# 15.1 Photoelectric effect  (Page 2/2)

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While solving problems we need to decide for ourselves whether we should consider the wave property or the particle property of light. For example, when dealing with interference and diffraction, light should be treated as a wave, whereas when dealing with photoelectric effect we consider the particle nature.

## Applications of the photoelectric effect

We have learnt that a metal contains electrons that are free to move between the valence and conduction bands. When a photon strikes the surface of a metal, it gives all its energy to one electron in the metal.

• If the photon energy is equal to the energy between two energy levels then the electron is excited to the higher energy level.
• If the photon energy is greater than or equal to the work function (energy needed to escape from the metal), then the electron is emitted from the surface of the metal (the photoelectric effect).

The work function is different for different elements. The smaller the work function, the easier it is for electrons to be emitted from the metal. Metals with low work functions make good conductors. This is because the electrons are attached less strongly to their surroundings and can move more easily through these materials. This reduces the resistance of the material to the flow of current i.e. it conducts well. [link] shows the work functions for a range of elements.

 Element Work Function ( $\mathrm{J}$ ) Aluminium $6,9×{10}^{-19}$ Beryllium $8,0×{10}^{-19}$ Calcium $4,6×{10}^{-19}$ Copper $7,5×{10}^{-19}$ Gold $8,2×{10}^{-19}$ Lead $6,9×{10}^{-19}$ Silicon $1,8×{10}^{-19}$ Silver $6,9×{10}^{-19}$ Sodium $3,7×{10}^{-19}$

## Interesting fact

The electron volt (eV) is the kinetic energy gained by an electron passing through a potential difference of one volt (1 $\mathrm{V}$ ). A volt is not a measure of energy, but the electron volt is a unit of energy. When you connect a $1.5\phantom{\rule{3.33333pt}{0ex}}\mathrm{V}$ battery to a circuit, you can give $1.5\phantom{\rule{3.33333pt}{0ex}}\mathrm{eV}$ of energy to every electron.

Ultraviolet radiation with a wavelength of 250 nm is incident on a silver foil (work function $\phi$ = $6,9×{10}^{-19}$ ). What is the maximum kinetic energy of the emitted electrons?

1. We need to determine the maximum kinetic energy of an electron ejected from a silver foil by ultraviolet radiation.

The photoelectric effect tells us that:

$\begin{array}{ccc}\hfill {E}_{k}& =& {E}_{photon}-\phi \hfill \\ \hfill {E}_{k}& =& h\frac{c}{\lambda }-\phi \hfill \end{array}$

We also have:

Work function of silver: $\phi =6,9×{10}^{-19}$ J

UV radiation wavelength = 250 nm = $250×{10}^{-9}$ m

Planck's constant: $h=6,63×{10}^{-34}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{\mathrm{m}}^{2}\mathrm{kg}{\mathrm{s}}^{-1}$

speed of light: $c=3×{10}^{8}\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}{\mathrm{s}}^{-1}$

2. $\begin{array}{ccc}\hfill {E}_{k}& =& \frac{hc}{\lambda }-\phi \hfill \\ & =& \left[6,63×{10}^{-34}×\frac{3×{10}^{8}}{250×{10}^{-9}}\right]-6,9×{10}^{-19}\hfill \\ & =& 1,06×{10}^{-19}\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}\hfill \end{array}$

The maximum kinetic energy of the emitted electron will be $1,06×{10}^{-19}\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}$ .

If we were to shine the same ultraviolet radiation ( $f=1,2×{10}^{15}\phantom{\rule{3.33333pt}{0ex}}\mathrm{Hz}$ ), on a gold foil (work function $=8,2×{10}^{-19}\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}$ ), would any electrons be emitted from the surface of the gold foil?

For the electrons to be emitted from the surface, the energy of each photon needs to be greater than the work function of the material.

1. $\begin{array}{ccc}\hfill {E}_{photon}& =& hf\hfill \\ & =& 6,63×{10}^{-34}×1,2×{10}^{15}\hfill \\ & =& 7,96×{10}^{-19}\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}\hfill \end{array}$

Therefore each photon of ultraviolet light has an energy of $7,96×{10}^{-19}\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}$ .

2. $\begin{array}{ccc}\hfill {\phi }_{gold}& =& 8,2×{10}^{-19}\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}\hfill \end{array}$
3. $\begin{array}{ccc}\hfill 7,96×{10}^{-19}\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}& <& 8,2×{10}^{-19}\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}\hfill \\ \hfill {E}_{photons}& <& {\phi }_{gold}\hfill \end{array}$

The energy of each photon is less than the work function of gold, therefore, the photons do not have enough energy to knock electrons out of the gold. No electrons would be emitted from the gold foil.

## Units of energy

When dealing with calculations at a small scale (like at the level of electrons) it is more convenient to use different units for energy rather than the joule (J). We define a unit called the electron-volt (eV) as the kinetic energy gained by an electron passing through a potential difference of one volt.

$E=q×V$

where $q$ is the charge of the electron and $V$ is the potential difference applied. The charge of 1 electron is $1,6×{10}^{-19}$  C, so 1 eV is calculated to be:

$1\phantom{\rule{0.277778em}{0ex}}\mathrm{eV}=\left(1,{610}^{-19}\phantom{\rule{0.277778em}{0ex}}\mathrm{C}×1\phantom{\rule{0.277778em}{0ex}}\mathrm{V}\right)=1,6×{10}^{-19}\phantom{\rule{0.166667em}{0ex}}\mathrm{J}$

You can see that $1,6×{10}^{-19}$ J is a very small amount of energy and so using electron-volts (eV) at this level is easier.

Hence, 1eV = $1.6×{10}^{-}19$ J which means that 1 J = $6.241×{10}^{18}$ eV

## Solar cells

The photo-electric effect may seem like a very easy way to produce electricity from the sun. This is why people choose to make solar panels out of materials like silicon, to generate electricity.In real-life however, the amount of electricity generated is less than expected. This is because not every photon knocks out an electron. Other processes such as reflection or scattering also happen. This means that only a fraction $\approx 10%$ (depends on the material) of the photons produce photoelectrons. This drop in efficiency results in a lower current. Much work is being done in industry to improve this efficiency so that the panels can generate as high a current as possible, and create as much electricity as possible from the sun. But even these smaller electrical currents are useful in applications like solar-powered calculators.

## The photoelectric effect

1. Describe the photoelectric effect.
2. List two reasons why the observation of the photoelectric effect was significant.
3. Refer to [link] : If I shine ultraviolet light with a wavelength of 288 nm onto some aluminium foil, what would be the kinetic energy of the emitted electrons?
4. I shine a light of an unknown wavelength onto some silver foil. The light has only enough energy to eject electrons from the silver foil but not enough to give them kinetic energy. (Refer to [link] when answering the questions below:)
1. If I shine the same light onto some copper foil, would electrons be ejected?
2. If I shine the same light onto some silicon, would electrons be ejected?
3. If I increase the intensity of the light shining on the silver foil, what happens?
4. If I increase the frequency of the light shining on the silver foil, what happens?

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