GR8677 #31


Problem



Quantum Mechanics}Photoelectric Effect
Perhaps one recalls the ``hot phrase" stopping potential. The used in the photoelectric equation is essentially the stopping potential (energy). This memoryrecall immediately narrows the choices down to just (A) and (C). Now, to find out if the potential is positive or negative.
FYI: The Photoelectric effect equation is basically just conservation of energy. One has , where is the kinetic energy of the electron as it accelerates through the medium between the cathode and collector, is the energy to free the electron from the metal, and is the energy given by the light source. Essentially, the energy from the photon first frees the electron from a sort of (valence) electron sea it lives in while in the metal, and then the excess energy propels it from cathode to collector in order to keep the current running. The minimum kinetic energy required to get the current going is the stopping potential (energy) . (According to the fight analogy below, the photon is the one who starts the fight, socking the electron off and away across the currents...)
The electron charge is negative and the potential must be a negative quantity in order to make positive overall.
(On the lighter side... The Photoelectric effect is also related to the Compton Effect. The effects can be seen as an arena fight. Think WWWF, but with electrons and photons. In PEE, the photon knocks the electron out, while in CE, the electron retaliates! It knocks the photon off course.)


Alternate Solutions 
Jovensky 20130214 01:11:57  The photon energy is hf and the work function of the photocathode is W. To knock electrons out of the photocathode, the photon energy has to be larger than W. The excess energy will contribute to the kinetic energy of the electron.
If there is zero potential between photocathode and collector, then the electron will just go to collector and generate a current. To counter this, we can generate a force opposing the electron moving direction. This force is given by F=qE. Since this force has points from collector to photocathode and q here is negative. then E has to point from photocathode to collector to make the force points from collector to photocathode. And a useful fact to remember is that electric field points in the direction of decreasing potential. Thus, the collector has to be at a lower potential than photocathode. Since initially the potential is zero, thus, to decrease collector potential, it has to be a negative value.
We can keep on decreasing the potential until at a point, the force is strong enough to repel the electron with the highest kinetic energy and that is the V (stopping potential) in the equation.  

Comments 
Jovensky 20130214 01:11:57  The photon energy is hf and the work function of the photocathode is W. To knock electrons out of the photocathode, the photon energy has to be larger than W. The excess energy will contribute to the kinetic energy of the electron.
If there is zero potential between photocathode and collector, then the electron will just go to collector and generate a current. To counter this, we can generate a force opposing the electron moving direction. This force is given by F=qE. Since this force has points from collector to photocathode and q here is negative. then E has to point from photocathode to collector to make the force points from collector to photocathode. And a useful fact to remember is that electric field points in the direction of decreasing potential. Thus, the collector has to be at a lower potential than photocathode. Since initially the potential is zero, thus, to decrease collector potential, it has to be a negative value.
We can keep on decreasing the potential until at a point, the force is strong enough to repel the electron with the highest kinetic energy and that is the V (stopping potential) in the equation.   Jovensky 20130214 01:11:28  The photon energy is hf and the work function of the photocathode is W. To knock electrons out of the photocathode, the photon energy has to be larger than W. The excess energy will contribute to the kinetic energy of the electron.rnrnIf there is zero potential between photocathode and collector, then the electron will just go to collector and generate a current. To counter this, we can generate a force opposing the electron moving direction. This force is given by F=qE. Since this force has points from collector to photocathode and q here is negative. then E has to point from photocathode to collector to make the force points from collector to photocathode. And a useful fact to remember is that electric field points in the direction of decreasing potential. Thus, the collector has to be at a lower potential than photocathode. Since initially the potential is zero, thus, to decrease collector potential, it has to be a negative value. rnrnWe can keep on decreasing the potential until at a point, the force is strong enough to repel the electron with the highest kinetic energy and that is the V (stopping potential) in the equation.   spacemanERAU 20091015 09:10:35  Why does eV have to be positive overall? doesnt the absolute value take care of that?
kroner 20091016 10:43:22 
There is no "negative value of at which the current stops." Any negative voltage would attract all the free electrons so there would always be a current.

his dudeness 20100818 13:12:00 
kroner, the situation is the exact opposite of what you said. If point B has a negative voltage with respect to point A, then the electric field points from A>B. This means that positive charges want to drop in voltage and electrons want to go up in voltage. So we need the detector to be at a negative value that's sufficiently high to stop the electron current which arises naturally (even at zero voltage). Tricky stuff!

  FortranMan 20081019 15:50:26  Was the question in error in saying hv instead of hf?
FortranMan 20081019 16:11:28 
my bad, latex fooled me into thinking it was v, and so did the test >:(

  phys2718 20081016 10:58:57  Logically, the value of the potential at which the current stops is the same as the value of the potential at which the current starts, depending only on whether you approach it from above or below. This is a stupid question.
justguessing 20091004 06:43:05 
right, but it says in the question "(...) is initially zero and is then (...) decreased" ... so you know youre coming from the right side. I missed that one, too; because I'm used to read from left to right. Totally agree: Stupid question!

mpdude8 20120415 20:57:23 
I don't think we'll see anything like this on a current GRE.

fermion 20120707 16:33:48 
@justguessing
Actually, it says monotonically increased or decreased.

cgrman1 20150410 07:35:55 
They mean the voltage at which there is NO current, not where the current starts or stops in the setup

cgrman1 20150410 08:02:37 
see Jovensky's solution

  erc 20051103 14:50:58  Do you perhaps mean "narrows the answer choices down to (A) or (C)" (i.e. not B)?
Thank you for this exceedingly useful site! Please, please post GR0177 solutions in the next few days!
yosun 20051103 20:07:56 
erc: thanks for the typo note! It has been corrected.

 

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