Electrons in atoms exist in discrete energy levels (quantised states). They cannot have
Quantum Physics - OCR A-Level Physics
- Electrons in atoms exist in discrete energy levelsThe discrete allowed energies that electrons can have in an atom. Transitions between levels emit/absorb photons. (quantisedExisting only in discrete values. ChargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). is quantised because it can only exist as whole-number multiples of the elementary chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). e. states). They cannot have energies between these levels.
- The ground stateThe lowest 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. of an atom. The state in which all electrons are in their lowest possible energyThe capacity to do work. Measured in joules (J). levels. is the lowest energy levelA discrete amount of energy that an electron in an atom can have. Electrons can only exist at specific energy levels, not between them. (n = 1). Higher levels are called excited statesEnergy levels above the ground state. An electron in an excited state will eventually drop to a lower level, emitting a photon..
- Energy levels are negative (measured relative to ionisation $energy = 0 eV). The ground$ state has the most negative energy.
- Ionisation energyThe minimum energy required to remove an electron completely from an atom in its ground stateThe lowest energy level of an atom. The state in which all electrons are in their lowest possible energy levels. to infinity. is the minimum energy needed to completely remove an electron from the atom (from the ground stateThe lowest energy level of an atom. The state in which all electrons are in their lowest possible energy levels. to $E = 0)$.
$$hf = E_{1} - E_{2}$$
- Emission spectrumA series of bright lines at specific wavelengths produced when excited atoms return to lower energy levels, emitting photons.: an electron drops from a higher to a lower energy levelA discrete amount of energy that an electron in an atom can have. Electrons can only exist at specific energy levels, not between them. and emits a photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. with energy equal to the gap. Produces bright lines on a dark background.
- Absorption spectrumA continuous spectrum with dark lines at specific wavelengths where photons have been absorbed by atoms, exciting electrons to higher energy levels.: a photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. with energy exactly matching an energy level gap is absorbed, promoting an electron to a higher level. Produces dark lines on a continuous background.
- Each element has a unique set of energy levels, so emission and absorption spectra act as fingerprintsEach element has a unique set of spectral lines (emission/absorption), acting as a 'fingerprint' to identify it. for identifying elements.
- The absorption lines appear at exactly the same wavelengths as the emission lines for a given element.
Worked Example [4 marks]
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 photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency..
Show Solution
1
Energy of photon
$\DeltaE = E_{1} - E_{2} = (-1.51) - (-3.40) = 1.89 eV$
[1]2
Convert to joules
$E = 1.89 \times 1.6 \times 10⁻¹⁹ = 3.024 \times 10⁻¹⁹ J$
[1]3
$\lambda = hc/E = (6.63 \times 10⁻^{3}⁴ \times 3.00 \times 10⁸) / (3.024 \times 10⁻¹⁹)$
[1]4
$\lambda = 6.58 \times 10⁻⁷\;\text{m} = 658 nm (red light - the hydrogen H-alpha line)$
[1]Answer
\lambda = 658 nm (red light)
Common Mistake
MEDIUM
Students often: Getting the sign wrong when subtracting energy levels. Writing (−1.51) − (−3.40) = −4.91 eV.
Instead: Be careful with double negatives: (−1.51) − (−3.40) = −1.51 + 3.40 = +1.89 eV. The photon energy must always be positive.
Instead: Be careful with double negatives: (−1.51) − (−3.40) = −1.51 + 3.40 = +1.89 eV. The photon energy must always be positive.