## Chapter 1 #3

$c=\lambda v$

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Glendy Gonzalez 1A
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### Chapter 1 #3

Can someone please explain what happens when the frequency of electromagnetic radiation decreases?
Based on the textbook, the answer is (c). the extent of the change in the electrical field at a given point decreases. But I don't know why or how to explain it.

MadisonFuentes1G
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### Re: Chapter 1 #3

We know that the electrical field (as opposed to the magnetic field) corresponds to the amplitude of a wave. Because the frequency is decreasing, this means that there are less cycles being completed per second, meaning that the waves must be getting longer since it will take a longer duration of time to complete a cycle. Because of this, there is less of a change in the amplitude of longer waves.

Esin Gumustekin 2J
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### Re: Chapter 1 #3

I'm still confused on this problem. Isn't the amplitude of the wave staying the same? I know that the frequency of the wave is decreasing but that doesn't mean that the amplitude is also decreasing. So, how can we say that the electric field is decreasing if the amplitude is staying the same?

Jingyi Li 2C
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### Re: Chapter 1 #3

As stated by the textbook, the electric field pushes the electron first in one direction and then pulls it in the opposite direction, over and over again. Electromagnetic radiation of frequency of 1 Hz pushes a charge in one direction, then the opposite direction, and returns to the original direction once per second. Therefore, if the electric field at a given point decreases, its ability to push and pull the charge also decreases. As a result, the frequency of the electromagnetic radiation decreases.

JamesAntonios 1E
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### Re: Chapter 1 #3

EM radiation can be seen through the particle lens. If the frequency is decreasing, the energy of the field/wave is decreasing as well. With relation to an electric field, if the frequency, hence the energy decreases.

kaushalrao2H
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### Re: Chapter 1 #3

λν=c

Just think of this formula when answering questions related to wavelength (λ) and frequency (v). The product of the two must always equal c (3.00 x 10^8), so if either the wavelength or frequency decreases, then the other must increase.

Alissa Stanley 3G
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### Re: Chapter 1 #3

I understand why a. and b. are incorrect, but can someone please explain why d. is wrong? Wouldn't energy increase if the frequency decreases (as in the equation E=hv).

Priyanka Bhakta 1L
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### Re: Chapter 1 #3

To answer your question Alissa, the energy would increase if the frequency increased because in the equation E=hv, h is a constant. Thus, the only thing that has an effect on the energy is the frequency. So, if the frequency increases and is multiplied by a constant, the energy will increase. If the frequency decreases and is multiplied by the same constant, that means the product (energy) will also decrease. It is as if the equation is E=5v; if v=3, E=15 - but if v increases to 5, E=25. In other words, energy and frequency have a direct relationship; when one increases, so does the other, and vice versa. Hope this helps and answers your question!

Jessica Nunez 1I
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### Re: Chapter 1 #3

With respect to Part C, is there a relationship between the extent of change in an electrical field and the frequency of electromagnetic radiation?

Angel R Morales Dis1G
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### Re: Chapter 1 #3

Jessica Nunez 1G wrote:With respect to Part C, is there a relationship between the extent of change in an electrical field and the frequency of electromagnetic radiation?

According to the solutions manual: "Yes, The Electrical Field corresponds to the amplitude; as the frequency decreases the waves broaden and the extent of the change (the slope of the wave) decreases."
Basically, the longer the wave, the smaller its slope and the smaller its amplitude.

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