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compared by wavelength and frequency, e(radio) < e(microwave) < e(infra…

Question

compared by wavelength and frequency, e(radio) < e(microwave) < e(infrared) < e(visible) < e(ultraviolet) < e(x - ray) < e(gamma ray) and for light within the visible region e(red) < e(orange) < e(yellow) < e(green) < e(blue) < e(indigo) < e(violet) there is significant practical value in knowing that, for example, infrared photons are less energetic than ultraviolet photons, and red light photons are less energetic than blue light photons. question 5.1.8 a. drag and drop the regions of the electromagnetic spectrum below so that they are in increasing energy order radio infrared visible x - ray b. drag and drop the colors of the visible region of the spectrum below so that they are in increasing energy order red yellow blue indigo

Explanation:

Step1: Recall energy - wavelength - frequency relationship

The energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength. Higher - frequency (shorter - wavelength) radiation has higher energy.

Step2: Arrange electromagnetic spectrum regions

Based on the known energy order of electromagnetic spectrum regions, radio waves have the lowest energy, followed by infrared, then visible light, and x - rays have relatively high energy. So the order is radio < infrared < visible < x - ray.

Step3: Arrange visible light colors

Among visible light colors, red has the lowest energy, then yellow, followed by blue, and indigo has relatively higher energy. So the order is red < yellow < blue < indigo.

Answer:

a. radio, infrared, visible, x - ray
b. red, yellow, blue, indigo