Photocurrent VS stopping voltage and intensity of light (1 Viewer)

[CHUDYMASTER]

If you look at a graph of photocurrent (emitted obviously during the photoelectric effect) VS stopping/retarding voltage, you'll find that the photocurrent (under the constant light frequency) is constant upto a critical stopping voltage where it just falls to zero. (you may find this graph in Jacaranda). Now I don't get how this is possible. Wouldn't you expect the rate of emitted electrons to go down as you RETARD them???



Furthermore, I'm confused about the difference between intensity of light and frequency of light. Now I know that one packet of photons has an energy of E = hf, but what makes some light more intense than other light at a quantum level? From here, I need an explanation of why frequency affects emission of photoelectrions but intensity doesn't.


Lord help us all. (but particularly me)

 

[Constip8edSkunk]

1st part, i think the passage in jacaranda refers to where lenard attempts to measure the energy of the photelectrons are related to the frequency of the light projected on to the cathose. He does this by connecting the cathode ray tube to a DC powersource so that the current produced from the power source is opposite to the direction of the conventional current produced by the photoelectric effect. ie the positive terminode of teh power source is connected to the cathode from which the photelectrons are emitted and the negative electrode is connected to teh anode where the photoelctrons will be collected. At a certain voltage, the current will be zero as the photoelectrons no longer have enough energy to reach the collector , a movement resisting the electric field. by varying teh light's frequency and intensity and analysing the varying resulting stopping voltage he was able to determine that the only frequency not intensity determines the stopping voltage, intensity determines the size of teh photocurrent

2nd part, Intensity of light = amplitude of the wave = number of photons

greater intensity = greater amplitude=greater number

frequency of light = the energy value of each photon

greater frequency = more energy in each photon

Einsteins explaination of the photo electric effect is that imagine each atom absorbs 1 photon. now the atom require a certain amount of energy b4 an electron can be released. As a photons energy is varied by its frequency, once the threshold frequency is reached the electron may be released. Intensity determines the number of photons ther4 greater intensity = more photons = more electrons(if the frewuency is high enuf) = greater curernt
as the energy of each photon exceeds the energy required for an atom 2 lose 1 electron the remaining energy is converrted 2 kinetic energy, thus the frequency past teh threshold frequency is proportional to the kinectic energy of teh phtoelectrons

 

[inasero]

as always john is trying to confuse us with his big words...so I shall restate concisely...

Fist point: YES you would expect "the rate of emitted electrons to go down as you RETARD them"...BUT, that's only according to the classical definition of physics, which has continuous energies...the reason why we get that curve experimentally is because of the fact that the energy is quantised ie. in discrete "packets". This is the predicament einstein was on about in his "all or nothing" principle...that you can't simply reduce the input voltage by a tiny notch and expect the current rate to drop proportionally...if i used an analogy...i guess it would be like saying cant I hold a light switch on it's halfway position if i wanted cool ambient lighting for that romantic dinner date? Haha obviously not, cos the switch is either on or off. If you hold it halfway, you'll find that the light turns off completely, which is deifinitely not cool...

Second point: Intensity is how bright the light is, related to number pf photons. Frequency on the other hand if the type of photons which are there in the first place- be it red or green or purple, u name it.
So I guess you could sum it up by saying- Frequency=Quantity, Intensity=Quality

Hope I've cleared up some issues there

 

[Halo]

The stopping voltage retards the photoelectron, not the photon. The rate of emission shouldnt be affected? (Im not sure)

The intensity of light is the number of photons. Thus, intensity affects how many photoelectrons are emitted (remember one photon can only dislodge one electron - 'all or nothing'). This photoemission can only happen if each photon carries sufficient energy to knock electrons out. And this is dependent on frequency, since E = h f. Once a photon's frequency (and hence energy) is greater than the threshold frequency (on collision with an electron), the excess energy transferred to the electron in the form of kinetic energy.

 

[inasero]

errr halo...i think not...the all or nothing principle is that you can't simply reduce the input voltage by a tiny notch and expect the current rate to drop proportionally(refer to the above post)...get it right

 

[Dangar]

Quote: So I guess you could sum it up by saying- Frequency=Quantity, Intensity=Quality


Uh..inasero i think thats the wrong way around. The greater the intensity the greater the quantity of photoelectrons emitted and the higher the frequency the greater the kinetic energy of each photoelectron emitted.
my teacher used this analogy for the way stoppping voltage is affected by intensity and frequency...it works well for me:
If you have five people (electrons) running, it will take a certain strength of wind (voltage) to stop them. If you increase the number of people to ten people running, the same strength wind will still stop them. i.e if you increase the intesity, the stopping voltage remains the same. However if you increase the speed at which they are running, a stronger wind is needed to stop them. i.e. by increasing the frequency, you are increasing the kinetic energy of each electron which means a greater stopping voltage.