# Find current induced in the photoelectric effect

• NihalRi
In summary: But as it is, the problem is just a matter of converting between power and current.In summary, a metal surface with dimensions of 1.5mm x 2mm is exposed to light with a wavelength of 420nm and an intensity of 4.5×10^-6 W/m^2. One electron is emitted for every 300 photons incident on the surface, and the work function is given as 3.4×10^-19. To determine the initial current leaving the metal, we can relate intensity to the number of photons incident per second and calculate the number of electrons emitted per unit time. This results in a current of 2.1×10^-14 A or 21fA. The

## Homework Statement

A metal surface had dimensions of 1.5mm x 2mm. The intensity of light incident on surface is 4.5×10^-6 W/m^2. On average one electron is emitted for every 300 photons incident on the surface. Determine the initial current leaving the metal.
●wavelength of light incident is 420nm
●work function is 3.4×10^-19

I=P/A
P=W/t
I=Q/t

## The Attempt at a Solution

I know the intensity is related to the number of electrons that are emitted but not how, that's the only way I can think of aproaching this problem. Is there another way?

NihalRi said:
I know the intensity is related to the number of electrons that are emitted
I don't think so..

Intensity is energy per unit time per unit area or power per unit area.
You can relate intensity to the number of photons incident per second just by replacing Power in the Formula of Intensity
by (Energy of 1 photon * n) where n is the no. of photons incident per second. Calculate for n.

It is given that 300 incident photons emit one electron, try calculating the current from this.

P. S current is charge emitted per unit time.

NihalRi
NihalRi said:

## Homework Statement

A metal surface had dimensions of 1.5mm x 2mm. The intensity of light incident on surface is 4.5×10^-6 W/m^2. On average one electron is emitted for every 300 photons incident on the surface. Determine the initial current leaving the metal.
●wavelength of light incident is 420nm
●work function is 3.4×10^-19

I=P/A
P=W/t
I=Q/t

## The Attempt at a Solution

I know the intensity is related to the number of electrons that are emitted but not how, that's the only way I can think of aproaching this problem. Is there another way?
First compute the number of photons per unit time incident on the metal.
The rest is a gimme.

NihalRi
AbhinavJ said:
I don't think so..

Intensity is energy per unit time per unit area or power per unit area.
You can relate intensity to the number of photons incident per second just by replacing Power in the Formula of Intensity
by (Energy of 1 photon * n) where n is the no. of photons incident per second. Calculate for n.

It is given that 300 incident photons emit one electron, try calculating the current from this.

P. S current is charge emitted per unit time.
Ok so,
I=P/A becomes, P=IA=1.35×10^-11 J/s
But the power = Ep *(n/t) , {Ep= energy of photon=work function=3.4×10^-19J}
n/t ≈ 39700000 {n = number of photons}
Now I divide this number by three hundred to get the number of eldctrons per time and then multiply by the charge of the electron to get the value of the current = 2.1 ×10^-14 A or 21fA.
So tiny did I do this right?

By the way the units for the work function was J

NihalRi said:
{Ep= energy of photon=work function=3.4×10^-19J}

Are you sure?
What is the energy of a photon dependant on?

AbhinavJ said:
Are you sure?
What is the energy of a photon dependant on?
AbhinavJ said:
Are you sure?
What is the energy of a photon dependant on?
hf, oh so I use this instead of the work function?

NihalRi said:
hf, oh so I use this instead of the work function?
Yes!

NihalRi
NihalRi said:
hf, oh so I use this instead of the work function?
The work function is what determines the minimum photon energy to get photoelectric emission, but the way the problem is stated its effect is included in the wording. You get 1 electron per 300 photons regardless of the work function. If the problem had given just the incident power and frequency (assuming monochromatic light) then you would have had to have taken the work function into account, which varies from metal to metal.

NihalRi

## 1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material surface when it is exposed to light of a certain frequency or energy. This effect was first discovered by Heinrich Hertz in 1887 and explained by Albert Einstein in 1905.

## 2. How is current induced in the photoelectric effect?

Current is induced in the photoelectric effect when electrons are emitted from the material surface and travel through a circuit, creating an electric current. This occurs because the photons of light that hit the surface transfer their energy to the electrons, giving them enough energy to break free from the material.

## 3. What factors affect the amount of current induced in the photoelectric effect?

The amount of current induced in the photoelectric effect is affected by the intensity of the incident light, the frequency of the light, and the work function of the material. Higher intensity and frequency of light result in more electrons being emitted, while a higher work function requires more energy for electrons to be emitted.

## 4. How is the photoelectric effect used in technology?

The photoelectric effect is used in various technologies, such as solar cells, photodiodes, and photomultiplier tubes. These devices use the principle of the photoelectric effect to convert light energy into electrical energy, making them essential in many modern technologies.

## 5. What are the implications of the photoelectric effect on our understanding of light?

The photoelectric effect played a crucial role in the development of quantum mechanics and our understanding of the dual nature of light as both a particle and a wave. It also provided evidence for Einstein's theory of relativity and paved the way for the development of new technologies that rely on the properties of light.