, where  is the speed of light,  is the wavelength, and  is frequency.

, where  is the speed of light,  is the wavelength, and  is frequency.

Amplitude of a wave is defined as the maximum peak of oscillation or vibration. Recall that frequency, wavelength, and speed are not dependent on the amplitude and are calculated using the wave equation:

Here,  is the speed of the wave (which is the speed of light for electromagnetic waves),  is the frequency, and  is the wavelength. Amplitude of a wave is not present in this equation and, therefore, cannot be used to determine relative values of frequency, wavelength, and speed.

The phenomenon described in this question is called redshift and blueshift. All stars emit electromagnetic waves. The perceived wavelength of these waves depends on the relative motion of the object. If an object is approaching the observer (in this case, the earth) then the apparent wavelength of emitted waves decreases whereas if the object moves away then the apparent wavelength increases. This is called the Doppler effect.

To answer this question, we need to know the relative wavelengths of blue and red light. Blue light has wavelength between  whereas red light has wavelength between . The question states that star A appears red and star B appears blue. Since red has the higher wavelength, star A is going away from the earth. Similarly, since blue has a lower wavelength, star B is approaching earth.

Recall that lower wavelength means higher frequency; therefore, blue has the higher frequency.

To solve this question, we need to first recall the electromagnetic spectrum. Microwaves have very large wavelengths. They are found between visible light and radio waves. Ultraviolet, on the other hand, is found below visible light; therefore, ultraviolet rays have smaller wavelength than microwaves.

To solve for energy we need to use the following equation.

Here,  is the energy,  is Planck’s constant,  is speed of light, and  is wavelength. Since energy is inversely proportional to wavelength, higher wavelengths will lead to lower energy. This means that microwaves will have lower energy than ultraviolet rays.

Gamma rays have the lowest wavelength and highest energy whereas radio waves typically have the highest wavelength and lowest energy.

The question states that there are two light sources and that both are placed equidistant from a detector. This means that any differences in detection speed is not related to the distance. The question also states that the detector picks up light from source A quicker that light from source B. This means that the speed of the light from source A is faster. Recall that light (or electromagnetic rays) can have varying wavelengths and frequencies; however, the speed of all electromagnetic rays is the same. It doesn’t matter if it is gamma rays or visible light, the speed of light in air is . The information given in this question contradicts this; therefore, these results are invalid.

We cannot determine the color of the light for the light sources because we don't have any information regarding their respective wavelength or frequency. The number of photons per unit time refers to the amplitude of the wave, which also cannot be determined from the given information.

To solve this problem, we use the following equation:

Where  is the wavelength and  is the frequency.

Wavelength is given, and as we are dealing with the electromagnetic spectrum, we know that  is equal to the speed of light, which is given. We need to solve for frequency, so we arrange the equation as follows: 

Plug in known values and solve.

To solve this problem, we use the following equation:

Where  is the wavelength and  is the frequency.

Wavelength is given, and as we are dealing with the electromagnetic spectrum, we know that  is equal to the speed of light, which is given. We need to solve for frequency, so we arrange the equation as follows: 

Plug in known values and solve.

To solve this problem, we use the following equation:

Where  is the wavelength and  is the frequency.

Rearrange the equation to solve for wavelength.

Plug in known values and solve.