Dual Nature of Matter and Radiation • Topic 2 of 3

Einstein's Photoelectric Equation & Photons

To explain the photoelectric effect, Albert Einstein (1905) proposed that light is not a continuous wave when it exchanges energy with matter but comes in discrete packets called photons. Each photon carries energy $E = h\nu$, where $h = 6.63\times10^{-34}\,\text{J s}$ is Planck's constant and $\nu$ is the frequency of the light. A beam of light is a stream of photons; its intensity is set by the number of photons crossing unit area per second, while the energy of each photon is fixed by frequency alone.

A photon also carries momentum $p = \dfrac{E}{c} = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}$, even though it has zero rest mass. Photons travel at the speed of light $c$ in vacuum, are electrically neutral, and are not deflected by electric or magnetic fields. In any photon-electron interaction, both energy and momentum are conserved.

Einstein argued that a single electron absorbs a single photon in an all-or-nothing event. Part of the photon energy frees the electron (the work function $\phi_0$) and the remainder appears as the kinetic energy of the ejected electron. This gives Einstein's photoelectric equation:

$K_{max} = h\nu - \phi_0$, the maximum kinetic energy of a photoelectron.
$h\nu_0 = \phi_0$, so equivalently $K_{max} = h(\nu - \nu_0)$.

This single equation explains every law. The kinetic energy grows linearly with frequency and is independent of intensity. If $\nu < \nu_0$ then $h\nu < \phi_0$ and $K_{max}$ would be negative, which is impossible — hence the threshold frequency. Increasing intensity supplies more photons, so more electrons are ejected (higher current) but each still carries the same $K_{max}$. And because absorption is a one-photon, one-electron event, emission is instantaneous.

The maximum kinetic energy is measured using the stopping potential $V_0$ — the reverse voltage that just stops the fastest photoelectron, so $eV_0 = K_{max}$. Combining with Einstein's equation: $eV_0 = h\nu - \phi_0$, hence $V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}$. A graph of stopping potential $V_0$ against frequency $\nu$ is therefore a straight line. Its slope is $\dfrac{h}{e}$ — the same for all metals — which let Millikan measure $h$ experimentally. The line meets the frequency axis at $\nu = \nu_0$ (threshold), and its negative intercept on the $V_0$ axis equals $-\dfrac{\phi_0}{e}$, giving the work function. This agreement won Einstein the 1921 Nobel Prize in Physics.

Solved Examples

Worked Example

Calculate the energy of a photon of light of wavelength $5000\,\text{\u00c5}$ in joules and in eV. (Take $h = 6.63\times10^{-34}\,\text{J s}$, $c = 3\times10^{8}\,\text{m/s}$.)

Solution

$\lambda = 5000\,\text{\u00c5} = 5\times10^{-7}\,\text{m}$.
$E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times10^{-34})(3\times10^{8})}{5\times10^{-7}}$.
$E = \dfrac{1.989\times10^{-25}}{5\times10^{-7}} = 3.98\times10^{-19}\,\text{J}$.
In eV: $E = \dfrac{3.98\times10^{-19}}{1.6\times10^{-19}} = 2.48\,\text{eV}$.

Answer: The photon energy is $3.98\times10^{-19}\,\text{J}$, i.e. about $2.48\,\text{eV}$.

Worked Example

Light of frequency $10^{15}\,\text{Hz}$ is incident on a metal of work function $2\,\text{eV}$. Find the maximum kinetic energy of the emitted photoelectrons in eV.

Solution

Photon energy: $h\nu = (6.63\times10^{-34})(10^{15}) = 6.63\times10^{-19}\,\text{J}$.
In eV: $h\nu = \dfrac{6.63\times10^{-19}}{1.6\times10^{-19}} = 4.14\,\text{eV}$.
Einstein's equation: $K_{max} = h\nu - \phi_0 = 4.14 - 2 = 2.14\,\text{eV}$.

