PhysicsDual nature of radiation and matterClass 12

Photoelectric effect and matter waves: interactive simulation

Light knocks electrons out of a metal only if each photon carries more energy than the metal’s work function. Make the light brighter and you get more electrons, not faster ones — the result that convinced physicists light comes in packets.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, AP Physics 2, A-level Physics, IB Physics · 3 modes in one lab

What you can explore

  1. 1

    Photoelectric tube

    Five metals, wavelength, intensity and collector potential with the live current–voltage graph.
  2. 2

    Stopping potential vs ν

    Millikan’s straight lines for five metals and their threshold frequencies.
  3. 3

    Matter waves

    de Broglie wavelength of electrons, protons and α particles, with Davisson–Germer diffraction.

Try this

Change the metal, colour and brightness, then pull the collector negative until the current stops. Next, plot that stopping potential against frequency for five metals, and finish by firing electrons at a crystal to see them behave as waves.

  1. 1

    Shine 400 nm light on sodium (φ = 2.75 eV).

    What you should see

    Photon energy ≈ 3.10 eV, so K_max ≈ 0.35 eV and the current stops at V₀ ≈ 0.35 V.

  2. 2

    Change to 500 nm without touching the intensity.

    What you should see

    No electrons at all: 2.48 eV is below the work function. Sodium’s threshold is about 451 nm.

  3. 3

    Keep 400 nm and double the intensity.

    What you should see

    The saturation current doubles but the stopping potential stays at 0.35 V.

  4. 4

    In Stopping potential vs ν, compare the lines for caesium and platinum.

    What you should see

    They are parallel (slope h/e ≈ 4.14 × 10⁻¹⁵ V s) but start at ν₀ = 5.17 and 13.66 × 10¹⁴ Hz.

  5. 5

    In Matter waves, accelerate electrons through 54 V.

    What you should see

    λ ≈ 1.67 Å and the diffraction peak appears near 51°, as in the Davisson–Germer experiment.

Formulas behind it

Einstein’s photoelectric equation

Kmax=hν−ϕ0=eV0K_{max} = h\nu - \phi_0 = eV_0

Photon energy

E=hν=hcλ≈1240λ (nm) eVE = h\nu = \frac{hc}{\lambda} \approx \frac{1240}{\lambda\,(\text{nm})}\ \text{eV}

Threshold

ν0=ϕ0h,λ0=hcϕ0\nu_0 = \frac{\phi_0}{h},\qquad \lambda_0 = \frac{hc}{\phi_0}

Stopping potential against frequency

V0=heν−ϕ0eV_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}

de Broglie wavelength

λ=hp=h2mqV\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}

Electron accelerated through V volts

λ=12.27V A˚\lambda = \frac{12.27}{\sqrt{V}}\ \text{Å}

Common mistakes

  • Thinking brighter light gives faster electrons. Intensity changes the number of electrons, not their maximum energy.
  • Using the photon energy as the electron’s kinetic energy without subtracting the work function.
  • Writing the stopping potential as negative in the equation eV₀ = K_max. V₀ is the magnitude of the retarding potential.
  • Using 12.27/√V Å for protons or α particles. That shortcut is only for electrons.
Practise 200 questionsCheck your answer instantly, with a worked solution.Modern Physics chapter guideFormula sheet, solved examples and PYQ patterns.Stuck on a photoelectric effect and matter waves question?Snap it and get a whiteboard video solution.

FAQ

Why is there no time delay in photoelectric emission?

Each electron absorbs a single photon’s whole energy at once, so emission starts within about 10⁻⁹ s even in very dim light.

Why don’t all emitted electrons have the maximum kinetic energy?

Electrons deeper in the metal lose some energy on the way out. Only those right at the surface leave with hν − φ₀.

Why don’t we see the wave nature of a cricket ball?

Its momentum is so large that λ = h/p is around 10⁻³⁴ m, far too small to produce any measurable diffraction.

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