Photoelectric Effect: Calculating Electron Energy

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Homework Help Overview

The discussion revolves around calculating the time required for an electron to gain an energy of 1 eV when exposed to sunlight with a specified intensity, using classical physics arguments. The context involves the photoelectric effect and comparisons to classical energy absorption models.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of a hemispherical model for the electron's surface area and question the validity of this assumption in the context of classical theory. There is also discussion about the implications of energy absorption in classical versus quantum mechanics.

Discussion Status

Participants are actively questioning the assumptions made about the electron's surface area and the nature of light's interaction with it. Some guidance has been provided regarding the concept of projected area versus surface area, but no consensus has been reached on the best model to use.

Contextual Notes

There is uncertainty about the classical treatment of electrons in materials and how energy is absorbed from light, particularly regarding whether to consider the electron as a point or a spherical object. The discussion also touches on the differences between classical and quantum mechanical perspectives.

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Homework Statement



3. The classical radius of an electron is 2.82 x 10^-15 m. If a material is radiated with sunlight with an intensity of 500W/m^2, calculate using classical arguments the time required for an electron to gain an energy of 1eV. How does this result compare with electron emission in the photo-electric effect?

Homework Equations



surface area of a sphere = 4 (pi) r^2
1 electron volt = 1.602 x 10^-19 V


The Attempt at a Solution



surface area of hemispheric electron (assuming the light falls on one side of the material) = 2 (pi) r^2
= 5.00 x 10^-29 m^2

power of light falling on electron = 500 x ( 5.00 x 10^-29 )
= 2.50 x 10^-26 W

time taken = ( 2.50 x 10^-26 ) / ( 1.602 x 10^-19 )
= 1.56 x 10^-7 s
= 156 ns

in the photo-electric effect, electrons are emitted instaneously, at times much less than 156 ns.

4. The thoughts

is it okay to use the hemisphere instead of the whole sphere as a model of the area where the light falls on the electron? after all I am pretty unsure of how the classical theory looks at the electron in a material.
 
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after all I am pretty unsure of how the classical theory looks at the electron in a material.
Think about QM vs classical. In classical physics, one could continually add energy (light) to an atom/electron until is escaped the atoms. For example, in classical theory one could add ten 0.1 ev photons to an electron to give it 1 eV. But what really happens?

Think about Lyman, Balmer and Paschen spectral lines.


is it okay to use the hemisphere instead of the whole sphere as a model of the area where the light falls on the electron?
That depends on whether the energy is incident from all directions, or one direction. Usually, if one assumes that light is from one direction (infinitely far away, or a long, long way away), then one has to use the projected area.
 
Last edited:
Think about QM vs classical. In classical physics, one could continually add energy (light) to an atom/electron until is escaped the atoms. For example, in classical theory one could add ten 0.1 ev photons to an electron to give it 1 eV. But what really happens?

Think about Lyman, Balmer and Paschen spectral lines.

classically, an electron is regarded as a sphere, and light is a wave. in that respect, light would fall evenly on the surface area of the sphere. that is what i think.

That depends on whether the energy is incident from all directions, or one direction. Usually, if one assumes that light is from one direction (infinitely far away, or a long, long way away), then one has to use the projected area.

in that case would it be better if i assumed that the light was incident on a point surface area of the electron 'sphere'? i don't get what you mean by projected area. which is a better assumption.
 
With respect to projected area, if one looks at a sphere at a distance, one does not see a sphere but rather a disc. The surface area of a hemisphere is 4 pi r2, but the projected (planar) area is just pi r2.
 
ah, thanks. it seems crystal clear now.
 

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