Atom and electron collisions are processes in which electrons can ionise an atom, removing electrons from its structure or excite electrons, thus moving them from one place to another in the areas in which they occur inside the atom. The initial state of the electrons, if they are stable, is named the ground state.
Electron collision with atoms and ionisation
Electrons can also initiate a beta decay to ionise or excite the electrons, a process that depends on:
- The energy of the incident electron, which is given by its kinetic energy.
- The energy of the electrons moving around the atom.
Figure 1. When an electron or a photon impacts an atom, part of their kinetic injects energy into the electrons orbiting around the atom. This moves them to another orbit. If the energy is high enough, it ejects the electrons from the atom.
What happens when an atom and electrons collide?
Electrons can be released by the atom when a photon impacts it or when an unstable isotope breaks into pieces. Electrons that fly free with larger kinetic energies can impact another atom, leading to three possible interactions after this impact:
- Excited level states: the electron colliding with the atom gives its kinetic energy to another electron already present in the atom. The electron receiving the energy moves into an excited state and thus to a new orbit.
- Atom ionisation by knocking out other electrons. If the atom already has a negative charge, this increases its positive charge even more.
- Proton–neutron conversion, i.e., the conversion of a proton into a neutron, which can occur if the nucleus captures an electron. This process is named beta plus decay, as it releases a positron (positive electrons).
What happens when an electron collision excites another electron?
The first type of interaction between an electron colliding with an atom is the excitation of another electron. When the electron impacting the atom gives its kinetic energy to another electron that is orbiting within the atom, you can assume the electron was in the ground state.
Figure 2. Electron or photon impacting and exchanging energy with an electron in a ground state.
As the electron that lives in the atom is bound to its location with some fixed energy, the energy excess given by the other electron makes it jump to another energy level.
Figure 3. Electron jumping to an excited state after absorbing energy from the incident electron or photon.
The electron will stay in an excited state for some time. However, as the atom seeks to be stable, the electron eventually releases its excess energy as a photon. After releasing the excess energy, the electron reverts to the ground state and its initial energy level.
Figure 4. Electron jumping back from an excited state and releasing energy as a photon.
Calculating the energy levels after electron collisions
You need to use the energy conservation law to calculate the energy of an electron that jumps energy levels inside an atom after a collision with an electron. The new energy level is equal to that given by the incoming electron.
\[E_a(electron)_{kinetic} = E_b(electron)_{level} + E_{residual}\]
An electron flying with a kinetic energy of 10.2 electron volts collides with a hydrogen atom. What happens to the electron of the hydrogen after the collision?
First, we need to realise that the electron colliding with the atom will inject some energy into the electron inside the atom, making it jump to a new energy level. Its energy will be equal to that provided by the electron colliding with the atom.
\(E_a(electron)_{kinetic} = E_b(electron)_{level} + E_{residual}\)
The energy level is calculated as the difference in the atom’s energy levels. See the following table for the first three energy levels for hydrogen:
Energy level | Name | Energy level ‘n’ | Energy |
1st | Ground state | n = 1 | -13.6 [eV] |
2nd | 1st excited state | n = 2 | -3.4 [eV] |
3rd | 2nd excited state | n = 3 | -1.5 [eV] |
The energy of the incoming electron is used to move the electron up. Its energy will be equal to the difference between the two levels.
\(E_b(electron)_{level} = E_{level \space up} - E_{level \space down}\)
In this case, we subtract E2 from E1.
\(E_b(electron)_{level} = E_{n=2} - E_{n=1} = 10.2 [eV]\)
The electron in the atom will move to the energy level n = 2, which is the first excited state or the first energy level.
But what if you want to know what happens to the electron later?
The first excited state is unstable, as the atom, seeking stability, will release the excess energy in the form of a photon. The released energy of the photon will be equal to the energy gained by the electron. To calculate this, we need to use the photon’s energy equation as below.
\[Photon's \space energy = h \cdot f\]
\(10.2[eV] = h \cdot f\)
Here, f is the photon’s frequency, while h is the Planck constant: h = 6.62 ⋅ 10−34 [j] per hertz. The frequency of the photon released after the electron goes back to its ground state is calculated as follows:
\(f = \frac{10.2 [eV]}{6.62 \cdot 10^{-34}} [joules/second]\)
As one electron volt is equal to 1.6 ⋅ 10-19 [j], 10.2 [eV] are equal to \(1.63 \cdot 10^{-18}\) [j].
\(f = \frac{1.63 \cdot 10^{-18} [joules]}{6.62 \cdot 10^{-34}} [joules/second]\)
\(f = 2.46 \cdot 10^{15} [1/s]\)
And what if you want to know whether the released photon belongs to the visible light, the x-ray spectrum, UV light, radio waves, or some other form of radiation?
You will need to use the photon-wavelength-frequency relationship to work this out. This relationship relates the photon’s energy to its frequency and the Planck constant.
\[E_{photon} = h \cdot f\]
We also know that the energy of a wave is inversely proportional to its wavelength, and the photon is a type of wave.
\[E_{photon} = \frac{h \cdot c}{\lambda}\]
The photon’s energy is equal to the electron’s energy when the electron jumps back to n = 1. The energy to jump back is also equal to the energy used to move up 1.63 ⋅ 10-18 [j] or 10.2 [eV].
\(E_{photon} = 1.63 \cdot 10^{-18}[j]\)
This is equal to the wavelength-energy relationship.
\(\frac{h \cdot c}{\lambda} = 1.63 \cdot 10^{-18} [j]\)
As the light velocity (c) is 3 ⋅ 108 [m / s] and the Plank constant is h = 6.62 ⋅ 10 −34 [Joules / Hertz], we can calculate the wavelength λ as follows:
\(\lambda = \frac{(6.62 \cdot 10^{-34} [j/Hz]) \cdot (3 \cdot 10^8 [m/s])}{1.63 \cdot 10^{-18}[j]}\)
\(\lambda = 1.22 \cdot 10^{-7}[m]\)
The length of this wavelength is around 0.12 micrometres. Looking at the electromagnetic spectrum, you find that the range for UV light is from around 0.01 [micrometres] to 0.38 [micrometres]. The released photon is within this range, which tells us that the photon released after the electron reverts to its ground state is ultraviolet light.
What happens to the electron that caused the other electron to jump? The electron that gave the energy ends up with a velocity of 0.
Wavelengths and frequencies for common types of photons
Photons can be classified according to their frequency ‘f’ (energy) and their respective wavelength ‘λ’. See the following table, which displays the values for a range of photons.
Photon | Wavelength (\(\lambda\)) | Frequency (hertz) |
Gamma rays | 0.001 | \(4.42 \cdot 10^{-30}\) |
| 1 | \(4.42 \cdot 10^{-33}\) |
| 100 | \(4.42 \cdot 10^{-35}\) |
Dark blue (visible) | 442-663 | \(1 \cdot 10^{-35} - 6.98 \cdot 10^{-36}\) |
Infrared | 1000 | \(4.42 \cdot 10^{-36}\) |
A good way to convert between the photon frequency and wavelength is the equation below, where c is the speed of light in a vacuum.
\[f = \frac{c}{\lambda}\]
Collisions of Electrons with Atoms - Key takeaways
- Electrons and photons can impact atoms.
- The impact of electrons and photons injects energy into the electrons orbiting the atom.
- The energy of the electrons changes, exciting them and making them jump to another orbit or even ejecting them.
- Electrons that are not excited are said to be in their ground state.
- When the electrons go back to their ground state, they release energy in the form of a photon.
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