Photoelectric Effect: Calculating Electron Energy

In summary, using classical arguments, it can be calculated that it would take 156 nanoseconds for an electron to gain an energy of 1 electron volt when radiated with sunlight at an intensity of 500 watts per square meter. This result is much longer than the instantaneous emission of electrons in the photo-electric effect. It is also important to consider whether the light is incident from all directions or just one, as this affects the calculation of the projected area of the electron. In classical physics, an electron is regarded as a sphere and light is a wave, so assuming even distribution of light on the surface area of the sphere is a valid assumption. However, when looking at a sphere from a distance, the projected area is just a disc
  • #1
xiankai
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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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  • #2
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:
  • #3
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.
 
  • #4
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.
 
  • #5
ah, thanks. it seems crystal clear now.
 

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to light of a certain frequency. It was first observed by physicist Heinrich Hertz in 1887.

2. How is the energy of electrons calculated in the photoelectric effect?

The energy of electrons in the photoelectric effect is calculated using the equation E = hf - Φ, where E is the energy of the emitted electrons, h is Planck's constant, f is the frequency of the incident light, and Φ is the work function of the material.

3. What is the work function of a material?

The work function of a material is the minimum amount of energy required to remove an electron from the surface of that material. It is different for different materials and is dependent on factors such as the material's composition and crystal structure.

4. How does the intensity of light affect the photoelectric effect?

The intensity of light does not directly affect the energy of the emitted electrons in the photoelectric effect. However, increasing the intensity of light can increase the number of electrons emitted from the material, as more photons will be hitting the surface and causing more electrons to be ejected.

5. What is the significance of the photoelectric effect in modern science?

The photoelectric effect has many practical applications in modern science and technology. It is the basis for devices such as photodiodes and solar cells, which are used in electronic devices and renewable energy systems. It also helped to confirm the particle nature of light and played a crucial role in the development of quantum mechanics.

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