## Increasing Intensity in Photoelectric Effect [ENDORSED]

$c=\lambda v$

Priscilla_Covarrubias_HL
Posts: 16
Joined: Sat Sep 24, 2016 3:00 am

### Increasing Intensity in Photoelectric Effect

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
Posts: 25
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Increasing Intensity in Photoelectric Effect

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
Posts: 17
Joined: Sat Jul 09, 2016 3:00 am

### Re: Increasing Intensity in Photoelectric Effect

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
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

### Re: Increasing Intensity in Photoelectric Effect

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
Posts: 19
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Increasing Intensity in Photoelectric Effect

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
Posts: 4
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Increasing Intensity in Photoelectric Effect

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
Posts: 22
Joined: Fri Jul 15, 2016 3:00 am

### Re: Increasing Intensity in Photoelectric Effect

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
Posts: 20
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Increasing Intensity in Photoelectric Effect

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
Posts: 16
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Increasing Intensity in Photoelectric Effect

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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Joined: Fri Jul 22, 2016 3:00 am
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### Re: Increasing Intensity in Photoelectric Effect  [ENDORSED]

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.