Energy levels in atoms

Instead of orbiting freely, electrons occupy specific discrete energy levels around the nucleus. These levels are given integers:

• n = 1 is the ground state - the lowest possible energy level
• Higher n values are higher energy excited states

Since energy levels are quantised, electrons can only occupy or jump between fixed allowed states.

Photon absorption and emission in transitions

Transitions between energy levels involve photons:

• Absorption - Electrons jump to higher levels by absorbing photon energy
• Emission - Electrons falling to lower levels and emitting photons

The photon energy equals the difference in energy between levels:

$ \Delta \text{E = E}_{\text{f}}\text{ - E}_{\text{i} }$

Where:

ΔE = photon energy (J)

E_f = final energy level (J)

E_i = initial energy level (J)

Worked example - Calculating photon frequency.

Calculate the frequency of a photon which causes an electron to move from the ground state to energy level n=3.

Step 1: Calculate Photon energy

$\Delta\text{E = E}_\text{f} -E_\text{i}$

$\Delta\text{E = -1.5 - -13.6 = 12.1 eV}$

Step 2: convert eV to J

To convert from eV to J, multiply by 1.6 x 10$^{-19}$

12.1 eV = 1.936 x 10$^{-18}$ J

Step 3: Calculate photon frequency

$\text{f = }\frac{\text{E}}{\text{h}}$

$\text{f = }\frac{\text{1.936 x 10}^{-18}}{\text{6.63 x 10}^{-34}}\text{ = 2.92 x 10}^{15}\text{ Hz}$

Ionisation energies

Ionisation removes an electron completely from an atom.

The ionisation energy is the minimum photon energy required to remove an electron from an atom in its ground state.

Worked example - Calculating ionisation energy.

Calculate the photon frequency needed to remove an electron from the ground state (-13.6 eV).

Step 1: calculate energy required to remove the electron.

$\Delta\text{E = E}_\text{f}-\text{E}_\text{i}$

$\Delta\text{E = 0}-\text{-13.6 = 13.6 eV}$

Step 2: Convert eV to J

to convert from eV to J, multiply by 1.6 x 10$^{-19}$

13.6 eV = 2.176 x 10$^{-18}$ J

Step 3: Calculate photon frequency

$\text{f = }\frac{\text{E}}{\text{h}}$

$\text{f = }\frac{2.176 \times10^{-18}}{6.63\times10^{-34}}\text{ = 3.28 x 10}^{15}\text{ Hz}$

Fluorescent tubes

Fluorescent tubes contain mercury vapour which can be excited by accelerated electrons.

How it works:

1. Electrons are accelerated by an applied voltage across the ends of the tube.
2. When electrons collide with the mercury vapour, it causes electrons in the vapour to be promoted to a higher energy level.
3. When these excited electrons fall back to a lower energy level, they emit photons in the UV region of the EM spectrum.
4. The fluorescent coating of the tube absorbs these UV photons causing electrons in the coating to become excited.
5. When electrons in the coating de-excite, photons in the visible region are emitted.