Energy and momentum of a photon

1. Overview

Quantum physics represents a paradigm shift from classical electromagnetism. In classical physics, light is treated as a continuous electromagnetic wave characterized by its amplitude and frequency. However, this model fails to explain phenomena such as the photoelectric effect or the discrete nature of atomic spectra. To resolve these inconsistencies, the particulate nature of electromagnetic radiation is introduced.

The central principle of this topic is that electromagnetic radiation is not a continuous stream of energy but is instead quantized. It consists of discrete "packets" of energy called photons. A photon is the fundamental quantum of electromagnetic energy. This means that energy can only be exchanged in specific, indivisible amounts. The energy of these photons is determined solely by the frequency of the radiation, while the intensity of the radiation is determined by the number of photons arriving per unit area per unit time. Furthermore, despite having no rest mass, photons possess momentum, a property traditionally reserved for particles with mass in classical mechanics. This dual nature—exhibiting both wave-like and particle-like properties—is the cornerstone of modern physics.

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Key Definitions

• Photon: A quantum of electromagnetic radiation. It is a discrete "packet" of energy that travels at the speed of light in a vacuum and has zero rest mass.
• Quantum: The smallest, indivisible unit of a physical quantity. In this context, it refers to the discrete amount of energy associated with electromagnetic radiation.
• Electronvolt (eV): The energy gained by an electron when it is accelerated from rest through a potential difference of one volt.
• Planck Constant ($h$): A fundamental physical constant ($6.63 \times 10^{-34} \text{ J s}$) that serves as the proportionality constant between the energy of a photon and its frequency.
• Monochromatic Light: Light consisting of a single frequency (and therefore a single wavelength and a single photon energy).
• Intensity: The power per unit area incident on a surface. In the photon model, for monochromatic light, intensity is proportional to the photon flux (number of photons per unit area per unit time).

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Content

3.1 The Particulate Nature of Electromagnetic Radiation

In the late 19th and early 20th centuries, classical wave theory could not explain why low-frequency light, no matter how intense, could not eject electrons from a metal surface, while high-frequency light could do so instantaneously. This led to the realization that EM radiation behaves as a stream of particles.

Key Concepts of the Particulate Model:

1. Quantization: Energy is not continuous. It comes in discrete units.
2. One-to-One Interaction: In processes like the photoelectric effect, one photon interacts with exactly one electron. The electron absorbs the entire energy of the photon or none at all.
3. Energy vs. Intensity:
   • The energy of an individual photon depends only on its frequency.
   • The intensity of the beam depends on the number of photons passing through a unit area per second.
   • Increasing the intensity of a monochromatic source increases the quantity of photons but has zero effect on the energy of each individual photon.

3.2 Photon Energy ($E = hf$)

The energy $E$ of a single photon is directly proportional to the frequency $f$ of the electromagnetic radiation. This relationship is defined by the Planck equation:

$$E = hf$$

Where:

• $E$ = Energy of the photon (Joules, $\text{J}$)
• $h$ = Planck constant ($6.63 \times 10^{-34} \text{ J s}$)
• $f$ = Frequency of the radiation (Hertz, $\text{Hz}$ or $\text{s}^{-1}$)

Since all electromagnetic waves travel at the speed of light $c$ in a vacuum, we use the wave equation $c = f\lambda$ to derive the relationship between energy and wavelength $\lambda$:

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

Where:

• $c$ = Speed of light in a vacuum ($3.00 \times 10^8 \text{ m s}^{-1}$)
• $\lambda$ = Wavelength (metres, $\text{m}$)

Physical Implications:

• Inverse Relationship: Energy is inversely proportional to wavelength. Therefore, Gamma rays (shortest $\lambda$) have the highest energy photons, while Radio waves (longest $\lambda$) have the lowest energy photons.
• Blue vs. Red: A photon of blue light (higher frequency) carries more energy than a photon of red light (lower frequency).

3.3 The Electronvolt (eV)

The Joule is an inconveniently large unit for measuring the energy of individual photons or subatomic particles (which are typically on the order of $10^{-19} \text{ J}$). Physicists use the electronvolt (eV) instead.

Derivation: Recall that Work Done $W = QV$. If we move one electron (charge $e = 1.60 \times 10^{-19} \text{ C}$) through a potential difference of $1 \text{ V}$: $$W = (1.60 \times 10^{-19} \text{ C}) \times (1.00 \text{ V}) = 1.60 \times 10^{-19} \text{ J}$$ Thus, $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$.

