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Work Function Formula

Modern physics formulasClass 12AP Physics 2Updated 1 Nov 2026

ϕ0=hν0=hcλ0\phi_0 = h\nu_0 = \frac{hc}{\lambda_0}
SI unit
J (often quoted in eV)
Dimensions
[M¹ L² T⁻²]
where
  • φ₀ work function of the metal surface J
  • h Planck's constant, 6.626 × 10⁻³⁴ J s J s
  • ν₀ threshold frequency Hz
  • c speed of light, 3.00 × 10⁸ m/s m/s
  • λ₀ threshold wavelength m

The work function formula is ϕ0=hν0=hcλ0\phi_0 = h\nu_0 = \frac{hc}{\lambda_0}: the work function of a metal equals Planck's constant times the threshold frequency, the smallest light frequency able to just pull an electron out of the surface. Light below that frequency frees no electrons, however bright it is.

When the light is above threshold, the extra photon energy becomes kinetic energy of the electron. That gives the second form most exam questions use, ϕ0=hν−Kmax=hν−eV0\phi_0 = h\nu - K_{max} = h\nu - eV_0, where V₀ is the stopping potential. The topic sits in the Class 12 chapter on photons and matter waves and the photoelectric effect in AP Physics 2. See all the related results on the modern physics formula sheet.

Quick facts

  • Formula

    ϕ0=hν0=hcλ0\phi_0 = h\nu_0 = \frac{hc}{\lambda_0}

  • From the photoelectric equation

    ϕ0=hν−eV0\phi_0 = h\nu - eV_0

  • SI unit

    joule (J); quoted in electronvolts, 1 eV = 1.602 × 10⁻¹⁹ J

  • Dimensional formula

    [M¹ L² T⁻²], the same as energy

  • Shortcut

    ϕ0 (eV)=1240λ0 (nm)\phi_0\,(\text{eV}) = \frac{1240}{\lambda_0\,(\text{nm})}

  • Common symbols

    φ₀, W₀ or W

  • Depends on

    the metal and its surface, not on the light

Meaning of each symbol

Quantities in φ₀ = hν₀ = hc/λ₀
SymbolWhat it isSI unit
ϕ0\phi_0Minimum energy to remove an electron from the surfaceJ
hhPlanck's constant, 6.626 × 10⁻³⁴J s
ν0\nu_0Threshold frequency, the cut off for emissionHz
λ0\lambda_0Threshold wavelength, the longest wavelength that worksm
KmaxK_{max}Kinetic energy of the fastest emitted electronsJ
V0V_0Stopping potential that halts those electronsV

Threshold frequency and threshold wavelength are tied by c=ν0λ0c = \nu_0\lambda_0, so a high work function means a high threshold frequency and a short threshold wavelength. Photon energy itself is E=hνE = h\nu; see the energy formula page for other forms of energy and the wavelength formula page for c=νλc = \nu\lambda.

Where the work function formula comes from

The photoelectric equation for the effect is an energy balance for one photon and one electron:

hν=ϕ0+Kmaxh\nu = \phi_0 + K_{max}

The photon brings energy hν. Part of it, φ₀, is spent getting the electron through the surface. The rest is the largest kinetic energy an emitted electron can have; electrons from deeper inside lose a little more and come out slower. At the threshold the electron just escapes with zero kinetic energy, so hν0=ϕ0h\nu_0 = \phi_0. The stopping potential measures K directly, because an electron that climbs through a potential difference V₀ loses energy eV₀, so Kmax=eV0K_{max} = eV_0.

  • Below ν₀: no emission at any intensity.
  • At ν₀: emission starts with zero kinetic energy.
  • Above ν₀: K_max rises in a straight line with ν.

Worked examples

Example 1: from threshold wavelength

The threshold wavelength of a metal is 540 nm. Find its work function in joules and electronvolts.

Given: λ₀ = 540 nm = 540 × 10⁻⁹ m.

ϕ0=6.626×10−34×3.00×108540×10−9=3.68×10−19 J\phi_0 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{540 \times 10^{-9}} = 3.68 \times 10^{-19}\ \text{J}

In eV: 3.68×10−191.602×10−19=2.30 eV\frac{3.68 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.30\ \text{eV}. The shortcut 1240 / 540 gives 2.30 eV as well.

