Photoelectric effect question (1 Viewer)

[moonpie91]

its about photocurrent produced by the photoelectric effect

assuming the original emr had enough energy to cause the photoelectric effect

wud the same emr (same intensity) with higher frequency increase or decrease the photocurrent?

one book says higher frequency wud result in less photocurrent as increasing

the frequency results in less photons per second (it also says: however they

do release photoelectrons with a greater kinetic energy)

and another says higher frequency wud increase the photocurrent as

electrons are emitted with increased kinetic energy

help plz

 

[darkchild69]

its about photocurrent produced by the photoelectric effect

assuming the original emr had enough energy to cause the photoelectric effect

wud the same emr (same intensity) with higher frequency increase or decrease the photocurrent?

one book says higher frequency wud result in less photocurrent as increasing

the frequency results in less photons per second (it also says: however they

do release photoelectrons with a greater kinetic energy)

and another says higher frequency wud increase the photocurrent as

electrons are emitted with increased kinetic energy

help plz

Good question. The answer depends on the definition of "intensity" -- both of which are kinda correct

I would think that "intensity" has been defined differently in the two texts.

For our sakes "Intensity" implies number of photons per unit time.

Example:
Wavelength A (lower wavelength) has intensity x incident on the metal surface

Wavelength B (higher wavelength)has intensity x incident on the metal surface

this means both wavelengths have the same number of photons hitting the surface of the metal per unit time.

So if wavelength A has 100 photons striking the metal surface, all of which created a photoelectron, wavelength B (which has higher wavelength) has the same number of photons striking the metal surface, creating the same number of photoelectrons.

Now:

We know that KEmax = hf - w

KEmax = maximum kinetic energy of the emitted photoelectron
hf = energy of the incident photon
w = work function of the metal

Now, the higher frequency wave (wave B) has a higher energy (hf), so hence will have a greater KEmax

This higher KEmax will result in the electron moving faster.

Remember, we define current as the rate of flow of charge through a conductor.

Now if the electrons are moving faster, then the rate of flow of charge will increase (the number of charge carrying particles passing a point in a second is increasing)

Hence, photocurrent will increase


Text A which says "equal intensity at higher frequency will result in less photons per second" has defined "intensity" in this case wrong. I think they are referring to "irradiance" (which at times is used interchangably with intensity! AAAAAAAArggggggh!)

They are referring to intensity as the rate of energy being emitted from a source being constant, which is wrong. Don't get me started on astrophysicists using intensity and radiance interchangably!

 

[darkchild69]

By the way, i forgot to mention that there is an equation for photocurrent.

I can't remember it off the top of my head, but in essence, the photocurrent generated depends on a few variables:

1. Optical power
2. Absorption coefficient
3. Reflection coefficient
4. Frequency of light

Photocurrent is proportional to optical power and inversely proportional to frequency of light

I just cant remember how they define optical power..

 

[moonpie91]

thanks for the reply first of all

but both texts are referring to the change in the incident light

here is the actual question tat started my confusion

A beam of monochromatic light is shone onto a device to study the photoelectric effect.
In the first part of the experiment, while the light source is on, the microammeter records a steady current.
While keeping all other factors constant, the light source was altered by:
- increasing the intensity of the original monochromatic light source;
- increasing the wavelength of the monochromatic light source;
- increasing the frequency of the monochromatic light source.

How many of the factors changed would have resulted in the microammeter showing a greater current than in the original part of the experiment?

the answer says only increasing the intensity will produce more photons per second to release electrons to produce the photocurrent. Note that increasing the frequency of the light, while maintaining a constant intensity, actually results in less photons per second, as each photon carries more energy. However, they do release photoelectrons with a greater kinetic energy.

am i misunderstanding something or is this book crap?

 

[darkchild69]

Yeah, you see that book is incorrectly defining "intensity" as radiant energy per unit time.

It would really depend on if we are defining "light" as a wave or a particle

Intensity of light when regarding it as a wave would be defined as radiant energy per unit time or the integral of radiant flux over time.

