41.6.2 Collisional Excitation

If a particle, such as an electron, with more than 2.10 eV of kinetic energy collides with a ground-state sodium atom, a portion of its energy can be used to excite the atom to its $3 p$ state. This process is called collisional excitation of the atom.

Collisional excitation differs from excitation by absorption in one very fundamental way. In absorption, the photon disappears. Consequently, all of the photon’s energy must be transferred to the atom. Conservation of energy requires $E_{photon} = Δ E_{atom}$ . In contrast, the particle is still present after collisional excitation and can carry away some kinetic energy. That is, the particle does not have to transfer its entire energy to the atom. If the particle has, say, an incident energy of 2.85 eV, it could transfer 2.10 eV to the sodium atom, thereby exciting it to the $3 p$ state, and depart the collision with an energy of 0.75 eV.

To excite the atom, the incident energy of the particle merely has to exceed $Δ E_{atom}$ . That is, $E_{particle} \geq Δ E_{atom}$ . There’s a threshold energy for exciting the atom, but no upper limit. It is all a matter of energy conservation. FIGURE 41.24 shows the idea graphically.

The figure consists of two parts that illustrates the collisional excitation by electrons.

FIGURE 41.24 Full Alternative Text

Collisional excitation by electrons is the predominant method of excitation in electrical discharges such as fluorescent lights, street lights, and neon signs. A gas is placed in a tube at reduced pressure $(\approx 1 mm of Hg)$ , then a fairly high voltage $(\approx 1000 V)$ between electrodes at the ends of the tube causes the gas to ionize, creating a current in which both ions and electrons are charge carriers. The mean free path of electrons between collisions is large enough for the electrons to gain several eV of kinetic energy as they accelerate in the electric field. This energy is then transferred to the gas atoms upon collision. The process does not work at atmospheric pressure because the mean free path between collisions is too short for the electrons to gain enough kinetic energy to excite the atoms.

NOTE There are no selection rules for collisional excitation. Any state can be excited if the colliding particle has sufficient energy.

EXAMPLE 41.7 Excitation of hydrogen

Can an electron traveling at $2.0 \times 10^{6} m / s$ cause a hydrogen atom to emit the prominent red spectral line $(λ = 656 nm)$ in the Balmer series?

MODEL The electron must have sufficient energy to excite the upper state of the transition.

SOLVE The electron’s energy is $E_{elec} = \frac{1}{2} m v^{2} = 11.4 eV$ . This is significantly larger than the 1.89 eV energy of a photon with wavelength 656 nm, but don’t confuse the energy of the photon with the energy of the excitation. The red spectral line in the Balmer series is emitted by an $n = 3$ to $n = 2$ quantum jump with $Δ E_{atom} = 1.89 eV$ . But to cause this emission, the electron must excite an atom from its ground state, with $n = 1$ , up to the $n = 3$ level. The necessary excitation energy is

\begin{matrix} Δ E_{atom} & = & E_{3} - E_{1} = (- 1.51 eV) - (- 13.60 eV) \\ = & 12.09 eV \end{matrix}

The electron does not have sufficient energy to excite the atom to the state from which the emission would occur.