Wize University Physics Textbook (Master) > Modern Physics

Photoelectric Effect

The Photoelectric Effect

When light is incident on some metals, electrons can be ejected from the metal surface. This is called the Photoelectric effect.

The incoming photon energy is used by the electron to break free from the metal (if there's enough of it in the first place). This energy is called the work function, and is specific to each metal.
If there's any energy left afterwards, it will become the electron's kinetic energy.

Image illustrating The Photoelectric Effect

Using conservation of energy we get:

\boxed{\ hf=W+\dfrac{1}{2} mv^2 \ }

$h$ is Planck's constant
$f$ is the frequency of the incoming photon
$W$ is the work function of the metal
$m$ is the mass of the electron
$v$ is the speed of the ejected electron

The work function is the amount of energy that holds an electron to the metal surface, i.e. the minimum photon energy that will cause electrons to be ejected. The corresponding photon frequency is called the threshold frequency $\bcth {f_0}$ and is given by:

\boxed{ \ hf_0=W\ }

If the frequency of the photons is below the threshold frequency, no electrons will be ejected. This happens no matter how many photons we have (i.e. the intensity of the light doesn't matter).

The experiment highlights the particle nature of light:

$\to$ Electrons are "knocked out" by one-on-one, photon-electron collisions

$\to$ The ejection is "instantaneous": the energy of light is highly localized in space, i.e. it's a particle of light (photon)

$\to$ The photons transfer momentum to the electrons, which is similar to a hard particle collision (similar to the Compton effect)

Example: Electron Speed

A certain metal has a work function of $1.45$ eV. If light of wavelength $600$ nm is incident on the metal, what is the maximum possible speed of the ejected electrons?

Using conservation of energy we can set up the photoelectric effect equation as:

$\dfrac{hc}{\lambda}=W+\dfrac{1}{2} mv^2$

Rearrange to get speed:

$\dfrac{1}{2} mv^2=\dfrac{hc}{\lambda}-W$

$v=\sqrt{\dfrac{2}{m} \ \bigg(\dfrac{hc}{\lambda}-W\bigg)}$

$=\sqrt{\dfrac{2}{9.11\times10^{-31}} \ \bigg[\dfrac{(6.624\times10^{-34})(3\times10^8)}{600\times10^{-9}}-1.45\times1.602\times10^{-19}\bigg]}$

$=4.66\times10^5$ (m/s)

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Practice: Two Metals

A beam of photons with frequency $7.20\times10^{14}$ Hz and total energy $5\times10^{-10}$ J hit two metals:

Gold ( $\phi=5.47\times10^{-19}$ J ) and Sodium ( $\phi=3.78\times10^{-19}$ J )

a) Which metal(s) will emit electrons?

b) For the metal(s) in part a), calculate the kinetic energy of each ejected electron.

c) For the metal(s) in part a), find how many electrons are ejected.

d) Given that the frequency of light is kept constant, if the total energy of the beam increases, how will the Photoelectric Effect of each metal be affected?

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Photoelectric Effect

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The Photoelectric Effect

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Example: Electron Speed

Practice: Two Metals

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