Ranking the energies of four photons

Problem

Ranking. Four photons have wavelength $\lambda$:

• (A) $\lambda = 600\;\text{nm}$ (orange light)
• (B) $\lambda = 1\;\mu\text{m}$ (infrared)
• (C) $\lambda = 0{,}1\;\text{nm}$ (X-ray)
• (D) $\lambda = 400\;\text{nm}$ (violet)

Order the energies $E = hc/\lambda$ of the photons from smallest to largest.

Solution

Principle. $E = hc/\lambda$: the energy is inversely proportional to $\lambda$. The shorter the wavelength, the more energetic the photon. So we just need to order by decreasing $\lambda$.

$E_B = \frac{1240}{1000} = 1{,}24\;\text{eV}, \quad E_A = \frac{1240}{600} = 2{,}07\;\text{eV}, \quad E_D = \frac{1240}{400} = 3{,}10\;\text{eV}, \quad E_C = \frac{1240}{0{,}1} = 1{,}24\cdot 10^4\;\text{eV}$

$\ev{E_B < E_A < E_D < E_C}$

Infrared (B) is the least energetic; the X-ray (C), with $\lambda$ ten thousand times shorter than visible light, is enormously more energetic — this is why it ionises and passes through tissue.

Links

Topics: Quantum physics Concepts: Photon Skills: Ranking reasoning