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# Lesson 6.2: Quantized Energy and Photons

← Lesson 6.1: Lesson 6.3: →

In the previous lesson we treated light as a wave – oscillating electric and magnetic fields. Such waves exhibit classical wave like behavior.

However, in some instances, light appears to behave more like a particle. How are we to make sense of this? The first problem we face is how to represent light. If light is a wave we can represent it as an oscillating field. How can we show it behaving as a particle at the same time? We could draw the wave in clusters – implying chunks of waves.

It seems less confusing at this stage to simply draw them as dots and say that each particle of light has certain wave properties associated with it that relate to the amount of energy it has per chunk.

To better understand how energy is transferred to atoms when exposed to light, we will consider the photoelectric effect. This effect refers to the fact that when a metal is exposed to light, electrons can be ejected from the surface.

When we try to make sense of this effect while thinking of light as a wave, we run into trouble. Thinking of light as a wave, we would expect waves of greater frequency to eject more electrons, and waves of greater intensity to eject electrons with higher kinetic energy. What we find instead is that no electrons are ejected if the frequency is too low, and once the frequency is increased beyond a certain threshold, light with greater frequency results in ejected electrons with higher kinetic energy.

We can best make sense of this by abandoning the idea that light is a wave, and thinking of light as being made up of particles. We will called such light particles photons and think of their energy as coming in chunks - think of their energy as quantized. Light with greater frequency has more energy per photon. When a photon collides with an electron in a metal, the photon is absorbed and transfers its chunk of energy to the electron. At low frequency, when each photon has lower energy, no electrons are emitted. As we increase the frequency of light we increase the energy per photon and eventually reach a point when the energy transferred to the electron is sufficient for it to overcome the attractive forces of the nucleus. The electron is ejected from the metal, but with zero kinetic energy. If the energy of each photon is greater than what is required to remove the electron, the extra energy appears in the form of kinetic energy of the emitted photoelectron.

In the diagram below we imagine a stream of photons (shown in blue) streaming towards an electron in a metal atom. If 5 eV of energy are required to remove the electron from the atom, and each photon has 2 eV of energy, no electron is ejected.

What happens if we increase the energy of the photons to 4 eV? Since 5 eV of energy are required to remove the electron from the atom, still no electron is ejected.

How about if we increase the energy of the photons to 5 eV? Since 5 eV of energy are required to remove the electron from the atom, an electron is ejected, but it has zero kinetic energy.

And finally, if we increase the energy of the photons to 8 eV? Since 5 eV of energy are required to remove the electron from the atom, an electron is ejected, and has 3 eV of kinetic energy.

Here is an article on the photoelectric effect. Photoelectric effect

How can we find the energy of a single photon?

The energy of a photon is given by $$E = h \nu$$ where E is the energy of a single photon, h is Planck's constant (6.626 x 10-34 J s), and ν the frequency of the light in Hz.

If the energy of the incident light is known and the kinetic energy of the resulting electron can be measured, we can find the energy required to remove an electron from a particular type of atoms. It is an experimental result so we look it up in a table. It is called the work function.

A list of work functions. Work Functions

Example 1: It takes 5 eV of energy to remove an electron from a cobalt atom. If the metal is exposed to photons of light with 7 eV per photon, what will be the kinetic energy of the ejected electrons? Answer: 2 eV

Example 2: It takes 8 x 10-19 J of energy to remove an electron from a cobalt atom. If the metal is exposed to photons of light with 11.2 x 10-19 J per photon, what will be the kinetic energy of the ejected electrons? Answer: This is the same problem as above but in less convenient units. 1 eV is equivalent to 1.6 x 10-19 J. So the difference in energy is 11.2 x 10-19 J – 8 x10-19 J = 3.2 x 10-19 J

Example 3: It takes 1.95 eV of energy to remove an electron from a cesium atom. If the metal is exposed to light with a frequency of 6.0 x 1014 Hz, what will be the kinetic energy of the ejected electrons? Answer: First we convert the energy for the work function from eV into joules. $$1.95 \, eV \times \frac{1.6 \times 10^{-19} \, J}{1\, eV} = 3.12 \times 10^{-19} \, J$$ Next we calculate the energy per photon at the frequency given. \begin{align*} E & = h \nu \\\\ E & = (6.626 \times 10^{-34} \, J \cdot s)\times \left(6.0 \times 10^{14} \frac{1}{s}\right) \\\\ E & = 3.98 \times 10^{-19} \, J \end{align*} The remaining energy after the electron has been removed is $$3.98 \times 10^{-19} \, J - 3.12 \times 10^{-19} \, J = 8.56 \times 10^{-20} \, J$$

Example 4: If copper metal is exposed to violet light with a wavelength of 400 nm, will electrons be ejected ? If so, with how much kinetic energy? If not, what type of light would be required? The work function for copper is 4.7 eV. Answer: First we convert the energy for the work function from eV into joules. $$4.7 \, eV \times \frac{1.6 \times 10^{-19} \, J}{1\, eV} = 7.5 \times 10^{-19} \, J$$ Next we calculate the energy per photon at for the wavelength given. We first need to find the frequency of the light which in turn requires the wavelength in meters. $$400 \, nm \times \frac{1 \, m}{10^9 \, nm} = 4.00 \times 10^{- 7} \, m$$ To find the frequency of the light we use \begin{align*} c & = \lambda \nu \\\\ \nu & = \frac{c}{\lambda} \\\\ \nu & = 3.00 \times 10^8 \, \frac{m}{s} \times \frac{1}{4.00 \times 10^{- 7} \, m} \\\\ \nu & = 7.5 \times 10^{14} \, \frac{1}{s} \end{align*} Next we calculate the energy per photon at the frequency given. \begin{align*} E & = h \nu \\\\ E & = (6.626 \times 10^{-34} \, J \cdot s)\times \left(7.5 \times 10^{14} \frac{1}{s}\right) \\\\ E & = 4.97 \times 10^{-19} \, J \end{align*} We can see that the violet light does not have enough energy per photon to eject the electron. We need 7.5 x 10-19 J but have only 4.97 x 10-19 J available. What type of light would be required? We need light with 7.5 x 10 -19 J of energy per photon. The frequency of such light is given by \begin{align*} E & = h \nu \\\\ \nu & = \frac{E}{h} \\\\ \nu & = (7.5 \times 10^{-19} \, J ) \times \frac{1}{6.626 \times 10^{-34} \, J \cdot s} \\\\ \nu & = 1.1 \times 10^{15} \, Hz \end{align*} which is not visible but ultraviolet light.

Example 5: If potassium metal is exposed to violet light with a wavelength of 400 nm, will electrons be ejected ? If so, with what velocity? If not, what type of light would be required? The work function for copper is 2.3 eV.. Answer: In the previous question we found the energy per photon of 400 nm light to be 4.97 x 10-19J. Converting the work function into joules gives 3.7 x 10-19 J. We have enough energy per photon. The extra energy will be in the form of kinetic energy; 1.3 x 10-19 J. How can we find the velocity of a particle with this much kinetic energy? \begin{align*} KE & = \frac{1}{2} \times mv^2 \\\\ v^2 & = \frac{2 \times KE}{m} \\\\ v^2 & = \frac{2 \times (1.3 \times 10^{-19} \, J)}{9.11 \times 10^{-31} \, kg} \\\\ v^2 & = 2.8 \times 10^{11} \,\frac{m^2}{s^2} \\\\ v & = 5.3 \times 10^5 \,\frac{m}{s} \end{align*}