The photoelectric effect is indisputable evidence for the particle nature of light. Particles of light are discrete and named photons. The energy of a photon depends not on intensity but frequency.
Key Concepts
There is a significant body of evidence for the wave model of light including interference and diffraction. Reflection and refraction are also wave properties, but are not exclusive to waves.
The photoelectric effect is evidence only for the quantum nature of light.
When light is shone on a metal plate, electrons in the plate are excited and may escape the surface (photoelectrons). The results of the experiment are:
- Below a certain frequency of light, no electrons are emitted, regardless of the intensity of the light. This is evidence against the wave theory (in which continuous waves would eventually build up sufficient energy for escape).
- When the frequency is increased (i.e. towards or into the ultraviolet region of the electromagnetic spectrum), a threshold frequency is reached at which electrons are emitted. These are the electrons closest to the surface of the metal. The energy required to just release the electrons is the work function. This observation indicates that a single particle of light gives its entire energy to a single electron.
- Above the threshold frequency, increasing the frequency gives the electrons a higher maximum kinetic energy. The range of kinetic energies comes from the variation in where the electrons were located in the metal. This graph shows how the energy of the photoelectrons varies with frequency; none are emitted before the threshold frequency but then kinetic energy increases linearly with frequency.
- Above the threshold frequency, increasing the intensity increases the number of electrons released in a given time, but not the maximum kinetic energy.
The main deficiency in the wave model of light for explaining the photoelectric effect is that an increased intensity should increase the energy of the electrons. Over time, enough energy would be built up for electrons to be released. And the intensity should increase the maximum kinetic energy of the electrons.
Use quizzes to practise application of theory.
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The graph represents the maximum kinetic energy of photoelectrons against the frequency of illuminating light.
Planck's constant in Js is:
\(h\) is the gradient. The only of these options to do is \(B\over C\).
The units of the energy axis are eV, so we must multiply by \(e\) to obtain Planck's constant in the required units.
The graph represents the maximum kinetic energy of photoelectrons against the frequency of illuminating light.
The work function in eV is:
The work function, the minimum energy that must be overcome, is the vertical intercept. The units are already eV.
The graph represents the maximum kinetic energy of photoelectrons against the frequency of illuminating light.
The maximum wavelength for electron emission is:
\(C\) is the minimum frequency needed. The equivalent wavelength, \({c\over f}={c\over C}\).
The graph represents the maximum kinetic energy of photoelectrons against the frequency of illuminating light.
The work function in eV is:
\(\Phi =hf_0 = \text{gradient} \times C\)
Other methods are possible, but this is the only correct response of these options.
Two sources of light, one red and one blue, have the same intensity.
The ratio \(\text{number of photons from the blue source}\over \text{number of photons from the red source}\) is:
It is the number of photons per unit area that is greater for the red light. We are not told the surface area of the light sources.
Photoelectric emission takes place when a green light source is shone on a sodium plate but not for yellow light. If the green light is replaced by a white light of the same intensity there will be:
White light is made of all wavelengths. Only those corresponding to green light or shorter will emit electrons.
Photoelectric emission takes place when a green light source is shone on a sodium plate but not for yellow light. If the green light is replaced by a blue light of the same intensity there will be:
Although the blue light has fewer photons per unit area they have more energy. More electrons will gain enough energy for emission.
Photoelectric emission takes place when a green light source is shone on a sodium plate but not for yellow light. If the wavelength of the green source is reduced while keeping the intensity constant, the number of electrons emitted will:
Electron current will reach saturation then reduce as maintaining the intensity beyond saturation will reduce the number of photons.
A photocell has a threshold frequency in the green region of the spectrum. To get current to flow for red light could be achieved with:
The frequency of red light is below the threshold frequency and so no electrons would have sufficient energy to be emitted
Light is shone on a metal plate at the threshold frequency for photoelectric emission.
The ratio of \(\text{photoelectrons} \over \text{photons}\) is:
Not all of the photons cause photoelectric emission as some electrons will be far beneath the surface
How much of Photoelectric effect and photons have you understood?
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