The energy of the monochromatic wave as stated in the book would be proportional to the intensity, hence if you had two wave sources of different frequencies, but same intensity, the one with higher frequency would contain less photons due to the fact that each individual photons contains more energy.

I have e-mailed a university lecturer on this question and will not rest until he provides his opinion

It is quite a simple question, but definately open to debate as you can define intensity a number of different ways! i.e., radiant intensity, irradiance, radiant exitance some of which are different to the others!

Also, different branches of physics define intensity differently, i.e., astrophysics, radiometery, photometry

 

[moonpie91]

thanks again

so the answer really depends on what the question means by the 'intensity'?

if the intensity used here means radiant energy per unit time, then higher frequency wud have less photons, thus less photocurrent?

and if the intensity means the number of photons in the light, then higher frequency wud produce higher photocurrent as photoelectrons produced wud have more kinetic energy (tho the number of photoelectrons produced is the same) ?

have i understood correctly about wat u explained?

damn if this question comes up in the hsc, what am i supposed to say :angry:

 

[darkchild69]

Yep, u have understood

If they ask that question in the HSC i would start my answer by stating "Assuming that intensity is referring to..." then go rom there

 

[moonpie91]

Thank you

But im screwed if its aksed in multiple choice. lol

I'll just have to assume the idensity as the number of photons in a wave then

thanks anyway

 

[hayner]

Alright...so there is an increase in the incident frequency but NOT the intensity right?

so the photoelectric effect requires the quantisation of light. So...the intensity is not the strength of the incident light (like in wave theory) but it is how many photons that are hitting the metal surface...

According to einstein and planck

hf=hfo + Kinetic Energy

Therefore, increase in incident frequency will increase to kinetic energy and have no effect on the photocurrent!
so the book is right because the greater the 'intensity' of the incident light, the more photons can hit the metal and the more electrons that can be ejected, causing a greater photocurrent


Hope that helps

 

[darkchild69]

Therefore, increase in incident frequency will increase to kinetic energy and have no effect on the photocurrent!
so the book is right because the greater the 'intensity' of the incident light, the more photons can hit the metal and the more electrons that can be ejected, causing a greater photocurrent


Hope that helps

Question i pose to you.

Seeing as Current is defined as the amount of "charge" past a point per second.

If you increase the frequency of the incident EMR, thus increasing KE of electron, thus increasing speed of the electron, there will be more electrons passing a defined point per second as they are moving faster --> is this not the definition of current?

I wish the uni lecturer would reply soon

 

[hayner]

...well, if you define current as the number of electrons passing a certain point...then if the electrons have an increased velocity, the number of electrons passing that 'point' is the same except they're travelling faster...so the photocurrent will still be the same if you increase the frequency/wavelength...right?

 

[darkchild69]

...well, if you define current as the number of electrons passing a certain point...then if the electrons have an increased velocity, the number of electrons passing that 'point' is the same except they're travelling faster...so the photocurrent will still be the same if you increase the frequency/wavelength...right?

Not electrons passing a point, rather electrons passing a point per unit time.

Imagine u r at the finish line of a quarter mile drag strip. Assume the drag car accelerates to max speed instantaneously. Left hand lane had car a, which does quarter mile in 15 secs, right is car b which travels twice as fast. It is set up so there is an infinite supply of car a in left and car b in right. A car leaves the start line each second. You stay at the finish line for a minute. For every one car a which passes, there will be two car b's.

Apply that to this situation

Faster electrons corresponds to more charge passing a point each unit time, hence higher current

 

[darkchild69]

Not electrons passing a point, rather electrons passing a point per unit time.

Imagine u r at the finish line of a quarter mile drag strip. Assume the drag car accelerates to max speed instantaneously. Left hand lane had car a, which does quarter mile in 15 secs, right is car b which travels twice as fast. It is set up so there is an infinite supply of car a in left and car b in right. A car leaves the start line each second. You stay at the finish line for a minute. For every one car a which passes, there will be two car b's.

