Key Photon Energy Formulas at a Glance
Quick reference for the most important photon energy formulas used in CBSE Class 12 and competitive exams.
- Photon energy (frequency form): \( E = hf \)
- Photon energy (wavelength form): \( E = \frac{hc}{\lambda} \)
- Planck’s constant: \( h = 6.626 \times 10^{-34} \) J·s
- Speed of light: \( c = 3 \times 10^{8} \) m/s
- Energy in electron volts: \( E(\text{eV}) = \frac{E(\text{J})}{1.6 \times 10^{-19}} \)
- Momentum of a photon: \( p = \frac{h}{\lambda} = \frac{E}{c} \)
- de Broglie wavelength: \( \lambda = \frac{h}{mv} \)
The Photon Energy Formula gives the energy carried by a single photon as \( E = hf \), where \( h \) is Planck’s constant and \( f \) is the frequency of the electromagnetic radiation. This formula is a cornerstone of modern physics and quantum mechanics. It is covered in CBSE Class 12 Physics, Chapter 11 — Dual Nature of Radiation and Matter. The concept is equally critical for JEE Main, JEE Advanced, and NEET aspirants. This article covers the formula expression, variable definitions, derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
What is the Photon Energy Formula?
The Photon Energy Formula describes the relationship between the energy of a photon and the frequency (or wavelength) of the electromagnetic wave it represents. A photon is a massless, discrete packet (quantum) of electromagnetic energy. In 1900, Max Planck proposed that energy is not continuous but comes in small, indivisible units called quanta. Albert Einstein extended this idea in 1905 to explain the photoelectric effect, for which he received the Nobel Prize.
According to NCERT Class 12 Physics, Chapter 11, the energy of a photon is directly proportional to its frequency. A higher-frequency photon (such as an X-ray) carries far more energy than a lower-frequency photon (such as a radio wave). The formula is expressed in two equivalent forms: one using frequency and one using wavelength. Both forms are equally valid and are used depending on what information is given in a problem.
The Photon Energy Formula is fundamental to understanding the photoelectric effect, atomic spectra, Compton scattering, and the wave-particle duality of light. It connects the classical wave description of light with the quantum particle description.
Photon Energy Formula — Expression and Variables
The photon energy formula in its two primary forms is:
\[ E = hf \]
Equivalently, since \( c = f\lambda \), we can write:
\[ E = \frac{hc}{\lambda} \]
Here, \( E \) is the energy of the photon in Joules (J), \( h \) is Planck’s constant, \( f \) (or \( \nu \)) is the frequency of the radiation, \( c \) is the speed of light in vacuum, and \( \lambda \) is the wavelength of the radiation.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( E \) | Energy of the photon | Joule (J) or electron volt (eV) |
| \( h \) | Planck’s constant | J·s (\( 6.626 \times 10^{-34} \) J·s) |
| \( f \) or \( \nu \) | Frequency of radiation | Hertz (Hz) or s¹ |
| \( c \) | Speed of light in vacuum | m/s (\( 3 \times 10^{8} \) m/s) |
| \( \lambda \) | Wavelength of radiation | metre (m) or nanometre (nm) |
Derivation of the Photon Energy Formula
Max Planck proposed that oscillating atoms emit or absorb energy in discrete packets called quanta. The energy of each quantum is proportional to the frequency of oscillation.
Step 1: Planck’s quantisation hypothesis states that \( E \propto f \).
Step 2: Introducing the constant of proportionality \( h \) (Planck’s constant), we get \( E = hf \).
Step 3: Since the speed of light relates frequency and wavelength by \( c = f\lambda \), we substitute \( f = c/\lambda \) to obtain the wavelength form: \( E = hc/\lambda \).
Step 4: Einstein used this result to explain the photoelectric effect, confirming that light itself consists of photons each carrying energy \( E = hf \).
This derivation appears directly in NCERT Class 12 Physics, Chapter 11, and is frequently tested in board exams.
