Problem — Ranking wavelengths

Problem

Four electromagnetic radiations are characterised by their frequency. Order them by increasing wavelength:

• AM radio waves ($f \sim 1\;\text{MHz} = 10^6\;\text{Hz}$);
• infrared ($f \sim 10^{13}\;\text{Hz}$);
• green visible light ($f \sim 5\cdot 10^{14}\;\text{Hz}$);
• X-rays ($f \sim 10^{18}\;\text{Hz}$).

Solution

Idea. All electromagnetic waves in vacuum travel at the same speed $c$, so wavelength and frequency are related by $\lambda = c/f$: a higher frequency corresponds to a shorter wavelength (inverse relation). We just need to estimate each $\lambda$.

AM radio.

$\lambda_\text{radio} = \frac{c}{f} = \frac{3\cdot 10^8}{10^6} = 3\cdot 10^{2}\;\text{m} \sim 300\;\text{m}$

Infrared.

$\lambda_\text{IR} = \frac{3\cdot 10^8}{10^{13}} = 3\cdot 10^{-5}\;\text{m}$

Visible (green).

$\lambda_\text{vis} = \frac{3\cdot 10^8}{5\cdot 10^{14}} = 6\cdot 10^{-7}\;\text{m}$

X-rays.

$\lambda_\text{X} = \frac{3\cdot 10^8}{10^{18}} = 3\cdot 10^{-10}\;\text{m}$

Ordering by increasing wavelength:

$\ev{\lambda_\text{X} < \lambda_\text{vis} < \lambda_\text{IR} < \lambda_\text{radio}}$

As expected, the order is exactly the reverse of that of the frequencies.

Links

Topics: Onde elettromagnetiche Concepts: Onda elettromagnetica Skills: Ragionamento di ranking