Photoelectric Effect – Learn


The photoelectric effect is observed when electromagnetic radiation of a particular frequency is shone on a shiny metallic surface and electrons are emitted from its surface. Electrons that are emitted this way are called photoelectrons

A common apparatus used to observe the photoelectric effect is illustrated below:

It consists of a shiny metal surface exposed to an external ultraviolet light source in a vacuum tube. The photoelectrons emitted from the metal surface would move toward the collector and form a current that is detected at the anode. This current is known as a photocurrent.

The operation of this apparatus was explored by German physicist Philipp von Lenard. He investigated how the energy of emitted photoelectrons varied with the intensity of the light used. Lenard was able to study the magnitude of the photocurrent as it would strike the collector to form a current that could be measured. To measure the energy of the electrons emitted, Lenard gave the collector a known negative charge to repel the photoelectrons. He found there was a well defined minimum voltage that was required to stop the photocurrent. When the photocurrent was reduced to zero, the repelling voltage was equal to the voltage of the photocurrent.

There were two main features of the light source that provided interesting results. These features were the frequency and intensity of the light that was incident on the metallic surface. It was found that:

  • There was a minimum frequency at which photoelectrons would be emitted from the surface of the metal
  • This minimum frequency varied for different types of metals
  • Increasing the intensity of the light increased the number of photoelectrons. It did not give them more energy.
  • Increasing the frequency of the light increased the kinetic energy of the photoelectrons. It did not increase the photocurrent.

The following graph shows the maximum kinetic energy with which the photoelectrons are emitted versus the frequency of light, for five different metals. Note that the gradient of all the lines is equal to Planck’s constant.


Photoelectric effect – Quantitative Analysis

Lenard used a filter which allowed specific frequencies of light to be incident on the metal cathode. He found that for each cathode metal there was a specific frequency of light below which no photoelectrons would be emitted. This is called the threshold frequency, f0, and is different for each metal. For frequencies above the threshold frequency, photoelectrons will be emitted from the cathode and a photocurrent will be detected. When a negative voltage is applied to the illuminated cathode, the photoelectrons are pushed back toward the cathode and the photocurrent is reduced. If the applied voltage is increased, the photocurrent will eventually reduce to 0. This voltage is known as the stopping voltage. The maximum kinetic energy of the photoelectrons can be determined from the work done to stop the photoelectrons being emitted: W=qeV0. If the stopping voltage is 1.5V then the maximum kinetic energy of the photoelectrons is 1.5eV.

Work Function and Threshold Frequency

The work function of a metal is the minimum energy value that must achieved before photoelectrons will be emitted from the metal. The corresponding frequency given by E=hf is the threshold frequency – remember that this is different for each metal. The work function is given by:

\phi =h{ f }_{ 0 }

where:

\phi is the work function of the metal (in J or eV)

h is Planck’s constant (6.626 × 10−34 Js or 4.14 × 10−15 eV s)

{ f }_{ 0 } is the threshold frequency for any particular metal (in Hz)

The Kinetic Energy or Photoelectrons

If the energy of the incident photon is greater than the work function of the metal, any energy greater than the work function will be converted to kinetic energy. The following is the photoelectric equation developed by Einstein:

{ { K }_{ max }=hf- }\phi

where:

{ { K }_{ max } is the maximum kinetic energy of any photoelectrons (in J or eV)

h is Planck’s constant (6.626 × 10−34 Js or 4.14 × 10−15 eV s)

f is the frequency of the incident light (in Hz)

\phi is the work function of the metal (in J or eV)


Example 1:

Calculate the energy that a photon with a wavelength of 480nm has in a) joules b) eV

Answer:

a) using E=hf and rearranging c=f\lambda to give f=\cfrac { c }{ \lambda }  

we can use: E=h\cfrac { c }{ \lambda }

E=(6.626\times { 10 }^{ -34 })\times (\cfrac { 3\times { 10 }^{ 8 } }{ { 480\times 10 }^{ -9 } } )

E=4.14\times { 10 }^{ -19 }\:J

b) E=\cfrac { 4.14\times { 10 }^{ -19 } }{ 1.602\times { 10 }^{ -19 } } eV

E=2.58\:eV

Example 2:

Calculate the work function (in J) for magnesium if it has a threshold frequency of 8.84 × 1014 Hz:

Answer:

using \phi =h{ f }_{ 0 }

\phi=6.626\times { 10 }^{ -34 }\times { 8.84\times { 10 }^{ 14 } }

\phi=5.86\times { 10 }^{ -19 }\:J

Example 3:

Calculate the kinetic energy (in J and eV) of photoelectrons emitted from aluminium when ultraviolet light with a frequency of 2.1 × 1015 Hz strikes its surface. Aluminium has a work function of 6.50 × 10−19 J.

Answer:

using { { K }_{ max }=hf- }\phi

{ K }_{ max }=(6.626\times { 10 }^{ -34 }\: \times \: { 2.1\times { 10 }^{ 15 }) }-6.5\times { 10 }^{ -19 }

{ K }_{ max }=7.4\times { 10 }^{ -19 } } J

{ K }_{ max }=\cfrac { 7.4\times { 10 }^{ -19 } }{ 1.602\times { 10 }^{ -19 } } eV

{ K }_{ max }\:=\:4.63 eV

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