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Wize High School Grade 12 Physics Textbook > Light as a Particle

Photoelectric Effect (Light = Particle)

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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.


Using conservation of energy we get:
 hf=W+12mv2 \boxed{\ hf=W+\dfrac{1}{2} mv^2 \ }

  • hh is Planck's constant
  • ff is the frequency of the incoming photon
  • WW is the work function of the metal
  • mm is the mass of the electron
  • vv is the speed of the ejected electron

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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 f0\bcth {f_0} and is given by:

 hf0=W \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).



Wize Concept
  • Increasing the intensity (more photons) only causes more electrons to be ejected, but doesn't increase their energy.
  • Increasing the frequency (more available energy) will increase the kinetic energy of the electrons, but not the number of electrons.



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)


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Exam Tip
We can re-write the equation at the top as:
 KE=hf−hf0 \boxed{\ KE=hf-hf_0 \ }
which can be graphed as a line:

 KE vs f   with   slope=h   and    y intercept=hf0 \boxed{\ \bf{KE} \textbf{\ vs\ }f\ \ \text{\ with\ }\ \ slope=h\ \ \text{\ and \ }\ \ y\ intercept=hf_0 \ }













Exam Tip
If the ejected electrons are then stopped by a potential difference, all their kinetic energy 12mv2\dfrac{1}{2}mv^2 will be converted to electric potential energy qVqV (where qq is the charge of the electron and VVis the stopping voltage). Then the equation at the top becomes:
 hf=W+qV \boxed{\ hf=W+qV \ }


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Example: Electron Speed


A certain metal has a work function of 1.451.45 eV. If light of wavelength 600600 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:

hcλ=W+12mv2\dfrac{hc}{\lambda}=W+\dfrac{1}{2} mv^2

Rearrange to get speed:

12mv2=hcλ−W\dfrac{1}{2} mv^2=\dfrac{hc}{\lambda}-W

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

=29.11×10−31 [(6.624×10−34)(3×108)600×10−9−1.45×1.602×10−19]=\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]}

=4.66×105=4.66\times10^5 (m/s)

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Practice: Two Metals


A beam of photons with frequency 7.20×10147.20\times10^{14} Hz and total energy 5×10−105\times10^{-10} J hit two metals:

Gold (ϕ=5.47×10−19\phi=5.47\times10^{-19} J ) and Sodium (ϕ=3.78×10−19\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?

Practice: Photoelectric Effect

Photons of light with a frequency 9.36×1014 Hz9.36\times10^{14}\ Hz hit two metals:

A. Which metal(s) will emit electrons?
Copper (ϕ=5.10×10−19J\phi=5.10\times10^{-19}J ) and Aluminium (ϕ=6.86×10−19J\phi=6.86\times10^{-19}J ).

B. What is the kinetic energy of the electrons being ejected from the metal(s)?
Fill in the answer here (two 2 decimal places) ______x10-19 J