(adsbygoogle = window.adsbygoogle || []).push({}); 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.

**Physics Forums - The Fusion of Science and Community**

# Photoelectric effect

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Photoelectric effect

Loading...

**Physics Forums - The Fusion of Science and Community**