Question

if the wavelength of electromagnetic radiation increases, then
a. the frequency of the wave will decrease.
b. the speed of the wave will increase in direct relation.
c. the frequency of the wave will increase.
d. the speed of the wave will change.

Explanation:

Step1: Recall the wave equation

The wave equation is \(v = f\lambda\), where \(v\) is the speed of the wave, \(f\) is the frequency, and \(\lambda\) is the wavelength. For electromagnetic radiation, the speed \(v = c\) (speed of light in vacuum, \(c= 3\times10^{8}\space m/s\)) is constant in a vacuum.

Step2: Analyze the relationship between \(f\) and \(\lambda\)

Since \(c = f\lambda\) (constant \(c\)), we can rewrite the formula as \(f=\frac{c}{\lambda}\). This shows an inverse - relationship between frequency \(f\) and wavelength \(\lambda\). When \(\lambda\) increases, \(f=\frac{c}{\lambda}\) will decrease.

Step3: Analyze options

• Option a: Since \(f=\frac{c}{\lambda}\) ( \(c\) is constant), when \(\lambda\) increases, \(f\) decreases. This option is correct.
• Option b: The speed of electromagnetic radiation in a vacuum is constant (\(c = 3\times 10^{8}\space m/s\)), so it will not increase. This option is incorrect.
• Option c: As \(f=\frac{c}{\lambda}\), when \(\lambda\) increases, \(f\) decreases, not increases. This option is incorrect.
• Option d: The speed of electromagnetic radiation in a vacuum is constant (for a given medium), so it will not change. This option is incorrect.

Answer:

A. the frequency of the wave will decrease.