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There is a vacuum photocell whose one electrode is made of cesium and the other of copper. Find the maximum velocity of photoelectrons approaching the copper electrode when the cesium electrode is subjected to electromagnetic radiation of wavelength 0.22 mu m and the electrodes are shorted outside the cell. |
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Answer» Solution :A simple APPLICATION of Einstein's EQUATION `(1)/(2)mv_(max)^(2) = hv-hv_(0) = (2pi cancelh c)/(lambda) -A_(cs)` givens incorrect result in this case because the photoelectrons emitted by the Cesium electroded are retarded by the small electric field that ecists between the cesium electrode and the copper electrode even in the absence of external emf. This small electric field is CAUSED by the contact potential difference whose magnitude equals the difference of work functions `(1)/(e) (A_(cu)-A_(cs))`volts. Its physical origin is explained below: The maximum velocity of the photoelectrons reaching the copper electrode is then `(1)/(2)mv_(m)^(2) = (1)/(2)mv_(0)^(2) - (A_(cu)-A_(cs)) = (2picancelh c)/(lambda) - A_(cu)` Here `v_(0)` is the maximum velocity of the photoelectrons immediately after emission. Putting the VALUES we get, on using `A_(cu) = 4.47eV, lambda = 0.22mu m`, `v_(m) = 6.41 XX 10^(5) m//s` The origin of contact potential difference is the following. Inside the metals free electrons can be through of as a Fermi gas which occupy enegry levels upto a maximum called the Fermi enegry `E_(P)`. The work function `a` measures the depth of the Fermi level.  When two metals `1*2` are in contact, electrons flow one to the other till theri Fermi levels are the same. This require the appearanceof contact potential difference of `A_(1)-A_(2)` between the two metals externally. |
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