Example 2: from stopping potential

Ultraviolet light of wavelength 250 nm falls on a metal and the stopping potential is 0.80 V. Find the work function.

Photon energy: E=1240250=4.96 eVE = \frac{1240}{250} = 4.96\ \text{eV}.

ϕ0=E−eV0=4.96−0.80=4.16 eV≈4.2 eV\phi_0 = E - eV_0 = 4.96 - 0.80 = 4.16\ \text{eV} \approx 4.2\ \text{eV}

In joules that is 4.16 × 1.602 × 10⁻¹⁹ = 6.67 × 10⁻¹⁹ J.

Example 3: all in SI units

Light of frequency 1.2 × 10¹⁵ Hz ejects electrons with maximum kinetic energy 2.4 × 10⁻¹⁹ J. Find φ₀.

ϕ0=hν−Kmax=6.626×10−34×1.2×1015−2.4×10−19\phi_0 = h\nu - K_{max} = 6.626 \times 10^{-34} \times 1.2 \times 10^{15} - 2.4 \times 10^{-19}

ϕ0=7.95×10−19−2.4×10−19=5.55×10−19 J≈3.5 eV\phi_0 = 7.95 \times 10^{-19} - 2.4 \times 10^{-19} = 5.55 \times 10^{-19}\ \text{J} \approx 3.5\ \text{eV}

Enter these values in the calculator above to check the answer.

Typical work function values

Approximate work functions of clean metals
Metalφ₀ (eV)λ₀ (nm)
Cesiumabout 2.1about 590
Potassiumabout 2.3about 540
Calciumabout 3.2about 390
Copperabout 4.7about 260
Platinumabout 5.6about 220

Measured values shift by a few tenths of an eV with surface cleanliness and crystal face, so different books quote slightly different numbers. Use the value printed in the question. Metals with a low work function, such as the alkali metals, respond to visible light. Most others need ultraviolet. The value has the units of energy because it is an energy; the dimensional formula of Planck's constant times that of frequency gives [M¹ L² T⁻²].

Graph of stopping potential against frequency

Rewrite the equation as V0=heν−ϕ0eV_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}. A plot of V₀ against ν is a straight line with:

  • Slope h/e, the same for every metal, about 4.14 × 10⁻¹⁵ V s.
  • Intercept on the frequency axis equal to ν₀.
  • Intercept on the V₀ axis (extended back) equal to −φ₀/e.

Lines for different metals are parallel. The one further to the right belongs to the metal with the larger work function. Questions often hand you two points on such a line and ask for h and φ₀: use the slope for h and the frequency intercept for the threshold.

Errors to avoid

  • Mixing eV and J. If hν is in joules, φ₀ must be in joules too. Convert with 1 eV = 1.602 × 10⁻¹⁹ J.
  • Forgetting nm to m. 540 nm is 5.40 × 10⁻⁷ m, not 5.40 × 10⁻⁹ m.
  • Thinking intensity matters. Brighter light gives more electrons per second, not a higher K_max, and it does not change φ₀.
  • Using the 1240 shortcut with metres. It needs λ in nm and returns eV.
  • Negative kinetic energy. If hν − φ₀ comes out negative, no electrons are emitted; do not report a negative speed.
Sanity check: work functions of metals sit between about 2 eV and 6 eV. An answer far outside that range points to a conversion slip.

Frequently asked questions

What is the formula of work function?

The work function is φ₀ = hν₀ = hc / λ₀, where ν₀ is the threshold frequency and λ₀ the threshold wavelength. From an experiment you can also use φ₀ = hν − eV₀.

What is the SI unit of work function?

Its SI unit is the joule, since it is an energy. In practice it is almost always given in electronvolts, and 1 eV = 1.602 × 10⁻¹⁹ J.

Does the work function depend on the frequency of light?

No, because it belongs to the metal surface itself. Changing the frequency changes the kinetic energy of the emitted electrons, not the energy needed to free them.

What is the formula for work function in terms of threshold wavelength?

φ₀ = hc / λ₀. With λ₀ in nanometres, φ₀ in eV is approximately 1240 / λ₀.

What is the difference between work function and threshold frequency?

The work function is an energy, in J or eV. The threshold frequency is the light frequency whose photons carry exactly that energy, ν₀ = φ₀ / h, measured in Hz.