Conversions:

• To convert Joules to eV: Divide the value by $1.60 \times 10^{-19}$.
• To convert eV to Joules: Multiply the value by $1.60 \times 10^{-19}$.

3.4 Photon Momentum ($p = E/c$)

Although photons have no mass (rest mass $m_0 = 0$), they carry momentum. This is a consequence of Einstein's mass-energy equivalence. For a particle with zero rest mass, the relationship between energy $E$ and momentum $p$ is:

$$p = \frac{E}{c}$$

Substituting $E = hf$ and $E = \frac{hc}{\lambda}$ gives: $$p = \frac{hf}{c}$$ $$p = \frac{h}{\lambda}$$

Radiation Pressure and Force: When photons strike a surface, they exert a force. According to Newton’s Second Law, force is the rate of change of momentum ($F = \frac{\Delta p}{\Delta t}$).

• Absorption: If a photon is absorbed by a surface, its final momentum is zero. The change in momentum is $\Delta p = p_{initial} - 0 = \frac{h}{\lambda}$.
• Reflection: If a photon is perfectly reflected (bounces back), its momentum changes direction. The change in momentum is $\Delta p = p - (-p) = 2p = \frac{2h}{\lambda}$.
• Force: The total force exerted by a beam of light is the total change in momentum of all photons striking the surface per second.

3.5 Photon Flux and Power

In many exam questions, you are asked to relate the power of a source to the number of photons it emits. Power ($P$) is the total energy emitted per unit time. If a source emits $n$ photons per second, and each photon has energy $E$:

$$P = nE = n(hf) = \frac{nhc}{\lambda}$$

Where $n$ is the photon emission rate (photons per second, $\text{s}^{-1}$).

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3.6 Worked Examples

Worked example 1 — Energy and Unit Conversion

A ultraviolet lamp emits radiation with a wavelength of $254 \text{ nm}$. Calculate the energy of a single photon in both Joules ($\text{J}$) and electronvolts ($\text{eV}$).

• Step 1: List known values in SI units $\lambda = 254 \text{ nm} = 254 \times 10^{-9} \text{ m}$ $h = 6.63 \times 10^{-34} \text{ J s}$ $c = 3.00 \times 10^8 \text{ m s}^{-1}$

• Step 2: Calculate energy in Joules $$E = \frac{hc}{\lambda}$$ $$E = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{254 \times 10^{-9}}$$ $$E = 7.8307... \times 10^{-19} \text{ J}$$ $E = 7.83 \times 10^{-19} \text{ J}$ (to 3 s.f.)

• Step 3: Convert to eV $$E_{\text{eV}} = \frac{7.8307 \times 10^{-19}}{1.60 \times 10^{-19}}$$ $E = 4.89 \text{ eV}$ (to 3 s.f.)

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Worked example 2 — Photon Flux (Number of photons)

A $5.0 \text{ mW}$ laser produces a beam of red light with a wavelength of $633 \text{ nm}$. Determine the number of photons emitted by the laser every second.

• Step 1: Identify Power and Photon Energy $P = 5.0 \times 10^{-3} \text{ W (J s}^{-1}\text{)}$ $\lambda = 633 \times 10^{-9} \text{ m}$

• Step 2: Calculate the energy of one photon $$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{633 \times 10^{-9}}$$ $$E = 3.142 \times 10^{-19} \text{ J}$$

• Step 3: Calculate the number of photons per second ($n$) $$P = n \times E \implies n = \frac{P}{E}$$ $$n = \frac{5.0 \times 10^{-3}}{3.142 \times 10^{-19}}$$ $n = 1.59 \times 10^{16} \text{ photons s}^{-1}$ (to 3 s.f.)

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Worked example 3 — Photon Momentum and Force

A beam of X-rays consists of photons each with a momentum of $1.2 \times 10^{-23} \text{ kg m s}^{-1}$. (a) Calculate the frequency of the X-rays. (b) If $5.0 \times 10^{12}$ of these photons are absorbed by a surface every second, calculate the average force exerted on the surface.

