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Public topic · GRE Physics

Photons, Work Functions, and Matter Waves

A free GRE Physics note on the two quantization facts atomic-physics questions lean on: light delivers energy in photons of energy hf, and a particle of momentum p behaves like a wave of wavelength h/p.

Concise answer

Atomic-physics questions rest on two quantization facts. Light arrives in photons of energy hfhf, so a photoelectric surface emits electrons only above a threshold frequency and the surplus energy hf−ϕhf - \phi becomes kinetic energy. Matter runs the same relation in reverse: a particle of momentum pp has wavelength h/ph/p, and confining that wave to a bound region is what forces a discrete energy ladder.

Definitions

Photon energy
The energy E=hf=hc/λE = hf = hc/\lambda carried by one quantum of electromagnetic radiation of frequency ff and wavelength λ\lambda.
Work function
The minimum energy ϕ\phi needed to remove an electron from a particular metal surface; it is a property of the material, not of the incident light.
Threshold frequency
The frequency f0=ϕ/hf_0 = \phi/h below which no electrons are emitted no matter how intense the light is.
De Broglie wavelength
The wavelength λ=h/p\lambda = h/p associated with a particle of momentum pp; it is what makes electron diffraction possible.

Intuition

The photoelectric equation is a budget statement. The photon pays a fixed entrance fee ϕ\phi to free one electron, and whatever is left over becomes kinetic energy. Brighter light sends more photons, not richer ones, which is why intensity moves the current and never the maximum energy.

The de Broglie relation and energy quantization are the same idea seen twice. If a particle is a wave, then boxing it in only allows wavelengths that fit the box, and each allowed wavelength names one allowed momentum and therefore one allowed energy.

Concept walkthrough

Start every photon calculation by converting to energy. With hc≈1240 eV⋅nmhc \approx 1240\,\text{eV}\cdot\text{nm}, a 248 nm248\,\text{nm} photon carries 1240/248=5.0 eV1240/248 = 5.0\,\text{eV}. Against a 2.0 eV2.0\,\text{eV} work function the fastest emitted electron leaves with Kmax⁡=3.0 eVK_{\max} = 3.0\,\text{eV}, and a 3.0 V3.0\,\text{V} stopping potential is exactly what halts it. Below the threshold frequency f0=ϕ/hf_0 = \phi/h nothing is emitted at all, which is the observation classical wave theory could not reproduce.

Matter waves invert the same relation. For a nonrelativistic particle of kinetic energy KK, momentum is p=2mKp = \sqrt{2mK} and the wavelength is λ=h/p\lambda = h/p. Because hh is tiny, everyday objects have wavelengths far below any measurable aperture, while an electron accelerated through a modest potential difference has a wavelength on the scale of atomic spacing — which is why electron diffraction from a crystal works at all.

Bound systems then quantize. Once a wave must fit inside a region, only a discrete set of standing patterns survives, so the allowed energies form a ladder rather than a continuum. That is the structural reason atoms emit and absorb at sharp wavelengths: each observed line corresponds to a difference between two allowed levels, so a transition energy and a photon wavelength are two ways of writing the same number.

After this page, you should be able to

  • Convert between photon wavelength, frequency, and energy using E=hf=hc/λE = hf = hc/\lambda.
  • Apply Kmax⁡=hf−ϕK_{\max} = hf - \phi and explain why intensity changes the emitted current but not Kmax⁡K_{\max}.
  • Compute a de Broglie wavelength from momentum, including the nonrelativistic case p=2mKp = \sqrt{2mK}.
  • Explain why confinement produces discrete energy levels rather than a continuous spectrum.

Formulas and assumptions

Photon energy

E=hf=hcλE = hf = \frac{hc}{\lambda}

Variables

  • E: photon energy
  • f: frequency
  • lambda: wavelength
  • h: Planck's constant
  • c: speed of light in vacuum

Assumptions

  • hc is approximately 1240 eV*nm, which keeps wavelength-in-nanometers arithmetic in electronvolts.

Photoelectric equation

Kmax⁡=hf−ϕK_{\max} = hf - \phi

Variables

  • K_max: maximum kinetic energy of an ejected electron
  • hf: energy of one incident photon
  • phi: work function of the surface

Assumptions

  • Each ejected electron absorbs one photon.
  • No emission occurs when hf is below phi, regardless of intensity.

De Broglie wavelength

λ=hp,p=2mK\lambda = \frac{h}{p}, \quad p = \sqrt{2mK}

Variables

  • lambda: de Broglie wavelength
  • p: particle momentum
  • m: particle mass
  • K: nonrelativistic kinetic energy

Assumptions

  • The momentum-energy substitution p = sqrt(2mK) holds only when the speed is well below c.

Worked example

One surface, two questions

Light of wavelength 310 nm310\,\text{nm} falls on a surface with work function ϕ=2.0 eV\phi = 2.0\,\text{eV} (use hc≈1240 eV⋅nmhc \approx 1240\,\text{eV}\cdot\text{nm}). Find the maximum photoelectron kinetic energy, then state what happens to it if the light's intensity is doubled.

  1. 1Convert the wavelength to photon energy: E=1240/310=4.0 eVE = 1240/310 = 4.0\,\text{eV}.
  2. 2Subtract the entrance fee: Kmax⁡=hf−ϕ=4.0−2.0=2.0 eVK_{\max} = hf - \phi = 4.0 - 2.0 = 2.0\,\text{eV}.
  3. 3Sanity-check the threshold: emission requires hf>ϕhf > \phi, and 4.0>2.04.0 > 2.0, so electrons are indeed emitted.
  4. 4Doubling the intensity doubles the number of photons per second, so twice as many electrons are emitted per second.
  5. 5Each individual photon still carries 4.0 eV4.0\,\text{eV}, so Kmax⁡K_{\max} is unchanged at 2.0 eV2.0\,\text{eV}.

Kmax⁡=2.0 eVK_{\max} = 2.0\,\text{eV}; doubling the intensity doubles the photocurrent and leaves Kmax⁡K_{\max} at 2.0 eV2.0\,\text{eV}.

Common traps

  • Reporting the full photon energy as the kinetic energy, which skips the work function entirely.
  • Expecting brighter light to produce faster electrons; intensity sets how many electrons leave, not how much energy each one keeps.
  • Using λ=h/p\lambda = h/p with a kinetic energy in place of a momentum — convert with p=2mKp = \sqrt{2mK} first.
  • Treating an emission wavelength as an energy level rather than as the difference between two levels.

Related pages and practice

GRE Physics overviewQuantum Mechanics FoundationsInterference and DiffractionSign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

  1. GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
  2. University Physics Volume 3, Section 6.2: Photoelectric Effect — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  3. University Physics Volume 3, Section 6.5: De Broglie's Matter Waves — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  4. University Physics Volume 3, Section 7.4: The Quantum Particle in a Box — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

Sources and corrections

Sources last checked 2026-08-15

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