Collisions of Electrons with Atoms

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:

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.