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Verbatim question for GR8677 #31
Quantum Mechanics}Photoelectric Effect

Perhaps one recalls the ``hot phrase" stopping potential. The |eV| used in the photoelectric equation is essentially the stopping potential (energy). This memory-recall 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 eV+W=hf, where eV is the kinetic energy of the electron as it accelerates through the medium between the cathode and collector, W is the energy to free the electron from the metal, and hf 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) eV. (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 V must be a negative quantity in order to make eV 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.)

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
Jovensky
2013-02-14 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.
Alternate Solution - Unverified
Comments
Jovensky
2013-02-14 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.
Alternate Solution - Unverified
Jovensky
2013-02-14 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. NEC
spacemanERAU
2009-10-15 09:10:35
Why does eV have to be positive overall? doesnt the absolute value take care of that?
kroner
2009-10-16 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
2010-08-18 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!
NEC
FortranMan
2008-10-19 15:50:26
Was the question in error in saying hv instead of hf?
FortranMan
2008-10-19 16:11:28
my bad, latex fooled me into thinking it was v, and so did the test >:(
Answered Question!
phys2718
2008-10-16 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
2009-10-04 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
2012-04-15 20:57:23
I don't think we'll see anything like this on a current GRE.
fermi-on
2012-07-07 16:33:48
@justguessing

Actually, it says monotonically increased or decreased.
cgrman1
2015-04-10 07:35:55
They mean the voltage at which there is NO current, not where the current starts or stops in the set-up
cgrman1
2015-04-10 08:02:37
see Jovensky's solution


NEC
erc
2005-11-03 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
2005-11-03 20:07:56
erc: thanks for the typo note! It has been corrected.
Fixed Typos!

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