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Photoelectric effect 
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#1
Feb2407, 08:08 AM

P: 31

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 photoelectric 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 photoelectric 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. 


#2
Feb2407, 08:16 AM

Admin
P: 21,887

Think about Lyman, Balmer and Paschen spectral lines. 


#3
Feb2407, 11:11 PM

P: 31




#4
Feb2507, 09:07 AM

Admin
P: 21,887

Photoelectric effect
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}.



#5
Feb2507, 09:36 PM

P: 31

ah, thanks. it seems crystal clear now.



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