(d) the equations ΔE = hf and ΔE = hc/λ

The idea

These two equations convert between the energy gap of a transition and the electromagnetic radiation produced or absorbed. ΔE = hf gives the photon energy from its frequency, where h is the Planck constant, about 6.63 × 10⁻³⁴ J s. Because for light in a vacuum c = fλ, substituting f = c/λ gives ΔE = hc/λ, which is useful when a wavelength is known.

For example: A photon of wavelength 656 nm has energy hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (656 × 10⁻⁹), which is about 3.03 × 10⁻¹⁹ J. Convert nanometres to metres before substituting, and remember that ΔE is the difference between two energy levels, so it is positive for an emitted photon.

What earns the marks

• States ΔE = hf, identifying ΔE as the photon energy and f as its frequency.
• States ΔE = hc/λ, identifying λ as the wavelength in metres and c as the speed of light in a vacuum.
• Uses c = fλ to show how the two forms are equivalent.
• Substitutes values with consistent SI units, converting nm to m by multiplying by 10⁻⁹.
• Calculates photon energy accurately, for example about 3.03 × 10⁻¹⁹ J for 656 nm.
• Relates the calculated ΔE to the difference between two discrete energy levels.

Common mistakes and how to put them right

• Substituting a wavelength in nanometres directly into hc/λ.

  Put it right: Convert to metres first, since 656 nm is 656 × 10⁻⁹ m.

• Using the frequency of the light as the frequency of the electron or vice versa.

  Put it right: f in ΔE = hf is the photon frequency.

• Treating ΔE as always the full ionisation energy.

  Put it right: ΔE is the difference between the two levels involved in that particular transition.

• Mixing up h and c values or their powers of ten.

  Put it right: Write both constants with units before substituting to catch slips.

Exam tips

• Write the equation, then substitute numbers with units, then evaluate, so unit conversions are visible to the examiner.
• If the answer is needed in eV, divide the joule value by 1.60 × 10⁻¹⁹ J eV⁻¹.
• Check the order of magnitude: visible photons have energies of a few times 10⁻¹⁹ J, so a wildly different value signals a conversion error.