Q1. The work function of a metal is 2 eV. Find the maximum KE of electrons when 400 nm light falls on it.
Photon energy E = 1240/400 = 3.1 eV
KE_max = E − φ₀ = 3.1 − 2
KE_max = 1.1 eV
Dual Nature of Radiation and Matter — Class 12
Light is both a wave and a particle — and so is matter. Learn the photoelectric effect and de Broglie waves, and compute photon energies instantly.
Photoelectric effectEmission of electrons from a metal when light of high enough frequency falls on it. Instantaneous, and depends on frequency (not intensity) for the energy.
Work function & thresholdThe minimum energy φ₀ needed to eject an electron. Below the threshold frequency, no emission occurs whatever the intensity.
Einstein's photoelectric equationA photon of energy hf gives KE_max = hf − φ₀ to the electron; stopping potential eV₀ = KE_max.
de Broglie wavesEvery moving particle has a wavelength λ = h/p = h/(mv). This wave nature is significant only for tiny particles like electrons.
Photon energy calculator Interactive
Find the energy of a photon from its wavelength using the handy shortcut E(eV) = 1240/λ(nm).
Photon energy, 1240/λ—eV
Shorter wavelength (UV, X-rays) ⇒ higher energy photons; that is why they are more damaging than visible light.
Numericals practice Interactive
Photon energy and photoelectric problems with steps.
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Solved numericals
Q2. Find the de Broglie wavelength of an electron moving at 10⁶ m/s (m = 9.1×10⁻³¹ kg, h = 6.63×10⁻³⁴).
λ = h / (mv)
λ = 6.63×10⁻³⁴ / (9.1×10⁻³¹ × 10⁶)
λ = 7.3×10⁻¹⁰ m = 0.73 nm
Formula sheet
| Quantity | Formula | SI unit |
|---|---|---|
| Photon energy | E = h f = h c / λ | joule (J) or eV |
| Einstein's equation | KE_max = h f − φ₀ | J |
| Stopping potential | e V₀ = KE_max | V |
| de Broglie wavelength | λ = h / (m v) | metre (m) |
| Photon energy shortcut | E (eV) = 1240 / λ(nm) | eV |
Common mistakes & exam wins
- Use the shortcut E(eV) = 1240/λ(nm) to get photon energy in seconds.
- The photoelectric effect proves the PARTICLE nature of light; interference proves the WAVE nature.
- Increasing intensity increases the NUMBER of photoelectrons, not their maximum energy.
- de Broglie wavelength is significant only for very small masses (electrons), negligible for everyday objects.
Frequently asked questions
What is the photoelectric effect?
The emission of electrons from a metal surface when light of frequency above a threshold value falls on it.
What is Einstein's photoelectric equation?
KE_max = hf − φ₀, where hf is the photon energy and φ₀ the work function of the metal.
What is de Broglie's hypothesis?
Every moving particle has an associated wavelength λ = h/(mv), showing that matter also has a wave nature.
Does increasing the intensity of light increase the energy of photoelectrons?
No. It increases the number of electrons emitted, but the maximum kinetic energy depends only on the frequency of the light.