C
CraftExam
medium difficulty10 min

Einstein's Photoelectric Equation

Einstein's photoelectric equation states that the energy h nu of an incident photon is used partly to overcome the work function phi0 of the metal and the remaining energy appears as the maximum kinetic energy of the emitted photoelectron.

In one line: Photon energy is split into escape energy plus motion energy.

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Understand it

  • The equation explains photoelectric observations by treating light energy as delivered in photons.
  • One photon gives its energy to one electron.
  • If h nu is less than phi0, no electron is emitted.
  • If h nu is equal to phi0, the electron just escapes.
  • If h nu is greater than phi0, the excess energy becomes maximum kinetic energy.

Example

For a metal of work function 2.0 eV, light of photon energy 3.5 eV can eject photoelectrons with maximum kinetic energy 1.5 eV.

Formula and rule

Einstein equation: Kmax = h nu - phi0.

Also Kmax = (1/2)mvmax2 = eV0 and photon energy E = h nu = hc/λ. h is Planck's constant in J s, nu is frequency in Hz, phi0 and Kmax are in J or eV, e is charge in C, V0 is in V, c is speed of light in m s−1, and λ is wavelength in m.

Use when radiation frequency is known or can be found and the metal work function is given.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Light of frequency 8.0 × 1014 Hz falls on a metal of work function 2.0 eV.

  1. 1Photon energy E = h nu = (6.63 × 10−34)(8.0 × 1014) = 5.304 × 10−19 J = 3.315 eV.
  2. 2Kmax = 3.315 eV - 2.0 eV = 1.315 eV.
  3. 3The fastest photoelectrons leave with maximum kinetic energy about 1.32 eV.

Method you can reuse

Convert wavelength to frequency if needed, calculate photon energy, convert work function to the same unit, subtract phi0 from h nu, check that the result is not negative, and convert Kmax to stopping potential or speed if asked.

Visual tip

A graph of Kmax or stopping potential versus frequency is useful. The frequency-axis intercept gives threshold frequency nu0, and the slope of Kmax versus nu is h. For V0 versus nu, the slope is h/e.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Why is there no photoelectric emission when h nu is less than phi0?

Think first, then show answer
There is no photoelectric emission because the energy of each photon is less than the minimum energy needed by a surface electron to escape the metal. Increasing the number of such photons does not help a single electron overcome the work function.

Common wrong answer

Wrong: Kmax = phi0 - h nu. Correct: Kmax = h nu - phi0 because work function is the energy spent first and the remainder becomes kinetic energy.

Don't mix this up

Students often substitute wavelength directly into h nu without converting using nu = c/λ. Another common mistake is mixing eV and joule in the same calculation.

Exam tip

Write the equation first, list units, and then substitute values in one energy unit throughout. For stopping potential problems, use Kmax = eV0 after finding the excess energy.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Appears in numericals on threshold frequency, stopping potential, maximum kinetic energy, wavelength, and graph slope. Derivation-style questions ask how photon energy is distributed between work function and kinetic energy.
  1. 1

    Enough for 1 mark

    Einstein's photoelectric equation states that the energy h nu of an incident photon is used partly to overcome the work function phi0 of the metal and the remaining energy appears as the maximum kinetic energy of the emitted photoelectron.

  2. 2

    Add for 2 marks

    Example: For a metal of work function 2.0 eV, light of photon energy 3.5 eV can eject photoelectrons with maximum kinetic energy 1.5 eV.

  3. 3

    Add for 3 marks

    The equation explains photoelectric observations by treating light energy as delivered in photons.

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