Question

complete the statement for em radiation in a vacuum. as the frequency of em radiation increases, its wavelength and the energy of its photons.

Explanation:

Step1: Relationship between frequency and wavelength

The speed of EM radiation in a vacuum \(c = \lambda f\), where \(c\) is a constant (\(c = 3\times10^{8}\space m/s\)). Rearranging for \(\lambda=\frac{c}{f}\). As \(f\) increases, \(\lambda\) decreases.

Step2: Relationship between frequency and photon energy

The energy of a photon \(E = hf\), where \(h\) is Planck's constant (\(h=6.63\times 10^{-34}\space J\cdot s\)). As \(f\) increases, \(E\) increases.

Answer:

As the frequency of EM radiation increases, its wavelength decreases and the energy of its photons increases.