1. The problem statement, all variables and given/known data 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? 2. Relevant equations surface area of a sphere = 4 (pi) r^2 1 electron volt = 1.602 x 10^-19 V 3. 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 im 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. 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.
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. 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 dont 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 r^{2}, but the projected (planar) area is just pi r^{2}.