recall and use hf = E1 – E2

Key Equation: $hf = E_1 - E_2$

Here $h$ is Planck's constant (≈ $6.626 \times 10^{-34}$ J·s), $f$ is the frequency of the photon, and $E_1$ and $E_2$ are the energies of the two atomic levels involved. When an electron jumps from a higher level $E_2$ to a lower level $E_1$, it emits a photon with energy $hf$.

Analogy: The Electron Ladder 🚶‍♂️

Think of an atom as a ladder with rungs (energy levels). An electron is a person climbing up or down. When the person jumps down, they drop a small amount of energy, like dropping a pebble that makes a sound. That sound is the photon we see as light. The higher the jump, the louder (more energetic) the sound!

Example: Hydrogen Balmer Line (Hα)

For hydrogen, the Balmer series involves transitions from $n=3$ to $n=2$.

Energy difference: $ΔE = E_2 - E_3 = (-3.40) - (-1.51) = -1.89$ eV (negative sign indicates emission).

Convert to frequency: $f = \frac{ΔE}{h}$, then wavelength: $\lambda = \frac{c}{f}$.

Result: $\lambda ≈ 656$ nm (red light).

Exam Tip 📚

• Always write the equation $hf = E_1 - E_2$ and remember that $E_1$ is the lower level.
• Use $ΔE = h f$ to find frequency, then $\lambda = \frac{c}{f}$.
• Check units: eV → J (multiply by $1.602\times10^{-19}$).
• Remember common series: Lyman ($n=1$), Balmer ($n=2$), Paschen ($n=3$).
• When given wavelength, you can reverse the steps to find $ΔE$.

Quick Formula Sheet ✨

1. $hf = E_1 - E_2$
2. $ΔE = h f$
3. $f = \frac{c}{\lambda}$
4. $\lambda = \frac{c}{f}$
5. $E_n = -\frac{13.6\,\text{eV}}{n^2}$ for hydrogen

Keep practising with different $n$ values to become comfortable!