3.2.2.3
Photons are emitted when electrons drop to lower energy levels
Energy Levels & Wave-Particle Duality — AQA A-Level Physics
- An excited electron is unstable. It quickly drops to a lower energy levelA discrete amount of energyThe capacity to do work. Measured in joules (J). that an electron in an atom can have. Electrons can only exist at specific energyThe capacity to do work. Measured in joules (J). levels, not between them..
- The energyThe capacity to do work. Measured in joules (J). lost equals the gap between the two levels: $\Delta E = E_2 - E_1$.
- This energy is emitted as a single photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency..
$$\begin{aligned}
\Delta E &= hf \\
&= \frac{hc}{\lambda}
\end{aligned}$$
- $ΔE$: energy difference between the two levels (J)
- $h$: Planck's constant (J s)
- $f$: frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). of emitted photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. (Hz)
- $λ$: wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). of emitted photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. (m)
- Larger energy gaps produce higher frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). (shorter wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).) photons.
- Transitions to the ground stateThe lowest energy level of an atom. The state in which all electrons are in their lowest possible energy levels. produce UV photons. Transitions between excited states can produce visible or IR photons.
- The reverse process: a photon can be absorbed if its energy exactly matches the gap between two levels, exciting the electron upwards.
Worked Example
An electron in a hydrogen atom drops from the $n = 3$ level ($-1.51$ eV) to the $n = 2$ level ($-3.40$ eV). Calculate the wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). of the emitted photon.
Show Solution
1
Calculate the energy gap
$$\Delta E = (-1.51) - (-3.40) = 1.89 \text{ eV}$$
2
Convert to joules
$$\Delta E = 1.89 \times 1.60 \times 10^{-19} = 3.02 \times 10^{-19} \text{ J}$$
3
Calculate wavelength
$$\lambda = \frac{hc}{\Delta E} = \frac{(6.63 \times 10^{-34}) \times (3.0 \times 10^{8})}{3.02 \times 10^{-19}}$$
$$= 6.59 \times 10^{-7} \text{ m} = 659 \text{ nm}$$Answer
$\lambda = 659$ nm (red light — this is the H-alpha line in the Balmer series)
Common Mistake
MEDIUM
Students often: Getting the sign wrong on the energy gap: writing $\Delta E = - 1.51 - (- 3.40) = - 4.91 eV$.
Instead: ΔE is always positive (energy is released). Take the higher level minus the lower level: (−1.51) − (−3.40) = +1.89 eV. Or use |E₂ − E₁|.
Instead: ΔE is always positive (energy is released). Take the higher level minus the lower level: (−1.51) − (−3.40) = +1.89 eV. Or use |E₂ − E₁|.