Is this formula given in the exam?

No. SPM Physics papers provide no formula sheet, so this must be recalled.

Why does increasing the light intensity not raise the electrons' kinetic energy?

Intensity increases the number of photons, so more electrons are emitted (larger current), but each photon still carries energy hf. The maximum kinetic energy hf − W depends only on frequency, not brightness.

What does the maximum in ½mv²max mean?

Electrons deeper in the metal lose extra energy escaping, so they emerge slower. Only electrons at the surface get the full hf − W, giving the maximum kinetic energy.

How is the threshold frequency linked to this equation?

At the threshold frequency f₀, the kinetic energy is zero, so hf₀ = W. Below f₀ the photon energy is less than W and no electrons are emitted.

Einstein's photoelectric equation Formula, SPM Physics

Einstein's photoelectric equation

hf = W + ½mv²max

Not given in the exam, recall it.

What it is for

Einstein's photoelectric equation is used in Quantum Physics (SPM Physics Form 5). Keep the SI units in every line of working. Quantum Physics

Symbols and SI units

Symbol	Meaning	SI unit
`h`	Planck constant	J s
`f`	frequency of incident light	Hz
`W`	work function	J
`m`	mass of electron	kg
`v_max`	maximum speed of photoelectron	m s⁻¹

Rearrangements

½mv²max = hf − W
W = hf − ½mv²max
hf = W + KEmax

Worked example

Photon energy hf = 5.0 × 10⁻¹⁹ J strikes a metal of work function W = 3.0 × 10⁻¹⁹ J. ½mv²max = hf − W = 5.0 × 10⁻¹⁹ − 3.0 × 10⁻¹⁹ = 2.0 × 10⁻¹⁹ J (maximum kinetic energy).

Common trap

The maximum kinetic energy of the photoelectron equals hf − W. Emission only occurs if hf ≥ W (that is, f ≥ f₀).

Understanding Einstein's photoelectric equation

The equation hf = W + ½mv²max is an energy balance for the photoelectric effect. hf is the incident photon energy in joules (J); W is the work function in joules (J); ½mv²max is the maximum kinetic energy of the emitted electron, where m is the electron mass in kilograms (kg) and v_max its maximum speed in metres per second (m s⁻¹).

It says one photon gives all its energy to one electron: part releases it (W) and the rest becomes kinetic energy. Energy is conserved, so no more can appear than the photon supplied.

Rearrange it as ½mv²max = hf − W to find the maximum kinetic energy, or W = hf − ½mv²max to find the work function from measured data.

A worked example, step by step

A photon of energy hf = 6.0 × 10⁻¹⁹ J strikes a metal of work function W = 3.6 × 10⁻¹⁹ J. Find the maximum speed of the emitted electron (m = 9.11 × 10⁻³¹ kg).

First find the maximum kinetic energy:

½mv²max = hf − W = 6.0 × 10⁻¹⁹ J − 3.6 × 10⁻¹⁹ J = 2.4 × 10⁻¹⁹ J

Now solve for v_max:

v²max = 2 × 2.4 × 10⁻¹⁹ J ÷ 9.11 × 10⁻³¹ kg = 5.27 × 10¹¹ m² s⁻²

v_max = √(5.27 × 10¹¹) = 7.26 × 10⁵ m s⁻¹ to three significant figures.

Common mistakes and how this is tested

Remember that ½mv²max is the maximum kinetic energy of the photoelectron, equal to hf − W. Emission only happens when hf ≥ W, that is when f ≥ f₀.

Frequent errors:

Forgetting to double before dividing when solving v²max = 2(hf − W) / m.
Forgetting the final square root, leaving v²max as the answer.
Adding W to the kinetic energy instead of subtracting it.

Under explain, state that increasing frequency raises the maximum kinetic energy while increasing intensity only raises the current. Under calculate, show each substitution line with units.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English), Sijil Pelajaran Malaysia: Format Pentaksiran mulai 2021, Fizik (4531) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Written by the spmphysics.com.my editorial team.· Updated 5 Sept 2026