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Ionisation, excitation and the electronvoltAQA A-Level Physics: Revision notes

Section 1

The electronvolt

The electronvolt (eV) is the kinetic energy gained by an electron accelerated from rest through a potential difference of 1 V. The energy gained is W=QVW = QV, so

1 eV=1.60×10−191 \text{ eV} = 1.60 \times 10^{-19} J

  • eV to J: multiply by 1.60×10−191.60 \times 10^{-19}
  • J to eV: divide by 1.60×10−191.60 \times 10^{-19}

An electron accelerated through 5.0 V gains 5.0 eV, which is 8.0×10−198.0 \times 10^{-19} J.

Key termselectronvolt
Common mistake

Convert eV to joules before using h or c in a calculation, and convert back only if the answer is asked for in eV.

Section 2

Excitation

Electrons in an atom can only have certain energies, called energy levels. In its lowest level an atom is in its ground state.

Excitation happens when an electron in an atom moves to a higher energy level. A colliding free electron can cause this by transferring energy to the atom, but only if it transfers exactly the difference between two levels.

If the colliding electron has less kinetic energy than the lowest excitation energy, no energy is transferred. The collision is elastic and the electron keeps its energy.

Key termsexcitationground stateenergy level

Section 3

Ionisation

Ionisation happens when an electron is removed completely from an atom, leaving a positive ion. The ionisation energy is the minimum energy needed to remove an electron from an atom in its ground state.

A colliding electron with kinetic energy greater than the ionisation energy can ionise an atom. The surplus energy remains as kinetic energy of the electrons.

Excited atoms return to lower levels by de-excitation, emitting a photon of energy hf=E1−E2hf = E_1 - E_2.

Key termsionisationionisation energyde-excitation

Section 4

The fluorescent tube

A fluorescent tube contains mercury vapour at low pressure.

  • A high potential difference accelerates free electrons along the tube
  • Collisions ionise some mercury atoms, giving more free electrons and ions so that the vapour conducts
  • Collisions excite mercury atoms
  • The excited atoms de-excite, emitting mostly ultraviolet photons
  • A phosphor coating absorbs the ultraviolet photons and re-emits lower-energy visible photons in several steps
Key termsfluorescent tubephosphor

Section 5

Worked example

An electron is accelerated through 5.0 V and excites a mercury atom (lowest excitation energy 4.9 eV).

  • Kinetic energy before: 5.0 eV = 5.0×1.60×10−19=8.0×10−195.0 \times 1.60 \times 10^{-19} = 8.0 \times 10^{-19} J
  • Kinetic energy after: 5.0 − 4.9 = 0.1 eV
  • Photon on de-excitation: E=4.9×1.60×10−19=7.8×10−19E = 4.9 \times 1.60 \times 10^{-19} = 7.8 \times 10^{-19} J
  • λ=hc/E=2.5×10−7\lambda = hc/E = 2.5 \times 10^{-7} m, which is ultraviolet

Must Know

  • 1 eV = 1.60 × 10⁻¹⁹ J; multiply to get joules, divide to get eV
  • Excitation moves an electron up a level; ionisation removes it
  • Atoms absorb only the exact energy differences between levels
  • De-excitation emits a photon with hf=E1−E2hf = E_1 - E_2
  • Fluorescent tube: ionisation, excitation, ultraviolet photons, phosphor, visible light

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Ionisation, excitation and the electronvolt

  1. Atomic physicists quote the energies of atoms in electronvolts, but calculations with the Planck constant use joules. The energy needed to ionise a hydrogen atom in its ground state is 13.6 eV.
    A photon has energy 4.8 × 10⁻¹⁹ J. Calculate its energy in electronvolts.2 marks
  2. A sodium vapour street lamp contains sodium atoms at low pressure. A potential difference across the lamp accelerates free electrons, which collide with the sodium atoms. The lowest excitation energy of a sodium atom is 2.1 eV and its ionisation energy is 5.1 eV.
    An electron with a kinetic energy of 1.5 eV collides with a sodium atom in its ground state. Explain why the atom is not excited.2 marks
  3. In a demonstration tube, electrons are accelerated from rest through a potential difference and then collide with mercury atoms in a vapour at low pressure. The lowest excitation energy of a mercury atom is 4.9 eV.
    An electron is accelerated from rest through a potential difference of 5.0 V and then excites a mercury atom by transferring the minimum energy possible. Calculate the kinetic energy of the electron before the collision in joules, and the kinetic energy it has after the collision in electronvolts.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).