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?
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.