AimMent
Foundation5 past questions

Photon Properties and Energy-Momentum Relations

Photons are massless particles of light characterized by their energy, frequency, momentum, and charge neutrality, and their interactions obey conservation laws.

Why this shows up in the exam

NEET often asks you to relate photon energy and momentum to frequency and wavelength, and to apply conservation laws in photon-matter interactions.

How NEET tests this

Direct recall · 3 QsStatement analysisNumerical

Learn the idea

Photons are mass‑less quanta whose energy E = hν = hc/λ and momentum p = h/λ (or p = E/c); they always travel at speed c in vacuum. The unlocking insight is that a single constant h ties frequency, wavelength, energy and momentum together.

🧠 Memory hook: Think of a photon as a light‑bulb ticket: h is the ticket number that tells you both the price (energy = h·ν) and the size (momentum = h/λ), and the ticket always rides at speed c.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Photon energy E = hν where ν = c/λ
  • Photon momentum p = h/λ = E/c
  • h ≈ 6.6×10⁻³⁴ J·s (NCERT value)
  • Photons are charge‑neutral, massless and move with speed c in free space
  • Power P = energy per second, so number of photons per second N = P/E_photon
  • Wavelength must be expressed in metres when using the formulas

How to approach it

  1. 1Write the photon energy using E = hc/λ (or E = hν)
  2. 2Convert every length unit to metres and frequency to s⁻¹
  3. 3If the question asks for a count, use N = P/E_photon (or N = Energy supplied / energy of one photon)
  4. 4For momentum‑related queries replace E by pc or use p = h/λ

Worked example — watch it click

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 × 10⁻³ watt will be (h = 6.6 × 10⁻³⁴ J s)

  • A)10¹⁵
  • B)10¹⁸
  • C)10¹⁷
  • ✅10¹⁶

The concept behind this problem

The worked example forces you to compute the energy of a single photon from its wavelength, then relate power to photon count – a direct application of E = hc/λ and N = P/E.

Step by step

  1. 1Energy of one photon: E = (hc)/(λ). h = 6.6×10⁻³⁴ J s, c = 3×10⁸ m/s, λ = 600 nm=6×10⁻⁷ m.
  2. 2E = 6.6×10⁻³⁴×3×10⁸6×10⁻⁷ = 1.98×10⁻²⁵6×10⁻⁷ = 3.3×10⁻¹⁹ J per photon.
  3. 3Number of photons emitted per second N = PowerE = 3.3×10⁻³ J/s3.3×10⁻¹⁹ J = 10¹⁶.
  4. 4A value of 10¹⁵ would result from using a larger photon energy (e.g., taking λ as 600 Å), while 10¹⁷ or 10¹⁸ would come from under-estimating the photon energy.
  5. 5So the correct answer is 10¹⁶.

Watch out

Students often forget to change 600 nm to 6×10⁻⁷ m, which makes the photon energy off by a factor of 10⁹ and gives the wrong photon‑per‑second count.

Common slip-ups that cost marks

  • •Forgetting to convert nm (or Å) to metres – a common 10⁹ error
  • •Mixing up ν and λ in the formula (using ν where λ is required)
  • •Assuming a photon can have a speed different from c in vacuum

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 5NEET 2021

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 × 10⁻³ watt will be (h = 6.6 × 10⁻³⁴ J s)

Push further

More challenging

5 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 5

An X-ray photon with a wavelength of 0.1 nm undergoes Compton scattering with a stationary electron. Which of the following statements is true regarding the scattered photon?

More from Dual Nature of Matter and Radiation