How Is the Frequency of an Emitted Photon Related to the Energy Level Change (E = hf)?

Direct Answer

The frequency of an emitted photon is directly proportional to the magnitude of the atom's energy-level change: $f = \frac{|\Delta E|}{h}$. A larger drop in energy produces a higher-frequency photon.

The Reasoning

This is part of E.1 Structure of the atom for both SL and HL. Atomic electrons occupy discrete energy levels, so an electron cannot have an energy between the allowed values.

When an electron moves from a higher energy level $E_i$ to a lower energy level $E_f$, the atom loses energy. That energy is emitted as one photon:

$$E_{\text{photon}} = hf = E_i-E_f = |\Delta E|$$

Here, $h$ is the Planck constant, $6.63 \times 10^{-34}\ \text{J s}$, and $f$ is the photon's frequency in hertz. Although the atom's energy change $\Delta E = E_f-E_i$ is negative during emission, the photon energy is positive, so its magnitude is used.

For example, suppose the energy-level drop is $3.00\ \text{eV}$. First convert electronvolts to joules:

$$|\Delta E|=(3.00)(1.60\times10^{-19})=4.80\times10^{-19}\ \text{J}$$

Then:

$$f=\frac{4.80\times10^{-19}}{6.63\times10^{-34}}=7.24\times10^{14}\ \text{Hz}$$

Because only particular energy changes are possible, only particular photon frequencies are emitted. This produces the discrete lines of an element's emission spectrum.

The relationship is a direct proportionality because $h$ is constant: doubling $|\Delta E|$ doubles $f$. Combining this with $c=f\lambda$ gives $\lambda=hc/|\Delta E|$, so a larger energy drop also produces a shorter-wavelength photon. The frequency is determined by the transition, not by how many atoms make that transition; more identical transitions increase the number of photons and therefore the radiation intensity, but they do not change each photon's frequency or energy.

A common misconception is that the electron itself becomes the photon. It does not: the electron remains in the atom at a lower energy level, while the lost energy is carried away by the photon.

Exam Technique

In a Calculate question, show the energy difference, convert eV to J if necessary, and substitute into $f=|\Delta E|/h$. In an Explain question, link discrete energy levels to discrete photon frequencies and therefore to line emission spectra.