Saved Bookmarks
| 1. |
Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2m.Assume that the metal surface has work function of 2eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å. (i)Estimate no. of photons emitted by the bulb per second.[Assume no other losses] (ii)Will much time would be required by the atomic disk to receive energy equal to work function (2eV)? (iv)How many photons would atomic disk receive within time duration calculated in (iii)above? (v) Can you explain how photoelectric effect was observed instantaneously? |
|
Answer» Solution :(i)`P=(E_(n))/(t)=(nhf)/(t)=(NHC)/(tlambda)` `therefore ((n)/(t))=(lambda)/(nc)` =`((20)(5000xx10^(-10)))/((6.625xx10^(-34)(3xx10^(8))` `=5.03xx10^(19)` photon/second (ii)Energy of one photon, `E_(1)=hf=(hc)/(lambda)` `therefore E_(1)=((6.625xx10^(-34))(3xx10^(8)))/((5000xx10^(-10))(1.6xx10^(-19)))eV` `therefore E_(1)=2.49 eV` Here `phi_(0)=2eV` `therefore E_(1)gtphi_(0)` `implies` Photo emission will take place. (III)  Intensity of radiation ,`I=(E_(0))/(A_(0)t_(0))=(P)/(A)` `therefore t_(0)=(E_(0)A)/(PA_(0))` `therefore t_(0)=((2xx1.6xx10^(-19))(4xx3.14xx(2)^(2)))/((20)(3.14xx(1.5xx10^(-10))^(2)))` `therefore t_(0)=11.38 s` (iv) Area `to` (No.of photons made incident in one second) `therefore piR^(2)to(?)` `implies` No. of photons made incident on atomic disc in one second, `=(piR^(2)xx((n)/(t)))/(4pir^(2))=(1)/(4)xx(R^(2))/(r^(2))xx(n)/(t)` `implies` No. of photons made incident in time `t_(0)`, `=(1)/(4)xx(R^(2))/(r^(2))xx((n)/(t))xxt_(0)` `=(1)/(4)xx((2)^(2))/((1.5xx10^(-10))^(2))xx5.03xx10^(19)xx11.38` (v)Above calculation is based upon the assumption "Photon behaves like a wave".If this assumption is true then we should get emission of photoelectrons,approximately after 11.38 second.But experimentally we get almost instantaneous emission of why we have to believe that in this phenomenon,photons must be behaving like particles. |
|