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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.

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Student-friendly explanation

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.

How to write this in exams

  1. 1

    Start with the exact idea

    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

    Then show how to use it

    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.

  3. 3

    Add one concrete 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.

  4. 4

    Avoid this incomplete 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.

Definition

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.

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.

Rule to remember

Einstein equation: Kmax = h nu - phi0. Also Kmax = (1/2)mvmax^2 = eV0 and photon energy E = h nu = hc/lambda. 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 lambda is wavelength in m. Use when radiation frequency is known or can be found and the metal work function is given.

Memory hook

Photon energy is split into escape energy plus motion energy.

Examples and method

Worked example

Light of frequency 8.0 x 10^14 Hz falls on a metal of work function 2.0 eV. Photon energy E = h nu = (6.63 x 10^-34)(8.0 x 10^14) = 5.304 x 10^-19 J = 3.315 eV. Kmax = 3.315 eV - 2.0 eV = 1.315 eV. The fastest photoelectrons leave with maximum kinetic energy about 1.32 eV.

Method to apply

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.

Diagram support

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.

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.

Avoid common mistakes

Common confusion

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

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.

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.

Quick check

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

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.

Answer writing and exam use

1-mark answer

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-mark answer

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. Einstein equation: Kmax = h nu - phi0. Also Kmax = (1/2)mvmax^2 = eV0 and photon energy E = h nu = hc/lambda. 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 lambda is wavelength in m. Use when radiation frequency is known or can be found and the metal work function is given. 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-mark answer

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. Einstein equation: Kmax = h nu - phi0. Also Kmax = (1/2)mvmax^2 = eV0 and photon energy E = h nu = hc/lambda. 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 lambda is wavelength in m. Use when radiation frequency is known or can be found and the metal work function is given. Light of frequency 8.0 x 10^14 Hz falls on a metal of work function 2.0 eV. Photon energy E = h nu = (6.63 x 10^-34)(8.0 x 10^14) = 5.304 x 10^-19 J = 3.315 eV. Kmax = 3.315 eV - 2.0 eV = 1.315 eV. The fastest photoelectrons leave with maximum kinetic energy about 1.32 eV. 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. Wrong: Kmax = phi0 - h nu. Correct: Kmax = h nu - phi0 because work function is the energy spent first and the remainder becomes kinetic energy.
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