Saved Bookmarks
| 1. |
Let an electron requires 5xx10^(-19) joule energy to just escape from the irradiated metal.If photoelectron us emited after 10^(-9) s of the incident light,calculate the rate of absorption of energy .If this process is considered clasically.The light energy is assumed to be continously distributed over the wave front.Now,the electron can only absorb the light incident withina small area,say 10^(-19)m^(2).What is the intensity of illumination in order to see the photoelectric effect? |
|
Answer» SOLUTION :The rate of ABSORPTION of energy (power) is `p(e )/(t)=(5xx10^(-19))/(10^(-9)=5xx10^(10)(J)/(S)` From the definition of the INTENSITY of light, `I=("Energy")/("time"xx"area")=(5xx10^(-10))/(10^(-19))=5xx10^(9)(J)/(s.m^(2))` `(i.e. 500 billion ("Watt")/(m^(2)))` Since, practically it is IMPOSSIBLY high energy,which suggests that explanation of the photoelectric effect in classical term is not POSSIBLE. |
|