Complete Modern Physics Formula Sheet
The table below covers all key formulas related to photons, the photoelectric effect, and dual nature of matter from NCERT Class 12 Physics.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Photon Energy (frequency) | \( E = hf \) | h = Planck’s constant, f = frequency | J | Class 12, Ch 11 |
| Photon Energy (wavelength) | \( E = hc/\lambda \) | c = speed of light, λ = wavelength | J | Class 12, Ch 11 |
| Photon Momentum | \( p = h/\lambda \) | h = Planck’s constant, λ = wavelength | kg·m/s | Class 12, Ch 11 |
| Einstein’s Photoelectric Equation | \( KE_{max} = hf – \phi \) | KE = max kinetic energy, φ = work function | J or eV | Class 12, Ch 11 |
| Work Function | \( \phi = hf_0 \) | f₀ = threshold frequency | J or eV | Class 12, Ch 11 |
| Stopping Potential | \( eV_0 = hf – \phi \) | e = electron charge, V₀ = stopping potential | V | Class 12, Ch 11 |
| de Broglie Wavelength | \( \lambda = h/(mv) \) | m = mass, v = velocity | m | Class 12, Ch 11 |
| de Broglie (kinetic energy) | \( \lambda = h/\sqrt{2mKE} \) | KE = kinetic energy of particle | m | Class 12, Ch 11 |
| Compton Shift | \( \Delta\lambda = \frac{h}{m_e c}(1 – \cos\theta) \) | mₑ = electron mass, θ = scattering angle | m | Class 12, Ch 11 |
| Energy in eV from Wavelength | \( E(\text{eV}) = 1240/\lambda(\text{nm}) \) | λ in nanometres | eV | Class 12, Ch 11 |
| Number of Photons | \( N = P/(hf) \) | P = power of source | photons/s | Class 12, Ch 11 |
Photon Energy Formula — Solved Examples
Example 1 (Class 9-10 / Basic Level)
Problem: Calculate the energy of a photon of visible light having a frequency of \( 6 \times 10^{14} \) Hz. Also express the answer in electron volts.
Given:
- Frequency, \( f = 6 \times 10^{14} \) Hz
- Planck’s constant, \( h = 6.626 \times 10^{-34} \) J·s
- 1 eV = \( 1.6 \times 10^{-19} \) J
Step 1: Write the photon energy formula: \( E = hf \)
Step 2: Substitute the values:
\[ E = 6.626 \times 10^{-34} \times 6 \times 10^{14} \]
Step 3: Calculate:
\[ E = 39.756 \times 10^{-20} = 3.976 \times 10^{-19} \text{ J} \]
Step 4: Convert to eV:
\[ E = \frac{3.976 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.49 \text{ eV} \]
Answer
Energy of the photon = \( 3.976 \times 10^{-19} \) J ≈ 2.49 eV
Example 2 (Class 11-12 Level)
Problem: A photon has a wavelength of 400 nm (violet light). Find its energy in Joules and in eV. Also find the momentum of the photon.
Given:
- Wavelength, \( \lambda = 400 \) nm \( = 400 \times 10^{-9} \) m
- \( h = 6.626 \times 10^{-34} \) J·s
- \( c = 3 \times 10^{8} \) m/s
Step 1: Use the wavelength form of the photon energy formula:
\[ E = \frac{hc}{\lambda} \]
Step 2: Substitute values:
\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{400 \times 10^{-9}} \]
Step 3: Calculate numerator:
\[ hc = 6.626 \times 10^{-34} \times 3 \times 10^{8} = 19.878 \times 10^{-26} \text{ J·m} \]
Step 4: Divide by wavelength:
\[ E = \frac{19.878 \times 10^{-26}}{400 \times 10^{-9}} = \frac{19.878 \times 10^{-26}}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J} \]
Step 5: Convert to eV:
\[ E = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.1 \text{ eV} \]
Step 6: Find photon momentum using \( p = h/\lambda \):
\[ p = \frac{6.626 \times 10^{-34}}{400 \times 10^{-9}} = 1.657 \times 10^{-27} \text{ kg·m/s} \]
Answer
Energy = \( 4.97 \times 10^{-19} \) J ≈ 3.1 eV; Momentum = \( 1.657 \times 10^{-27} \) kg·m/s
Example 3 (JEE/NEET Level)
Problem: A metal surface has a work function of 2.0 eV. Ultraviolet light of wavelength 200 nm is incident on it. Find (a) the energy of each incident photon in eV, (b) the maximum kinetic energy of the emitted photoelectrons, and (c) the stopping potential.
Given:
- Work function, \( \phi = 2.0 \) eV
- Wavelength, \( \lambda = 200 \) nm
- Use the shortcut: \( E(\text{eV}) = 1240/\lambda(\text{nm}) \)
Step 1: Calculate photon energy using the convenient shortcut formula:
\[ E = \frac{1240}{200} = 6.2 \text{ eV} \]
Step 2: Apply Einstein’s photoelectric equation:
\[ KE_{max} = E – \phi = 6.2 – 2.0 = 4.2 \text{ eV} \]
Step 3: Find the stopping potential. The stopping potential \( V_0 \) satisfies \( eV_0 = KE_{max} \):
\[ V_0 = \frac{KE_{max}}{e} = \frac{4.2 \text{ eV}}{e} = 4.2 \text{ V} \]
Step 4: Verify the threshold frequency. The threshold wavelength is:
\[ \lambda_0 = \frac{1240}{\phi(\text{eV})} = \frac{1240}{2.0} = 620 \text{ nm} \]
Since 200 nm < 620 nm (incident photon has higher energy), emission does occur. This is consistent with our result.
