Read · Light quanta: from entropy to an energy scale

§8 · photoelectric energy and counts

Follow all nine sections: the classical allocation problem, the Wien entropy calculation, independent configurations, the heuristic move, and three energy-transfer applications.

Newly authored explanatory preview in modern notation; editorial and physics review are pending. The introduction and all nine numbered sections have explanatory treatments below, but this is not a German transcription, an aligned English translation, or a complete critical edition. Source faces remain in preparation. The headings and argument units are editorial, not a verified paragraph-by-paragraph source inventory.

Read the whole available argument →

Show me one example before the notation →

§8 · photoelectric energy and counts

derivation · Within the stated model

More electrons is not more energy per electron

What changes when frequency rises, and what changes when power rises?

The most energetic electron receives a complete quantum and escapes with the least modeled energy loss. Subtracting the escape work Φ gives the maximum kinetic energy, provided emission is energetically allowed.

Kmax=hνΦ,ν0=ΦhK_{\mathrm{max}}=h\nu-\Phi,\qquad\nu_0=\frac{\Phi}{h}

Maximum kinetic energy equals h frequency minus escape work, with threshold frequency Phi over h.

For a hypothetical work function of 2 electron volts and a frequency of 600 terahertz, the maximum is about 0.4814 electron volts using modern SI constants. Doubling optical power while retaining that frequency and escape work leaves the maximum unchanged.

Power is energy arriving per second. At fixed frequency, dividing it by hν gives the incoming quantum rate. Under a fixed yield, twice the power can therefore produce twice the electron count per second. Below threshold, negative hν − Φ is an energy deficit, not a negative kinetic energy of an emitted electron.

Show every step here: More electrons is not more energy per electron
  1. Choose frequency and escape work independently of optical power.
  2. Calculate the energy hν available to one electron under the one-quantum hypothesis.
  3. Subtract the escape work; an additional loss only lowers the outgoing kinetic energy.
  4. If the budget is below zero, report no energetically allowed emission, not a negative emitted-electron energy.
  5. At fixed frequency, divide optical power by hν to obtain the incoming quantum rate.
  6. Vary power without altering hν or Φ, then compare energy and rate in separate readouts.
  7. A real experiment can test the model; repeating its own programmed equation cannot independently confirm it.
Assumptions and limits: More electrons is not more energy per electron

Assumed here

  • One electron can receive one quantum of energy hν.
  • Escape requires positive work Φ, with any additional loss nonnegative.
  • Keep frequency, work function, and yield assumptions fixed when comparing power.

What this does not establish

  • The threshold and linear relation are programmed assumptions and consequences, not independent experimental proof.
  • The work function used in a hypothetical example is not a sourced value for a named metal.
  • Real collection and emission rates require additional material and apparatus assumptions.

Earlier step: A coefficient suggests an energy element

Source context: German source · English · Interlinear gloss · Facsimile

qualification · Within the stated model

A stopping voltage is a magnitude with a sign convention

Why does retarding an electron not make its charge positive?

eVs=Kmax=hνΦeV_s=K_{\mathrm{max}}=h\nu-\Phi

The positive charge magnitude times the stopping-voltage magnitude equals maximum kinetic energy, h frequency minus escape work.

V_s is a nonnegative magnitude. The sign of a laboratory electrode voltage must also specify which electrode is the reference. Using e for a positive magnitude does not change the negative charge of the electron.

If the electron receives only part of the available energy or loses energy inside matter, its outgoing kinetic energy is at most hν − Φ. These losses do not reverse the frequency dependence or turn extra intensity into extra energy for one electron.

An inverse energy-budget question asks for the smallest accelerating-potential magnitude capable of supplying a quantum hν. Within that idealized conversion, eV must be at least hν. This is an energetic lower bound, not a promise that every collision produces light.

Show every step here: A stopping voltage is a magnitude with a sign convention
  1. Choose a voltage reference and state which direction is retarding.
  2. Keep the electron charge −e separate from the positive magnitude e.
  3. Use Kmax/e for the positive stopping magnitude only when an emitted-electron endpoint exists.
  4. For partial transfer or losses, replace an equality by the appropriate upper bound.
  5. For the inverse light-production question, require at least hν of supplied energy before discussing conversion efficiency.
Assumptions and limits: A stopping voltage is a magnitude with a sign convention

Assumed here

  • e denotes a positive charge magnitude; the electron charge is −e.
  • A retarding potential removes kinetic energy according to the explicitly chosen electrode convention.

What this does not establish

  • No stopping endpoint for emitted electrons is supplied when emission is not allowed.
  • A negative energy deficit and a signed collector voltage are different quantities.

Earlier step: More electrons is not more energy per electron

Source context: German source · English · Interlinear gloss · Facsimile

References and source status

Newly authored explanatory preview in modern notation; editorial and physics review are pending. The introduction and all nine numbered sections have explanatory treatments below, but this is not a German transcription, an aligned English translation, or a complete critical edition. Source faces remain in preparation. The headings and argument units are editorial, not a verified paragraph-by-paragraph source inventory.

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

A. Einstein, Does the inertia of a body depend upon its energy content?. Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s reviewed translation or pinned facsimile.

18 foundation readings sit behind this argument.

Keep the explanation for offline reading

Each HTML file contains the available explanation at every detail level, linked foundations, source references, and available build-time scalar results. It includes no private notes or running simulations. This remains an explanation preview, not a reviewed source edition.

Open the downloaded file in a browser. Online source links still need a connection. Interactive plots are not included. Browse saved-chapter downloads.