The Momentum Of Photon Formula, expressed as p = h/λ, defines the linear momentum carried by a single photon — a massless quantum of light. This concept is central to NCERT Class 12 Physics, Chapter 11 (Dual Nature of Radiation and Matter), and appears regularly in CBSE board exams, JEE Main, JEE Advanced, and NEET. In this article, we cover the formula expression, variable definitions, step-by-step derivation, a complete physics formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET application patterns.
Key Momentum Of Photon Formulas at a Glance
Quick reference for the most important photon momentum formulas.
- Photon momentum: \( p = \dfrac{h}{\lambda} \)
- Using frequency: \( p = \dfrac{hf}{c} \)
- Using energy: \( p = \dfrac{E}{c} \)
- De Broglie wavelength: \( \lambda = \dfrac{h}{p} \)
- Photon energy: \( E = hf = \dfrac{hc}{\lambda} \)
- Energy-momentum relation: \( E = pc \)
- Radiation pressure: \( P = \dfrac{I}{c} \) (complete absorption)
What is the Momentum Of Photon Formula?
The Momentum Of Photon Formula quantifies the momentum associated with a single photon of electromagnetic radiation. Classical mechanics defines momentum as the product of mass and velocity. However, a photon is massless. So how does it carry momentum?
The answer lies in quantum mechanics and Einstein’s special theory of relativity. Albert Einstein and Arthur Compton established that photons carry momentum despite having zero rest mass. This was confirmed experimentally by the Compton Effect (1923), where X-rays scattered off electrons and transferred momentum — just like billiard balls.
According to NCERT Class 12 Physics, Chapter 11 (Dual Nature of Radiation and Matter), the momentum of a photon is given by:
\[ p = \frac{h}{\lambda} \]
Here, p is the photon momentum, h is Planck’s constant, and λ is the wavelength of the photon. This formula also forms the foundation of Louis de Broglie’s hypothesis, which extended the idea of wave-particle duality to all matter. Understanding this formula is essential for scoring well in CBSE Class 12 board exams and for cracking JEE and NEET.
Momentum Of Photon Formula — Expression and Variables
The standard expression for photon momentum is:
\[ p = \frac{h}{\lambda} \]
This can also be written in terms of frequency and the speed of light:
\[ p = \frac{hf}{c} = \frac{E}{c} \]
| Symbol | Quantity | SI Unit | Value / Notes |
|---|---|---|---|
| p | Momentum of photon | kg·m/s | Always positive |
| h | Planck’s constant | J·s | 6.626 × 10⁻³⁴ J·s |
| λ | Wavelength of photon | metre (m) | Often given in nm or Å |
| f | Frequency of photon | Hertz (Hz) | f = c/λ |
| c | Speed of light in vacuum | m/s | 3 × 10⁸ m/s |
| E | Energy of photon | Joule (J) | E = hf = hc/λ |
Derivation of the Momentum Of Photon Formula
The derivation follows from Einstein’s energy-momentum relation for a relativistic particle:
\[ E^2 = (pc)^2 + (m_0 c^2)^2 \]
Step 1: For a photon, the rest mass \( m_0 = 0 \). So the equation simplifies to:
\[ E^2 = (pc)^2 \implies E = pc \]
Step 2: The energy of a photon is given by Planck’s quantum theory:
\[ E = hf = \frac{hc}{\lambda} \]
Step 3: Substitute E = pc into E = hc/λ:
\[ pc = \frac{hc}{\lambda} \]
Step 4: Divide both sides by c to get the final formula:
\[ p = \frac{h}{\lambda} \]
This derivation shows that photon momentum arises from its wave nature (wavelength λ) and Planck’s constant. The result is consistent with the de Broglie hypothesis, which applies the same formula to material particles.
Complete Physics Formula Sheet — Dual Nature and Quantum Physics
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Momentum of Photon | \( p = h/\lambda \) | h = Planck’s constant, λ = wavelength | kg·m/s | Class 12, Ch 11 |
| Photon Energy (wavelength) | \( E = hc/\lambda \) | h = Planck’s constant, c = speed of light, λ = wavelength | Joule (J) | Class 12, Ch 11 |
| Photon Energy (frequency) | \( E = hf \) | h = Planck’s constant, f = frequency | Joule (J) | Class 12, Ch 11 |
| Energy-Momentum Relation | \( E = pc \) | p = momentum, c = speed of light | Joule (J) | Class 12, Ch 11 |
| de Broglie Wavelength | \( \lambda = h/p \) | h = Planck’s constant, p = momentum | metre (m) | Class 12, Ch 11 |
| de Broglie (kinetic energy) | \( \lambda = h/\sqrt{2mK} \) | m = mass, K = kinetic energy | metre (m) | Class 12, Ch 11 |
| Photoelectric Effect (Einstein) | \( KE_{max} = hf – \phi \) | f = frequency, φ = work function | Joule (J) | Class 12, Ch 11 |
| Threshold Frequency | \( f_0 = \phi / h \) | φ = work function, h = Planck’s constant | Hertz (Hz) | Class 12, Ch 11 |
| Stopping Potential | \( eV_0 = hf – \phi \) | e = electron charge, V₀ = stopping potential | Volt (V) | Class 12, Ch 11 |
| Radiation Pressure (absorption) | \( P = I/c \) | I = intensity, c = speed of light | Pascal (Pa) | Class 12, Ch 11 |
| Compton Shift | \( \Delta\lambda = \frac{h}{m_e c}(1 – \cos\theta) \) | mₑ = electron mass, θ = scattering angle | metre (m) | Class 12, Ch 11 |
Momentum Of Photon Formula — Solved Examples
Example 1 (Class 9-10 / Basic Level)
Problem: Calculate the momentum of a photon of visible light with a wavelength of 500 nm.