Answer
(a) Photon energy = 6.2 eV (b) Maximum KE of photoelectrons = 4.2 eV (c) Stopping potential = 4.2 V
CBSE Exam Tips 2025-26
- Memorise both forms: Always learn \( E = hf \) and \( E = hc/\lambda \) together. CBSE 2025-26 papers frequently give wavelength, so you must be comfortable with both.
- Use the 1240 shortcut: We recommend memorising \( E(\text{eV}) = 1240/\lambda(\text{nm}) \) for rapid calculations. This saves significant time in board and competitive exams.
- State the formula first: In 3-mark and 5-mark CBSE questions, always write the formula before substituting values. Examiners award one mark for the formula alone.
- Unit conversion is a must: Convert nanometres to metres (multiply by \( 10^{-9} \)) before substituting into SI-unit formulas. Forgetting this is the most common source of wrong answers.
- Know Planck’s constant precisely: Use \( h = 6.626 \times 10^{-34} \) J·s for calculations. NCERT uses this value throughout Chapter 11.
- Link to the photoelectric effect: In 2025-26 board exams, photon energy questions are almost always linked to Einstein’s photoelectric equation. Practice combining \( E = hf \) with \( KE_{max} = hf – \phi \).
Common Mistakes to Avoid
- Confusing frequency and wavelength: Students sometimes substitute wavelength directly into \( E = hf \). Remember, \( f \) is frequency (Hz), not wavelength. If wavelength is given, use \( E = hc/\lambda \) or first calculate \( f = c/\lambda \).
- Wrong unit for wavelength: Wavelength is often given in nm or Ångströms in exam problems. Always convert to metres before using SI formulas. 1 nm = \( 10^{-9} \) m; 1 Å = \( 10^{-10} \) m.
- Forgetting to convert J to eV: Many problems ask for energy in eV. Always divide the energy in Joules by \( 1.6 \times 10^{-19} \) to get the answer in eV. Leaving the answer in Joules when eV is asked costs marks.
- Using wrong value of Planck’s constant: Some students write \( h = 6.6 \times 10^{-34} \) without the correct exponent. Always double-check the power of 10. A wrong exponent gives an answer that is off by several orders of magnitude.
- Confusing photon momentum with photon energy: Photon momentum is \( p = h/\lambda \), not \( E/\lambda \). Remember that \( p = E/c \) is the correct relationship between momentum and energy for a photon.
JEE/NEET Application of the Photon Energy Formula
In our experience, JEE aspirants encounter the Photon Energy Formula in at least 2–3 questions per paper, either directly or as part of a multi-concept problem. NEET typically includes one direct question on photon energy and one application in the photoelectric effect context.
Pattern 1: Photoelectric Effect Problems
The most common JEE/NEET pattern combines the Photon Energy Formula with Einstein’s photoelectric equation. A typical problem gives the wavelength of incident light and the work function of the metal. Students must calculate photon energy, maximum kinetic energy of emitted electrons, and stopping potential in a single multi-step solution. Mastery of \( E = hc/\lambda \) is essential here.
Pattern 2: Number of Photons from a Source
JEE Main frequently asks: “A laser source of power P emits light of wavelength λ. How many photons does it emit per second?” The answer uses \( N = P/E = P\lambda/(hc) \). This tests whether students can rearrange the Photon Energy Formula correctly under time pressure.
Pattern 3: Compton Scattering and Photon Momentum
JEE Advanced has tested Compton scattering, where a photon collides with a free electron. The Compton shift formula \( \Delta\lambda = (h/m_e c)(1 – \cos\theta) \) is a direct extension of the photon momentum concept \( p = h/\lambda \). Our experts suggest understanding this derivation conceptually, not just memorising the formula.
In our experience, students who understand the physical meaning of \( E = hf \) — that higher frequency means more energetic photons — perform significantly better on these conceptual questions than those who simply memorise the formula.
FAQs on Photon Energy Formula
To strengthen your understanding of quantum physics and related topics, explore our detailed articles on the Induced Voltage Formula, the Critical Velocity Formula, and the Superposition Formula. For a complete collection of physics formulas aligned with the NCERT curriculum, visit our Physics Formulas hub. For the official NCERT syllabus and textbooks, refer to ncert.nic.in.