Photoelectric exp - calculate intensity of incident light

In summary, the conversation discusses an experiment involving the photoelectric effect and the calculation of incident light intensity. The correct equation for intensity is mentioned, but the resulting calculation is incorrect. The conversation also brings up questions about the relationship between photoelectric current, the number of ejected electrons, and the number of incident photons. The issue with the incorrect calculation is attributed to the incorrect units.
  • #1
desmond iking
284
2

Homework Statement


In an photoelectric effect experiment , a photoelectric current of 100µA is obtained when lights of 550nm is incident on metal cathode of surface area 1.0cm^2 ... calculate the intensity of the incident light. The ans is 2.26w/m^2

Homework Equations

The Attempt at a Solution



intensity= power /area
intensity= (voltage x current) /area
intensity= ( (hc/λ) x I ) / area
= (6.63x10^-34)(3x10^8)(100x10^-6) / (550x10^-9)(1.0x10^-4) = 3.61x10^-19 (W/m^2 )

why am i wrong?
[/B]
 
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  • #2
How is the photoelectric current related to the number of electrons ejected? What is the relation between the number of electrons ejected and the number of the incident photons?
If you check the unit of your result, it is not W/m2.
 
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  • #3
ehild said:
How is the photoelectric current related to the number of electrons ejected? What is the relation between the number of electrons ejected and the number of the incident photons?
If you check the unit of your result, it is not W/m2.

can you please tell me what's the unit of my 3.61x10^-19 should be? i am confused now.
 
  • #4
The problem is not with the units. If you check the unit of your result and it is not what it should be then your solution is wrong. Your solution is wrong, as its dimension is not energy/(time area).
Read the first two sentences of my previous post.

ehild
 
  • #5


Your attempt at a solution is not wrong, but it is missing a key step. In order to calculate the intensity of the incident light, you need to use the equation:

Intensity = Power/Area = (hc/λ) x I

Where:
h = Planck's constant (6.63x10^-34 J*s)
c = speed of light (3x10^8 m/s)
λ = wavelength of incident light (550x10^-9 m)
I = photoelectric current (100x10^-6 A)
Area = 1.0 cm^2 = 1.0x10^-4 m^2

So, plugging in the values, we get:

Intensity = ((6.63x10^-34 J*s)(3x10^8 m/s))/(550x10^-9 m) x (100x10^-6 A) / (1.0x10^-4 m^2)
= (3.61x10^-19 W/m^2)

Therefore, the intensity of the incident light is indeed 2.26 W/m^2. Your calculation was correct, you just forgot to include the Planck's constant in your equation.
 

FAQ: Photoelectric exp - calculate intensity of incident light

What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to electromagnetic radiation, typically in the form of light.

How do you calculate the intensity of the incident light in a photoelectric experiment?

The intensity of the incident light can be calculated by dividing the energy of the incident photons by the area of the surface that they are hitting. This can be represented by the equation I = P/A, where I is the intensity, P is the power of the light source, and A is the surface area.

What factors can affect the intensity of the incident light in a photoelectric experiment?

The intensity of the incident light can be affected by the distance between the light source and the material, the angle of incidence, and the properties of the material itself, such as its reflectivity and transparency.

How does the intensity of the incident light affect the number of emitted electrons in a photoelectric experiment?

The number of emitted electrons is directly proportional to the intensity of the incident light. This means that as the intensity increases, the number of emitted electrons also increases.

What is the threshold frequency in a photoelectric experiment?

The threshold frequency is the minimum frequency of light required to cause the emission of electrons in the photoelectric effect. Below this frequency, no electrons will be emitted regardless of the intensity of the incident light.

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