Problem — Ranking photon energy

Problem

Four electromagnetic radiations have the following wavelengths:

• (A) gamma rays, $\lambda=10^{-12}$ m;
• (B) ultraviolet, $\lambda=10^{-7}$ m;
• (C) microwaves, $\lambda=10^{-2}$ m;
• (D) AM radio, $\lambda=300$ m.

Rank them by photon energy, from most to least energetic.

Solution

Relation to use. The energy of a photon is $E=hf=\frac{hc}{\lambda}$ Since $h$ and $c$ are constants, energy is inversely proportional to wavelength: $E\propto\frac{1}{\lambda}$

Comparison. A smaller wavelength means greater energy. Ordering the $\lambda$ values from smallest to largest: $10^{-12}<10^{-7}<10^{-2}<300$ m, i.e. $A<B<C<D$ in $\lambda$, hence $A>B>C>D$ in energy.

Numerical check (for photon A, the most energetic): $E_A=\frac{6{,}63\cdot 10^{-34}\cdot 3{,}00\cdot 10^8}{10^{-12}}\approx 2\cdot 10^{-13}\ \text{J}$ enormous compared with radio D, for which $E_D\approx 7\cdot 10^{-28}$ J.

Final ranking: $\ev{A>B>C>D}$

Connections

Topics: Electromagnetic waves Concepts: Photon Skills: Ranking reasoning