Answer: The maximum kinetic energy of the photoelectrons is about $2.14\,\text{eV}$.

Worked Example

The stopping potential for a metal is $1.5\,\text{V}$ when illuminated by light of energy $3.5\,\text{eV}$. Find the work function of the metal.

Solution

Stopping potential gives $K_{max} = eV_0 = 1.5\,\text{eV}$.
Einstein's equation: $K_{max} = h\nu - \phi_0$.
$\phi_0 = h\nu - K_{max} = 3.5 - 1.5$.
$\phi_0 = 2.0\,\text{eV}$.

Answer: The work function of the metal is $2.0\,\text{eV}$.

Worked Example

Find the momentum of a photon of wavelength $6.6\times10^{-7}\,\text{m}$. (Take $h = 6.63\times10^{-34}\,\text{J s}$.)

Solution

Photon momentum: $p = \dfrac{h}{\lambda}$.
$p = \dfrac{6.63\times10^{-34}}{6.6\times10^{-7}}$.
$p = 1.0\times10^{-27}\,\text{kg m/s}$.

Answer: The momentum of the photon is about $1.0\times10^{-27}\,\text{kg m/s}$.

Worked Example

In a photoelectric experiment, the slope of the stopping potential versus frequency graph is $4.14\times10^{-15}\,\text{V s}$. Determine Planck's constant. (Take $e = 1.6\times10^{-19}\,\text{C}$.)

Solution

From $eV_0 = h\nu - \phi_0$, the slope of the $V_0$-$\nu$ line is $\dfrac{h}{e}$.
So $h = e \times \text{slope}$.
$h = (1.6\times10^{-19})(4.14\times10^{-15})$.
$h = 6.62\times10^{-34}\,\text{J s}$.

Answer: Planck's constant comes out as about $6.62\times10^{-34}\,\text{J s}$, matching the accepted value.

Worked Example

A metal has a threshold wavelength of $6000\,\text{\u00c5}$. Light of wavelength $4000\,\text{\u00c5}$ falls on it. Find the maximum kinetic energy of the photoelectrons in eV. (Take $hc = 1240\,\text{eV nm}$.)

Solution

Work function: $\phi_0 = \dfrac{hc}{\lambda_0} = \dfrac{1240}{600} = 2.07\,\text{eV}$.
Photon energy: $h\nu = \dfrac{hc}{\lambda} = \dfrac{1240}{400} = 3.10\,\text{eV}$.
$K_{max} = h\nu - \phi_0 = 3.10 - 2.07$.
$K_{max} = 1.03\,\text{eV}$.

Answer: The maximum kinetic energy of the photoelectrons is about $1.03\,\text{eV}$.

Key Points

Light is emitted and absorbed as photons of energy $E = h\nu = \dfrac{hc}{\lambda}$.
A photon has zero rest mass but momentum $p = \dfrac{h}{\lambda}$.
Einstein's photoelectric equation: $K_{max} = h\nu - \phi_0 = h(\nu - \nu_0)$.
Stopping potential measures $K_{max}$ since $eV_0 = h\nu - \phi_0$.
The $V_0$ vs $\nu$ graph is a straight line of slope $\dfrac{h}{e}$ that intercepts the $\nu$-axis at $\nu_0$.

Quick Check — 4 MCQs

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Q1.The energy of a photon of frequency $\nu$ is given by:

Explanation: A photon carries energy $E = h\nu$, where $h$ is Planck's constant.

Q2.Einstein's photoelectric equation is:

Explanation: The photon energy $h\nu$ splits into the work function $\phi_0$ and the kinetic energy $K_{max}$.

Q3.The stopping potential $V_0$ is related to maximum kinetic energy by:

Explanation: The retarding voltage that just stops the fastest electron satisfies $eV_0 = K_{max}$.

Q4.The slope of the stopping potential versus frequency graph equals:

Explanation: Since $V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}$, the slope is $\dfrac{h}{e}$, the same for all metals.

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