Given:
- Wavelength, λ = 500 nm = 500 × 10⁻⁹ m = 5 × 10⁻⁷ m
- Planck’s constant, h = 6.626 × 10⁻³⁴ J·s
Step 1: Write the Momentum Of Photon Formula:
\[ p = \frac{h}{\lambda} \]
Step 2: Substitute the given values:
\[ p = \frac{6.626 \times 10^{-34}}{5 \times 10^{-7}} \]
Step 3: Perform the division:
\[ p = \frac{6.626}{5} \times 10^{-34+7} = 1.325 \times 10^{-27} \text{ kg·m/s} \]
Answer
The momentum of the photon is 1.325 × 10⁻²⁷ kg·m/s.
Example 2 (Class 11-12 Level)
Problem: A photon has an energy of 3.0 eV. Find its momentum and wavelength. (Given: 1 eV = 1.6 × 10⁻¹⁹ J, h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s)
Given:
- Energy, E = 3.0 eV = 3.0 × 1.6 × 10⁻¹⁹ J = 4.8 × 10⁻¹⁹ J
- h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s
Step 1: Use the energy-momentum relation \( E = pc \) to find momentum:
\[ p = \frac{E}{c} = \frac{4.8 \times 10^{-19}}{3 \times 10^{8}} \]
\[ p = 1.6 \times 10^{-27} \text{ kg·m/s} \]
Step 2: Use \( \lambda = h/p \) to find the wavelength:
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.6 \times 10^{-27}} \]
\[ \lambda = 4.14 \times 10^{-7} \text{ m} = 414 \text{ nm} \]
Step 3: This wavelength (414 nm) falls in the violet region of the visible spectrum. This is consistent with a 3.0 eV photon energy.
Answer
Momentum = 1.6 × 10⁻²⁷ kg·m/s; Wavelength = 414 nm (violet light).
Example 3 (JEE/NEET Level)
Problem: In a Compton scattering experiment, an X-ray photon of wavelength 0.0200 nm is scattered at an angle of 90° by a free electron. Find (a) the wavelength of the scattered photon and (b) the momentum transferred to the electron. (Given: h = 6.626 × 10⁻³⁴ J·s, mₑ = 9.11 × 10⁻³¹ kg, c = 3 × 10⁸ m/s)
Given:
- Initial wavelength, λ₀ = 0.0200 nm = 2.00 × 10⁻¹¹ m
- Scattering angle, θ = 90°
- Compton wavelength, h/(mₑc) = 6.626 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 3 × 10⁸) = 2.426 × 10⁻¹² m
Step 1: Apply the Compton shift formula:
\[ \Delta\lambda = \frac{h}{m_e c}(1 – \cos\theta) = 2.426 \times 10^{-12}(1 – \cos 90°) \]
\[ \Delta\lambda = 2.426 \times 10^{-12}(1 – 0) = 2.426 \times 10^{-12} \text{ m} \]
Step 2: Find the scattered photon wavelength:
\[ \lambda’ = \lambda_0 + \Delta\lambda = 2.000 \times 10^{-11} + 0.2426 \times 10^{-11} \]
\[ \lambda’ = 2.2426 \times 10^{-11} \text{ m} \approx 0.02243 \text{ nm} \]
Step 3: Find momentum of the incident photon using \( p_0 = h/\lambda_0 \):
\[ p_0 = \frac{6.626 \times 10^{-34}}{2.00 \times 10^{-11}} = 3.313 \times 10^{-23} \text{ kg·m/s} \]
Step 4: Find momentum of the scattered photon using \( p’ = h/\lambda’ \):
\[ p’ = \frac{6.626 \times 10^{-34}}{2.2426 \times 10^{-11}} = 2.954 \times 10^{-23} \text{ kg·m/s} \]
Step 5: Since θ = 90°, the incident and scattered photon momenta are perpendicular. Apply vector momentum conservation. The magnitude of momentum transferred to the electron is:
\[ p_e = \sqrt{p_0^2 + p’^2} = \sqrt{(3.313)^2 + (2.954)^2} \times 10^{-23} \]
\[ p_e = \sqrt{10.976 + 8.726} \times 10^{-23} = \sqrt{19.702} \times 10^{-23} \approx 4.44 \times 10^{-23} \text{ kg·m/s} \]
Answer
(a) Scattered photon wavelength = 0.02243 nm. (b) Momentum transferred to electron = 4.44 × 10⁻²³ kg·m/s.
