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: Recall the relationship between frequency and wavelength

The speed of EM radiation \(c = \lambda f\) (where \(c\) is the speed of light in vacuum, \(\lambda\) is wavelength, \(f\) is frequency). Since \(c\) is constant (\(c = 3\times10^{8}\ m/s\)), \(\lambda=\frac{c}{f}\). As \(f\) increases, \(\lambda\) decreases.

Step2: Recall the relationship between frequency and photon energy

The energy of a photon \(E = hf\) (where \(h\) is Planck's constant, \(h= 6.63\times10^{-34}\ J\cdot s\)). As \(f\) increases, \(E\) increases.

Answer:

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