$
photon energy: $\quad \mathrm{E}=\mathrm{hf} \rightarrow \mathrm{E}=\frac{h c}{\lambda}$
Step 3: $\quad$ Calculate the energy of one photon
$
E=\frac{h c}{\lambda}=\frac{\left(6.63 \times 10^{-34}\right) \times\left(3.0 \times 10^8\right)}{490 \times 10^{-9}}=4.06 \times 10^{-19} \mathrm{~J}
$
Step 4:
Calculate the number of photons hitting the surface every second
$
\frac{\text { Power of light source }}{\text { Energy of one photon }}=\frac{3.6 \times 10^{-3}}{4.06 \times 10^{-19}}=8.9 \times 10^{15} \mathrm{~s}^{-1}
$
Step 5:
Calculate the number of photons that hit the surface in $2 \mathrm{~s}$
$
\left(8.9 \times 10^{15}\right) \times 2=1.8 \times 10^{16}
$
Exam Tip
The values of Planck’s constant and the speed of light will always be given to you in an exam, however, it helps to memorise them to speed up calculation questions!
Photon Momentum
- Einstein showed that a photon travelling in a vacuum has momentum, despite it having no mass
- The momentum (p) of a photon is related to its energy (E) by the equation:
$
p=\frac{E}{c}
$
- Where $c$ is the speed of light
Worked example: Calculating photon momentum
A $5.0 \mathrm{~mW}$ laser beam is incident normally on a fixed metal plate. The cross-sectional area of the beam is $8.0 \times 10^{-6} \mathrm{~m}^2$. The light from the laser has frequency $5.6 \times 10^{14} \mathrm{~Hz}$. Assuming all the photons are absorbed by the plate, calculate the momentum of the photon, and the pressure exerted by the laser beam on the metal plate.
Answer/Explanation
Step 1:
Write down the known quantities
Power, $P=5.0 \mathrm{~mW}=5.0 \times 10^{-3} \mathrm{~W}$
Frequency, $f=5.6 \times 10^{14} \mathrm{~Hz}$
Cross-sectional area, $A=8.0 \times 10^{-6} \mathrm{~m}^2$
Step 2:
Write the equations for photon energy and momentum
photon energy: $E=$ hf
photon momentum: $\quad \mathrm{p}=\frac{\underline{E}}{c} \rightarrow \mathrm{p}=\frac{h f}{c}$
Step 3:
Calculate the photon momentum
$$
\mathrm{p}=\frac{h f}{c}=\frac{\left(6.63 \times 10^{-34}\right) \times\left(5.6 \times 10^{14}\right)}{3.0 \times 10^8}=1.24 \times 10^{-27} \mathrm{~N} \mathrm{~s}
$$
Step 4:
Calculate the number of photons incident on the plate every second
$\frac{\text { Power of light source }}{\text { Energy of one photon }}=\frac{5.0 \times 10^{-3}}{h f}=\frac{5.0 \times 10^{-3}}{\left(6.63 \times 10^{-34}\right) \times\left(5.6 \times 10^{14}\right)}=1.35 \times 10^{16} \mathrm{~s}^{-1}$
Step 5:
Calculate the force exerted on the plate in a $1.0 \mathrm{~s}$ time interval
$
\begin{aligned}
& \text { Force }= \text { rate of change of momentum } \\
&= \text { number of photons per second } \times \text { momentum of each } \\
& \quad \text { photon } \\
&=\left(1.35 \times 10^{16}\right) \times\left(1.24 \times 10^{-27}\right) \\
&=1.67 \times 10^{-11} \mathrm{~N}
\end{aligned}
$
Step 6:
Calculate the pressure
$
\text { Pressure }=\frac{\text { Force }}{\text { Area }}=\frac{1.67 \times 10^{-11}}{8.0 \times 10^{-6}}=2.1 \times 10^{-6} \mathbf{P a}
$
THE ELECTRONVOLT
- The electronvolt is a unit which is commonly used to express very small energies
- This is because quantum energies tend to be much smaller than 1 Joule
- The electronvolt is derived from the definition of potential difference:
$
V=\frac{E}{Q}
$
- When an electron travels through a potential difference, energy is transferred between two points in a circuit, or electric field
- If an electron, with a charge of $1.6 \times 10^{-19} \mathrm{C}$, travels through a potential difference of $1 \mathrm{~V}$, the energy transferred is equal to:
$
E=Q V=1.6 \times 10^{-19} \mathrm{C} \times 1 \mathrm{~V}=1.6 \times 10^{-19} \mathrm{~J}
$
- Therefore, an electronvolt is defined as:
The energy gained by an electron travelling through a potential difference of one volt
$
1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}
$
Relation to kinetic energy
- When a charged particle is accelerated through a potential difference, it gains kinetic energy
- If an electron accelerates from rest, an electronvolt is equal to the kinetic energy gained:
$
e V=1 / 2 m v^2
$
- Rearranging the equation gives the speed of the electron:
$
v=\sqrt{\frac{2 e V}{m}}
$
Worked example: Electronvolt conversion
Show that the photon energy of light with wavelength $700 \mathrm{~nm}$ is about $1.8 \mathrm{eV}$.
Answer/Explanation
Step 1: Write the equations for wave speed and photon energy
wave speed:
$
c=f \lambda \rightarrow f=\frac{c}{\lambda}
$
photon energy:
$
E=h f \rightarrow E=\frac{h c}{\lambda}
$
Step 2:
Calculate the photon energy in Joules
$
E=\frac{h c}{\lambda}=\frac{\left(6.63 \times 10^{-34}\right) \times\left(3.0 \times 10^8\right)}{700 \times 10^{-9}}=2.84 \times 10^{-19} \mathrm{~J}
$
Step 3:
Convert the photon energy into electronvolts
$
1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \quad \mathrm{~J} \rightarrow \mathrm{eV} \text { : divide by } 1.6 \times 10^{-19}
$
$
E=\frac{2.84 \times 10^{-19}}{1.6 \times 10^{-19}}=1.78 \mathrm{eV}
$
Exam Tip
- To convert between $\mathrm{eV}$ and J:
- $\mathrm{eV} \rightarrow \mathrm{J}$ : multiply by $1.6 \times 10^{-19}$
- $\mathrm{J} \rightarrow \mathrm{eV}$ : divide by $1.6 \times 10^{-19}$