Is this formula given in the exam?

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

Does a brighter light source give photons with more energy?

No. Brightness increases the number of photons per second, but the energy of each photon depends only on frequency through E = hf. A dim blue lamp emits higher-energy photons than a bright red one.

What value of the Planck constant should I use?

Use h = 6.63 × 10⁻³⁴ J s. Because there is no formula sheet in the paper, memorise this value along with the equation E = hf.

How do I convert a photon energy in joules to electronvolts?

Divide the joule value by 1.6 × 10⁻¹⁹, since 1 eV = 1.6 × 10⁻¹⁹ J. For example, 3.98 × 10⁻¹⁹ J becomes about 2.49 eV.

Photon energy Formula, SPM Physics

Photon energy

E = hf

Not given in the exam, recall it.

What it is for

Photon energy 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
`E`	photon energy	J
`h`	Planck constant	J s
`f`	frequency	Hz

Rearrangements

f = E / h
h = E / f

Worked example

Light of frequency f = 5.0 × 10¹⁴ Hz (h = 6.63 × 10⁻³⁴ J s). E = hf = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.32 × 10⁻¹⁹ J.

Common trap

Use h = 6.63 × 10⁻³⁴ J s and f in hertz; the answer is in joules. Divide by 1.6 × 10⁻¹⁹ to convert to eV.

Understanding photon energy

The equation E = hf gives the energy carried by a single photon of light. Here E is the photon energy in joules (J), h is the Planck constant measured in joule-seconds (J s), and f is the frequency of the light in hertz (Hz).

The idea that light comes in discrete packets is the heart of quantum theory: energy rises in direct proportion to frequency, so blue light (higher f) carries more energy per photon than red light. Brightness only changes the number of photons, not the energy each one carries.

You can rearrange the relationship in two ways: f = E / h to find the frequency from a known energy, and h = E / f if an experiment measures both E and f. Use h = 6.63 × 10⁻³⁴ J s throughout.

A worked example, step by step

Find the energy of a photon of green light whose frequency is f = 6.0 × 10¹⁴ Hz. Take h = 6.63 × 10⁻³⁴ J s.

Start from the formula:

E = hf

Substitute the values with their units:

E = 6.63 × 10⁻³⁴ J s × 6.0 × 10¹⁴ Hz

Multiply the numbers and add the powers of ten:

E = 3.978 × 10⁻¹⁹ J

Rounded to three significant figures, the photon energy is E ≈ 3.98 × 10⁻¹⁹ J. Because a hertz is one per second, J s × Hz gives joules, which confirms the unit of the answer is correct.

Common mistakes and how this is tested

The most common slip is using the wrong value of h. Always take h = 6.63 × 10⁻³⁴ J s and keep f in hertz, so the answer emerges in joules.

To convert to electronvolts, divide the joule value by 1.6 × 10⁻¹⁹.

Two more errors to avoid: forgetting to add the exponents when multiplying powers of ten, and confusing frequency with wavelength. This formula needs frequency directly.

Do not assume brighter light means higher energy per photon.
Do not drop the negative sign in 10⁻³⁴.

Under the command word calculate you must show the substitution line. Under define, state that photon energy is the energy of one quantum of radiation, equal to hf.

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