## Chapter 1, problem 3 [ENDORSED]

$c=\lambda v$

Stephanie H
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### Chapter 1, problem 3

If the frequency of electromagnetic radiation decreases, will the speed of electromagnetic radiation also decrease?
I was thinking the speed will not decrease because the speed of light is constant (c= wavelength*frequency), are there other reasons?

Paul Wong1B
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### Re: Chapter 1, problem 3

I believe you are correct in that the speed of light is a constant so it should be the same for all photons regardless of the frequency. However, the case would be different with particles that have rest mass as De Broglie's equation includes velocity to solve for the wavelength of a particle, which could then be used to find the frequency.

Stephanie H
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### Re: Chapter 1, problem 3

I think this question is referring to light and not particles, so I think we are on track.

Sarah_Wilen
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### Re: Chapter 1, problem 3

As Paul said, speed is a constant, therefore, cannot change. If the frequency of EMR decreases, then the wavelength would increase. In the speed of light equation, the frequency is inversely proportional to wavelength. That means as one value decreases, the other increases. If wavelength decreases, frequency increases and vice versa.

Typed on tiny keys, just for you.

Peter Dis1G
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### Re: Chapter 1, problem 3

wait,so what's the answer?is that D? I found from the internet that the shorter the frequency,the larger the radiation energy.

Rakhi Ratanjee 1D
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### Re: Chapter 1, problem 3

For this question, my reasoning was that the answer would be A) the speed of the radiation decreases as frequency of electromagnetic radiation decreases, because a higher frequency means that more waves are traveling per second or more cycles per second. The solutions manual, however, states that A) is incorrect and that C, which is "the extent of the change in the electrical field at a given point decreases" is the answer. What exactly does this mean?

Sean Monji 2B
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### Re: Chapter 1, problem 3

The electric, or magnetic, fields are the reason we draw light as waves. We do this to simplify the image of light so that we do not have to draw both fields, as it is much easier to draw one wave rather than two in a 2D plane (since the electric and magnetic fields are perpendicular to each other). Essentially, the wavelength of the electric field is the wavelength of the light. So if the frequency of light decreases, wavelength increases, and thus the wavelengths of the electric fields increase. If the waves are elongated, the slopes of the wave decrease on each point of the wave, thus the change per each point of the wave is less. (Think like graphing sin or cos waves on a calculator). Please correct me if I'm wrong

Like others said, the answer will probably never be speed decrease, unless the light is passing through a medium, like water.

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### Re: Chapter 1, problem 3  [ENDORSED]

Rakhi Ratanjee 1B wrote:For this question, my reasoning was that the answer would be A) the speed of the radiation decreases as frequency of electromagnetic radiation decreases, because a higher frequency means that more waves are traveling per second or more cycles per second. The solutions manual, however, states that A) is incorrect and that C, which is "the extent of the change in the electrical field at a given point decreases" is the answer. What exactly does this mean?

Note that speed will always be a constant and that waves traveling per second only relates to frequency. The speed of radiation refers to the speed of light, not the "speed" of the wavelength.

Gurvardaan Bal1L
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### Re: Chapter 1, problem 3

The answer to this problem is (c) due to how electrical fields are directly related to the frequency. If you go to page 4, on the third paragraph directly before the definition of 1 Hertz, the book talks about how the passing of an light ray by an electron pushes the electrical field of an electron in one direction and then pulls it the other direction. The number of cycles of this back and forth motion that occur per second are called the frequency of the radiation. Because of this, as the frequency decreases, the change in the electrical field at a given point is decreasing.

Michelle Dong 1F
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### Re: Chapter 1, problem 3

What does answer choice D mean when it says the energy of the radiation increases? Does energy have anything to do with frequency or amplitude/intensity of the wave?

Gurvardaan Bal1L
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### Re: Chapter 1, problem 3

Michelle Dong 1H wrote:What does answer choice D mean when it says the energy of the radiation increases? Does energy have anything to do with frequency or amplitude/intensity of the wave?

What D means is that the energy of the oscillations of the electric and magnetic field is increasing. These oscillations all transfer energy from one space to the other, so the increasing of energy of radiation would mean that the waves are oscillating much more.