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c059741
NotesPhysics HLTopic 5.1The electronvolt
Back to Physics HL Topics
5.1.32 min read

The electronvolt (Physics HL)

IB Physics • Unit 5

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Contents

  • What the electronvolt is
  • Converting between eV and joules
  • Exam-style question
The big idea: You wouldn't price a single sweet in millions of dollars — the unit would be absurdly big for the job. In the same way the joule (J) is far too big for the energy of one particle, 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.

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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

E (J) on top, E (eV) and 1.60 × 10⁻¹⁹ below. Cover the one you want: two side by side → multiply; one above the other → divide.

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IB-style questionCalculate[2 marks]

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

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IB-style questionCalculate[2 marks]

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

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

Paper 1A

  • Photon and transition energies are quoted in eV — convert to joules before using E = hf or E = hc/λ.

Paper 2

  • A part often asks you to state, in eV, the energy of a particle, or quotes a nuclear energy in MeV.
The 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 questionCalculate[3 marks]

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.)

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the electronvolt. [2 marks]

Related Physics HL Topics

Continue learning with these related topics from the same unit:

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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