Photoelectric Effect: Calculating Electron Energy

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SUMMARY

The discussion focuses on calculating the time required for an electron to gain an energy of 1 eV when exposed to sunlight with an intensity of 500 W/m². Using classical physics, the surface area of a hemispherical electron is calculated as 5.00 x 10^-29 m², leading to a power of 2.50 x 10^-26 W and a time of 156 ns for the energy gain. This result starkly contrasts with the instantaneous emission of electrons observed in the photoelectric effect, highlighting the limitations of classical theories in explaining quantum phenomena.

PREREQUISITES
  • Understanding of classical physics concepts, particularly energy transfer.
  • Familiarity with the photoelectric effect and its implications.
  • Knowledge of basic quantum mechanics principles.
  • Ability to perform calculations involving surface area and power.
NEXT STEPS
  • Explore the principles of the photoelectric effect in detail.
  • Learn about quantum mechanics and its differences from classical physics.
  • Study the Lyman, Balmer, and Paschen spectral lines and their significance.
  • Investigate the concept of projected area in physics and its applications.
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Students of physics, educators teaching classical and quantum mechanics, and researchers exploring the boundaries between classical theories and quantum phenomena.

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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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