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v0.1.1065
NotesPhysics HLTopic 5.1The electronvolt
Back to Physics HL Topics
5.1.32 min read

The electronvolt

IB Physics • Unit 5

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Contents

  • What the electronvolt is
  • Converting between eV and joules
  • Exam-style question
The big idea: The joule (J) is huge for a single particle. Atomic and nuclear energies are tiny fractions of a joule, so physicists use a smaller unit: the electronvolt (eV).

One electronvolt is the energy gained by one electron (charge e) when it is pushed through a potential difference of one volt.

That works out to a fixed amount of energy:

1 eV = 1.60 × 10⁻¹⁹ J
Spot it: The eV is just a unit of energy, like the joule — not a new kind of energy.

The number 1.60 × 10⁻¹⁹ is the elementary charge e in coulombs. That is no coincidence: energy = charge × voltage, so e coulombs × 1 volt = 1.60 × 10⁻¹⁹ J.
Bigger eV units: Like metres → kilometres, the eV has bigger cousins:

- 1 keV (kilo-electronvolt) = 10³ eV - 1 MeV (mega-electronvolt) = 10⁶ eV

Atomic-transition energies are a few eV; nuclear (decay, binding) energies are a few MeV.

Everything rests on one given number: 1 eV = 1.60 × 10⁻¹⁹ J. To go each way:

The two directions

  • eV → J: multiply by 1.60 × 10⁻¹⁹ (you have many eV, each worth a tiny number of joules)
  • J → eV: divide by 1.60 × 10⁻¹⁹ (you are counting how many eV fit into the energy)
The conversion is given in the data booklet (Unit conversions). Multiply electronvolts by 1.60 × 10⁻¹⁹ to get joules; divide to go back.
the energy expressed in joules (J)
the same energy expressed in electronvolts (eV)
the joules in one electronvolt — the elementary charge e in coulombs

[Diagram: phys-formula-triangle] - Available in full study mode

Worked example — eV to joules

An electron in an atom drops between two energy levels and emits a photon of energy 2.5 eV. Express this energy in joules.

Solution

  1. Going eV → J, so use the given conversion:
  2. Put in 2.5 eV:
  3. Work it out — keep the unit:

Final answer

E = 4.0 × 10⁻¹⁹ J. (A few eV is a tiny fraction of a joule — that is why we use the eV.)

Worked example — joules to MeV

A nuclear decay releases 8.0 × 10⁻¹³ J of energy. Express this in MeV.

Solution

  1. Going J → eV, so divide by the given conversion:
  2. Put in 8.0 × 10⁻¹³ J:
  3. Convert eV → MeV (divide by 10⁶):

Final answer

E = 5.0 MeV — a typical nuclear-decay energy.

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How this is tested: The electronvolt rarely gets its own question — instead it is the unit you answer in across Theme E and the fields topics.

- Paper 1A: photon energies and transition energies are quoted in eV; you may need to convert to joules before using E = hf or E = hc/λ. - Paper 2: a part often says 'state, in eV' the energy of a particle, or quotes a nuclear energy in MeV.

Classic trap: multiplying when you should divide. eV → J multiply by 1.60 × 10⁻¹⁹; J → eV divide.
Convert first, then use the formula: Planck's equation E = hf gives the energy in joules. If the answer is wanted in eV, work in joules, then divide by 1.60 × 10⁻¹⁹ at the end.
Both given in the data booklet (E.1). Photon energy comes out in joules; convert to eV at the end if asked.
energy of the photon (J)
Planck constant, 6.63×10⁻³⁴ J s (given)
frequency of the light (Hz)
speed of light, 3.00×10⁸ m s⁻¹ (given)
wavelength of the light (m)

IB-style question — photon energy in eV

An atom emits a photon of light of wavelength 4.4 × 10⁻⁷ m (violet light). Find the energy of the photon in electronvolts. (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹.)

Solution

  1. The wavelength is given, so use the given photon equation:
  2. Put in the numbers (this gives the energy in joules):
  3. Work out the energy in joules:
  4. Now convert J → eV by dividing by 1.60 × 10⁻¹⁹:

Final answer

E ≈ 2.8 eV. (Visible-light photons carry a few eV — right for an atomic transition.)

Try an IB Exam Question — Free AI Feedback

Test yourself on The electronvolt. Write your answer and get instant AI feedback — just like a real IB examiner.

An alpha particle (charge +2e) is accelerated from rest through a potential difference of 1.5 × 10³ V between two parallel plates.

, in electronvolts, the kinetic energy it gains.
[1 mark]

Related Physics HL Topics

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5.1.1Nuclear model and atomic structure
5.1.2Energy levels and atomic spectra
5.1.4Quantisation of charge
5.1.5Properties of nuclei and high-energy scattering (HL)
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