Quantum mechanically, I'm not sure what would be different..

[bon]

Homework Statement

If the energy flux associated with a light beam of wavelength 3 x10^-7 m is 10 W/m^2, estimate how long it would take, classically, for sufficient energy to arrive at a potassium atom of radius 2 x 10^-10 m in order that an electron be ejected. What would be the average emission rate of photoelectrons if such light fell on a piece of potassium 10^-3 m^2 in area? Would you expect your answer to the latter questions to be significantly affected by quantum mechanical considerations?

Homework Equations

The Attempt at a Solution

Ok so I'm guessing we can assume the work function of potassium to be 3 x 10^-19 J (calculated in answer to a previous question)

So thinking about the first question, I'm just not sure how to work it out..

So classically, we imagine photons are arriving at the surface of this potassium atom at 10 W/m^2, so how do i work out energy arriving at the atom per second? Do I work out the surface area of the atom, and assume light is arriving at all directions?

 

[hikaru1221]

In classical physics, we don't have the notion of "quantum packet" (or photon). Wave is something continuous, and energy is transmitted continuously. Electron can be excited under any wavelength, as long as there is enough energy "received". You can assume that the effective area of the atom to receive energy is pi*r^2 where r is atomic radius (Just imagine that the atom is a sphere; there is a light beam coming onto it, then how much power does it receive?). You have the power of the light beam per area & the energy required to excite an electron, you can calculate the needed time in classical viewpoint.

 

[bon]

Thanks for your help.

Why can we assume to effective area to receive energy is pi r^2? The total surface area is 4 pi r^2.. so why would it be a quarter of this value? What if the light beam was much wider than the atom - surely the max effective area would be 2pi r^2?

Also...is the energy needed to excite an electron just the work function of the atom?

 

[hikaru1221]

Thanks for your help.

Why can we assume to effective area to receive energy is pi r^2? The total surface area is 4 pi r^2.. so why would it be a quarter of this value? What if the light beam was much wider than the atom - surely the max effective area would be 2pi r^2?

Light beam, as I understand, is a straight beam propagating in only one direction (see the 1st picture). Laser is an example.
Then have a look at this: http://en.wikipedia.org/wiki/Flux and this: http://en.wikipedia.org/wiki/Poynting_flux
Although those are not relevant, they may help you understand how the sphere receives the energy. Anyway, look at the 2nd picture. There are 2 surfaces "catching" the flow of energy (or energy flux, or energy per unit area). Intuitively we can see that they catch the same amount of flux, though their areas are different. So however slanted the surface is, the real effective area is only the one that faces directly and perpendicularly to the flow. In the case of a sphere, that's surely be pi*r^2.

Also...is the energy needed to excite an electron just the work function of the atom?

I think so. As you may know, potassium is an active metal which can easily detach its most outermost electron, but holds the other electrons quite firmly. Have a look at the ionization energy of potassium: http://en.wikipedia.org/wiki/Ionization_energies_of_the_elements. Therefore, we can assume that only the outermost electron can be excited. The energy to excite that electron corresponds to the work function of potassium.

 

[bon]

Light beam, as I understand, is a straight beam propagating in only one direction (see the 1st picture). Laser is an example.
Then have a look at this: http://en.wikipedia.org/wiki/Flux and this: http://en.wikipedia.org/wiki/Poynting_flux
Although those are not relevant, they may help you understand how the sphere receives the energy. Anyway, look at the 2nd picture. There are 2 surfaces "catching" the flow of energy (or energy flux, or energy per unit area). Intuitively we can see that they catch the same amount of flux, though their areas are different. So however slanted the surface is, the real effective area is only the one that faces directly and perpendicularly to the flow. In the case of a sphere, that's surely be pi*r^2.



I think so. As you may know, potassium is an active metal which can easily detach its most outermost electron, but holds the other electrons quite firmly. Have a look at the ionization energy of potassium: http://en.wikipedia.org/wiki/Ionization_energies_of_the_elements. Therefore, we can assume that only the outermost electron can be excited. The energy to excite that electron corresponds to the work function of potassium.

Thank you so much for such a comprehensive reply!

I understand now that the effective area is only the one that faces perpendicularly to the flow. I'm just curious as to what the working would be to show that this is pi r^2?

Thank you again.

 

[granpa]

its just the area of a circle of radius equal to the spheres radius

 

[bon]

its just the area of a circle of radius equal to the spheres radius

Yeh I realized that - I just wasn't sure as to the derivation of why that is the area perp. to the beam..I know this one is quite easy to see just by visualising it, but want to learn the method for more complicated examples...

 

[hikaru1221]

Yeh I realized that - I just wasn't sure as to the derivation of why that is the area perp. to the beam..I know this one is quite easy to see just by visualising it, but want to learn the method for more complicated examples...

Hehe, I'm not good at math, so I don't know if there is a general method. You can find the area of the projection of the whole surface/thing onto a plane which is perpendicular to the propagation direction of the light beam.

In the case of a sphere, regardless of direction, the projection of the sphere onto ANY plane is a circle of the same radius.

Hope that helps

 

[bon]

Ok thanks. So I worked out the emission rate to be 0.24s, is this right?

Now asked: What would be the average emission rate of photoelectrons if such light fell on a piece of potassium 10^-3 m^2 in area?

Not sure how to do this? Do i imagine the area is a square and see how many potassium atoms i'd be able to fit in? (I work this out to be 6.25 x 10^16). Thus one is emitted every 0.24s/6.25 x 10^6 = every 3.84 x 10^-18 s..?

This doesn't seem right to me.. Any help would be great please.. :S

 

[bon]

Hehe, I'm not good at math, so I don't know if there is a general method. You can find the area of the projection of the whole surface/thing onto a plane which is perpendicular to the propagation direction of the light beam.

In the case of a sphere, regardless of direction, the projection of the sphere onto ANY plane is a circle of the same radius.

Hope that helps

Anyone? (Sorry hikaru I quoted you in case you see this, because you helped so much last time!)

 

[hikaru1221]

Ok thanks. So I worked out the emission rate to be 0.24s, is this right?

Yep. I got that too.

Now asked: What would be the average emission rate of photoelectrons if such light fell on a piece of potassium 10^-3 m^2 in area?

Not sure how to do this? Do i imagine the area is a square and see how many potassium atoms i'd be able to fit in? (I work this out to be 6.25 x 10^16). Thus one is emitted every 0.24s/6.25 x 10^6 = every 3.84 x 10^-18 s..?

This doesn't seem right to me.. Any help would be great please.. :S

So the assumption here is that only those atoms on the surface of the potassium piece absorb all the light (I'm not sure about this; scattering must be taken into account, even if it's under classical view, but let's just assume so). If so, then the result is not so surprising to me, though it conflicts with the fact that the average time for an electron to stay in the excited state is around 10^-8 s. First, in our model, each atom receives energy from light wave GRADUALLY, but all the atoms receive light simultaneously, so the situation will be that they emit electrons at the same time! That means, the number 3.84 x 10^-18 s here is quite meaningless; in our model, one atom needs 0.24s to absorb enough energy. Moreover, due to the above assumption, our model turns out to be not so true at all, so again the result doesn't show anything.
Anyway I'm not good at these topics, so don't rely much on my reply Just my 2 cents