Part (a):

• Step 1: Use the momentum-frequency relationship $$p = \frac{hf}{c} \implies f = \frac{pc}{h}$$
• Step 2: Substitute and solve $$f = \frac{(1.2 \times 10^{-23}) \times (3.00 \times 10^8)}{6.63 \times 10^{-34}}$$ $f = 5.43 \times 10^{18} \text{ Hz}$

Part (b):

• Step 1: Identify the change in momentum for one photon Since the photons are absorbed, $\Delta p = p_{initial} = 1.2 \times 10^{-23} \text{ kg m s}^{-1}$.
• Step 2: Calculate total change in momentum per second Total $\Delta p$ per second = (Number of photons per second) $\times$ ($\Delta p$ per photon) Total $\Delta p / \Delta t = (5.0 \times 10^{12}) \times (1.2 \times 10^{-23})$
• Step 3: State the force $F = \frac{\Delta p}{\Delta t} = 6.0 \times 10^{-11} \text{ N}$ $F = 6.0 \times 10^{-11} \text{ N}$

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Key Equations

Equation	Symbols & Units	Data Sheet?
$E = hf$	$E$: Energy ($\text{J}$), $f$: Frequency ($\text{Hz}$)	Yes
$E = \frac{hc}{\lambda}$	$\lambda$: Wavelength ($\text{m}$), $c$: Speed of light ($\text{m s}^{-1}$)	No (Derivable)
$1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$	$e$: Elementary charge (conversion factor)	Yes (as $e$)
$p = \frac{E}{c}$	$p$: Momentum ($\text{kg m s}^{-1}$), $E$: Energy ($\text{J}$)	Yes
$p = \frac{h}{\lambda}$	$p$: Momentum ($\text{kg m s}^{-1}$), $\lambda$: Wavelength ($\text{m}$)	Yes (as de Broglie)
$P = n \frac{hc}{\lambda}$	$P$: Power ($\text{W}$), $n$: Photons per second ($\text{s}^{-1}$)	No

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Common Mistakes to Avoid

• ❌ Wrong: Using $E = \frac{1}{2}mv^2$ or $p = mv$ for a photon.
• ✓ Right: Photons are massless. You must use $E = hf$ and $p = E/c$. The classical kinetic energy and momentum formulas do not apply to quanta of light.
• ❌ Wrong: Forgetting to convert wavelength from nanometres ($\text{nm}$) or picometres ($\text{pm}$) to metres ($\text{m}$) before using $E = hc/\lambda$.
• ✓ Right: Always convert to SI base units immediately. $1 \text{ nm} = 10^{-9} \text{ m}$; $1 \text{ pm} = 10^{-12} \text{ m}$.
• ❌ Wrong: Thinking that "Intensity" affects the energy of the electrons in the photoelectric effect.
• ✓ Right: Intensity only affects the number of photons. Only frequency affects the energy of individual photons.
• ❌ Wrong: Confusing the Planck constant ($h = 6.63 \times 10^{-34}$) with the elementary charge ($e = 1.60 \times 10^{-19}$).
• ✓ Right: Check the data sheet carefully. $h$ is for energy-frequency; $e$ is for Joule-eV conversions.
• ❌ Wrong: Using $v$ instead of $c$ for the speed of a photon in a vacuum.
• ✓ Right: Photons always travel at $c = 3.00 \times 10^8 \text{ m s}^{-1}$ in a vacuum.

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Exam Tips

1. The "Show your working" Rule: In 9702 Paper 4, if you are asked to "show that" an energy is a certain value in eV, you must show the calculation in Joules first, then show the division by $1.60 \times 10^{-19}$.
2. Significant Figures: Always provide your final answer to the same number of significant figures as the least precise data value given in the question (usually 2 or 3 s.f.). Never leave answers as fractions or with recurring decimals.
3. Defining the Photon: If asked to define a photon, always use the phrase "quantum of electromagnetic radiation". This is the specific phrase required by the Cambridge mark scheme.
4. Momentum Change on Reflection: Be extremely careful with questions involving photons reflecting off a mirror. The change in momentum is $2p$, not $p$. This factor of 2 is a frequent source of lost marks in force/pressure calculations.
5. Power and Photon Rate: If a question gives you the power of a source and asks for the number of photons, remember that $P = \frac{\text{Total Energy}}{\text{time}}$. For $n$ photons per second, $P = n \times (\text{energy of one photon})$.
6. Proportionality: Remember $E \propto f$ and $E \propto \frac{1}{\lambda}$. If the wavelength of light is doubled, the energy of each photon is halved. This conceptual understanding helps you quickly check if your calculated answers make sense.