CBSE Exam Tips 2025-26
- Memorise key values: Always keep h = 6.626 × 10⁻³⁴ J·s and c = 3 × 10⁸ m/s at your fingertips. CBSE frequently gives numerical problems where you must substitute these directly.
- Unit conversions are critical: Wavelengths are often given in nanometres (nm) or angstroms (Å). Convert to metres before substituting. 1 nm = 10⁻⁹ m; 1 Å = 10⁻¹⁰ m. We recommend writing the conversion step explicitly in your answer.
- Know all three forms: CBSE 2025-26 papers may give energy, frequency, or wavelength. Be fluent in all three forms: \( p = h/\lambda \), \( p = hf/c \), and \( p = E/c \).
- Link to de Broglie: The de Broglie formula \( \lambda = h/p \) is the same formula rearranged. CBSE often asks you to connect photon momentum with matter waves in 3-mark questions.
- Compton Effect questions: For 5-mark problems, practise the Compton shift formula. Show the derivation steps clearly. The examiner awards marks for each logical step.
- Express answers in scientific notation: Photon momenta are of the order 10⁻²⁷ to 10⁻²³ kg·m/s. Always express your final answer in proper scientific notation to avoid careless errors.
Common Mistakes to Avoid
- Forgetting to convert wavelength units: The most common error is substituting λ in nm directly into \( p = h/\lambda \) without converting to metres. This gives an answer that is 10⁹ times too small. Always convert to SI units first.
- Confusing photon momentum with classical momentum: Classical momentum is \( p = mv \). A photon has zero rest mass, so this formula does not apply. Use \( p = h/\lambda \) or \( p = E/c \) for photons only.
- Using the wrong formula for energy vs. momentum: \( E = hf \) gives energy, not momentum. Students sometimes write \( p = hf \), which is dimensionally incorrect. Remember that \( p = hf/c \), not \( hf \).
- Ignoring the direction of photon momentum: In Compton scattering and radiation pressure problems, momentum is a vector. Failing to account for direction leads to wrong answers in vector-based questions.
- Mixing up Planck’s constant values: h = 6.626 × 10⁻³⁴ J·s (in SI). Some students mistakenly use h in eV·s (4.136 × 10⁻¹⁵ eV·s) when the problem requires SI units. Match the unit of h to the unit of energy used.
JEE/NEET Application of the Momentum Of Photon Formula
In our experience, JEE aspirants encounter the Momentum Of Photon Formula in at least 2-3 questions per year across JEE Main and Advanced. NEET also tests this concept in its Modern Physics section. Here are the key application patterns:
Pattern 1: Radiation Pressure and Force
JEE problems frequently ask about the force exerted by light on a surface. If n photons per second strike a perfectly absorbing surface, each photon transfers momentum \( p = h/\lambda \). The total force is:
\[ F = \frac{n h}{\lambda} \]
For a perfectly reflecting surface, the momentum change is doubled: \( F = 2nh/\lambda \). This distinction between absorbing and reflecting surfaces is a classic JEE trap. Our experts suggest always reading the problem statement carefully for the word “absorbing” or “reflecting.”
Pattern 2: Photoelectric Effect Combined with Momentum
NEET and JEE Main combine the photoelectric effect with photon momentum. A typical question gives the stopping potential and asks for the momentum of the incident photon. The approach is:
- Find photon energy: \( E = eV_0 + \phi \) (stopping potential + work function)
- Find momentum: \( p = E/c \)
- Find wavelength: \( \lambda = h/p \)
Practise this three-step chain until it becomes automatic. It appears in nearly every NEET Modern Physics section.
Pattern 3: de Broglie Wavelength Comparison
JEE Advanced loves comparing the de Broglie wavelength of a particle with the wavelength of a photon of the same energy. For a photon: \( \lambda_{photon} = hc/E \). For a particle of mass m and kinetic energy K: \( \lambda_{particle} = h/\sqrt{2mK} \). Setting these equal and solving for the ratio \( E/K \) is a standard multi-concept JEE problem. In our experience, students who master the Momentum Of Photon Formula first find de Broglie problems significantly easier.
For further reading on the NCERT curriculum, visit the official NCERT website to download Class 12 Physics Chapter 11 PDFs and exemplar problems.
FAQs on Momentum Of Photon Formula
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