Increasing Intensity in Photoelectric Effect  [ENDORSED]


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Priscilla_Covarrubias_HL
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Increasing Intensity in Photoelectric Effect

Postby Priscilla_Covarrubias_HL » Fri Sep 30, 2016 2:56 am

In the last lecture (9/28), the Professor mentioned that increasing the intensity of the light source does not remove more electrons from the metal. I also heard that increasing the intensity of the light source actually increases the number of photons but not the amount of energy each photon carries. I am wondering if what I heard was actually correct or if I confused what the Professor said.

Jenny_Thompson_3I
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Re: Increasing Intensity in Photoelectric Effect

Postby Jenny_Thompson_3I » Fri Sep 30, 2016 7:55 am

The main concept was that Dr. Lavelle said that increasing the intensity of white light (the light source) does not lead to electrons being removed. In order to remove electrons from the metal you must have a light source with a shorter wavelength. Then in your increase the intensity of this shorter wavelength it will result in more electrons being removed. There is a proportional relationship between intensity and the number of photons found in light. Since a higher intensity of the shorter wavelength light has more photons, more electrons will be removed. I hope this helps!

Katelynn Luansing 2B
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Re: Increasing Intensity in Photoelectric Effect

Postby Katelynn Luansing 2B » Fri Sep 30, 2016 10:40 am

I had the same question, but reading this post definitely helped!
Also during that same lecture on 9/28, when going through the worked example in our Course Readers I was confused on where the values for "h" and "c" came from. We were trying to calculate energy and wavelength of the light utilized in the photoelectric experiment. Are "h" and "c" constants?

Samuel_Vydro_1I
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Re: Increasing Intensity in Photoelectric Effect

Postby Samuel_Vydro_1I » Fri Sep 30, 2016 11:10 am

Katelynn Luansing 1J wrote:I had the same question, but reading this post definitely helped!
Also during that same lecture on 9/28, when going through the worked example in our Course Readers I was confused on where the values for "h" and "c" came from. We were trying to calculate energy and wavelength of the light utilized in the photoelectric experiment. Are "h" and "c" constants?


Yes, h and c are constants and found on our constant sheets that will be given during tests. h is known as Planck's constant and assumes the value of 6.62607004x10^-34 m^2kg/s while c is the speed of light (3x10^8 m/s)

Xiaoman_Kang_2J
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Re: Increasing Intensity in Photoelectric Effect

Postby Xiaoman_Kang_2J » Fri Sep 30, 2016 12:23 pm

Jenny_Thompson_1I wrote:The main concept was that Dr. Lavelle said that increasing the intensity of white light (the light source) does not lead to electrons being removed. In order to remove electrons from the metal you must have a light source with a shorter wavelength. Then in your increase the intensity of this shorter wavelength it will result in more electrons being removed. There is a proportional relationship between intensity and the number of photons found in light. Since a higher intensity of the shorter wavelength light has more photons, more electrons will be removed. I hope this helps!

So does this mean that one photon can only remove one electron if the energy of this photon actually reaches the minimum energy required by the removal?

Ruthie Jia 1L
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Re: Increasing Intensity in Photoelectric Effect

Postby Ruthie Jia 1L » Fri Sep 30, 2016 12:38 pm

Xiaoman_Kang_1L wrote:
Jenny_Thompson_1I wrote:The main concept was that Dr. Lavelle said that increasing the intensity of white light (the light source) does not lead to electrons being removed. In order to remove electrons from the metal you must have a light source with a shorter wavelength. Then in your increase the intensity of this shorter wavelength it will result in more electrons being removed. There is a proportional relationship between intensity and the number of photons found in light. Since a higher intensity of the shorter wavelength light has more photons, more electrons will be removed. I hope this helps!

So does this mean that one photon can only remove one electron if the energy of this photon actually reaches the minimum energy required by the removal?


Yes, 1 photon ejects 1 electron; they should be proportional. The energy of the photon must be greater than the energy needed to remove an electron in order for an electron to be emitted.

Amy_Shao_2D
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Joined: Fri Jul 15, 2016 3:00 am

Re: Increasing Intensity in Photoelectric Effect

Postby Amy_Shao_2D » Fri Sep 30, 2016 1:24 pm

Going off of this, he also mentioned threshold value a few times. Is that the value that the wavelength has to be shortened to in order for increasing intensity to actually remove electrons?

Jessie_Chen_2L
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Joined: Wed Sep 21, 2016 2:59 pm

Re: Increasing Intensity in Photoelectric Effect

Postby Jessie_Chen_2L » Fri Sep 30, 2016 3:16 pm

Hi, from what I know, the threshold value is the minimum amount of energy it requires to remove an electron from the metal. It is also called the work function. And yes, the wavelength must be shorter to increase the frequency, which in turn increases the energy so that there is enough energy to remove the electron based on the equation E = h x ν.

Jared Gao 3G
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Joined: Wed Sep 21, 2016 2:58 pm

Re: Increasing Intensity in Photoelectric Effect

Postby Jared Gao 3G » Fri Sep 30, 2016 3:25 pm

So for further clarification, the intensity of the light has no effect until the energy of the photon reaches the work function? Also, I'm confused on what happens to the excess energy from the light.

TiengTum2D
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Re: Increasing Intensity in Photoelectric Effect  [ENDORSED]

Postby TiengTum2D » Mon Oct 03, 2016 2:19 am

Jared Gao 3G wrote:So for further clarification, the intensity of the light has no effect until the energy of the photon reaches the work function? Also, I'm confused on what happens to the excess energy from the light.


Yes, in the photoelectric effect, light acts as a particle so the intensity of the light only means the number of photons, not the energy that the photons have. The energy of electromagnetic radiation depends on the frequency v, according to E= hv. Excess energy from the light converts to the kinetic energy of the ejected electron.


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