Apply that to this situation

Faster electrons corresponds to more charge passing a point each unit time, hence higher current

This. Attempted example has now succeeded in confusing me further..

 

[darkchild69]

Found the following equation in a reliable book:



where



n = quantum efficiency
e = charge on electron
Io = intensity
A = surface area of detector
phi = incident photon flux density (# photons per sqcm per s)

So what this says for equal intensity light (which has been defined as radiant energy per unit time):
- the higher the frequency, the lower the incident photon flux density (i.e., less photons arriving at surface), the lower the photocurrent.

Still not comfortable with this definition of intensity though...

 

[mmo]

thanks for the reply first of all

but both texts are referring to the change in the incident light

here is the actual question tat started my confusion

A beam of monochromatic light is shone onto a device to study the photoelectric effect.
In the first part of the experiment, while the light source is on, the microammeter records a steady current.
While keeping all other factors constant, the light source was altered by:
- increasing the intensity of the original monochromatic light source;
- increasing the wavelength of the monochromatic light source;
- increasing the frequency of the monochromatic light source.

How many of the factors changed would have resulted in the microammeter showing a greater current than in the original part of the experiment?

the answer says only increasing the intensity will produce more photons per second to release electrons to produce the photocurrent. Note that increasing the frequency of the light, while maintaining a constant intensity, actually results in less photons per second, as each photon carries more energy. However, they do release photoelectrons with a greater kinetic energy.

am i misunderstanding something or is this book crap?

This is a very good question:
because it is targeted at a band 5-6 student.
I brought this up with my student not long ago.
now with this question I don't think the main intention of asking it is about the real definition of intensity, instead it wants you to find out what are the conditions at which the photocurrent is increased.

u see in this question for the photoelectric effect to be analyzed completely u need to understand that the the electrons will not be deflected due to the increasing frequency. according to Lenard's analysis, he found that under a certain frequency u could not increase the kinetic energy and so it was the hypothesis that if the "intensity increased so did the photocurrent". in this case there is a critical frequency cutoff that must be met for the photocurrent to be increased.

So the only factor that could increase the photocurrent is the increased intensity rather than the frequency(ONLY this question requires such an analysis) higher frequency does have an impact but since there is no definitive contradiction in the stem of the question it is the only valid conclusion.

 

[hayner]

Imagine u r at the finish line of a quarter mile drag strip. Assume the drag car accelerates to max speed instantaneously. Left hand lane had car a, which does quarter mile in 15 secs, right is car b which travels twice as fast. It is set up so there is an infinite supply of car a in left and car b in right. A car leaves the start line each second. You stay at the finish line for a minute. For every one car a which passes, there will be two car b's.

I see where you're coming from but in the case of the photoelectric effect I don't think that you have an infinite supply of electrons and in the case of you example, cars. The number of electrons ejected is dependent upon how many photons hit the surface of the metal so...technically, in case of your example, two car b's can only pass the line if two photons hit metal b and in the case of car 'a', only one photon has hit so car 'b' still has a greater intensity therefore, greater photocurrent. (hmm....wow, the cars did complicate things).....

 

[moonpie91]

ok, thanks for your applies

but can we make it clear?

it seems like everyone agrees on the fact that increased intensity of light hitting a metal increases emitted photocurrent

what about increased frequency of light hitting a metal?

does it increase or have no effect (no difference) on produced photocurrent?

 

[homijoe]

I believe both an increase in intensity and frequency would yield a larger photocurrent:

Increase in intensity; would mean more electrons are being librerated as there are more photons bombarding the metal surface. Since current is a flow of electrons, the more electrons that a liberated the greater the amount of electrons that are flowing hence the magnitude of the current would be greater.

Correct me if i am wrong in this.

Increase in frequency: since K.E=hf-@ The greater the frequency of the incident photon the greater the kinetic energy of the photo-electron hence the current would be larger.