B The wave solution to the Photoelectric effect

[in2infinity]

TL;DR Summary

(λW)² = c/2

I was recently examining the relationship between the work function of a material and its threshold wavelength. It was clear to me that the relationship is expressed as:

(λW)² = c/2

Where λ is the threshold wavelength, W is the work function, and c is the speed of light. However, I am unable to find the name of this solution online.

Does anyone know what this relationship is called, and a link that explains it?

Thanks in advance for your help.

 

[Vanadium 50]

Not every equation has a name.

 

[in2infinity]

Thanks.

But do you know where I can find the equation in the literature. As this provides a wave solution to the photoelectric effect, I am sure it must be an important concept in science, right?

I mean, regardless of the name, I just want to find out more about it.

 

[gleem]

(λW)² = c/2 is dimensionally inconsistent.

 

[Vanadium 50]

Goof point: but one can replace f with c/lambda or whatever, Doesn't need a name,,

 

[Doc Al]

TL;DR Summary: (λW)² = c/2

I was recently examining the relationship between the work function of a material and its threshold wavelength.

The work function would be the energy of a photon with that threshold wavelength:
W = hc/λ

 

[gleem]

(λW)² = (hc)² ≠ c/2

 

[in2infinity]

OK, maybe I should supply some data. here is a list of some of the elements i checked.

Element	W (Ev)	λ	Multiplied
Cs 55 (s)	212	580	122.9
Na 11 (s)	227	540	122.5
Ba 56 (s)	251	490	122.9
Mg 12 (s)	346	350	121.1
Zn 30 (D)	374	330	123.4
Al 14 (P)	374	330	123.4
Pb 82 (P)	402	300	120.6

Now the result is squared and multiplied by 2 to get the speed of light. I think that is relativly straight forward.

Can you explain what you mean by 'dimensionally inconsistent'?

 

[gleem]

Basically, you had apples equal to oranges

 

[in2infinity]

Right sorry, i d

(λW)² = (hc)² ≠ c/2

Sorry, I didn't have that sign on my laptop. We can go with that. :-)

 

[in2infinity]

Sorry, I am using Electronvolts, so you need to divide by 1.6 to get the work function in EVs.

The work function would be the energy of a photon with that threshold wavelength:
W = hc/λ

see that example dataset I posted

 

[Doc Al]

Sorry, I am using Electronvolts

Use whatever units you like. The work function is an energy term, so eV is fine.

 

[Ibix]

This is just numerology, isn't it? If you take the quoted energies in eV and multiply them by the wavelengths in nm (post #8), divide by 1.6 because reasons (post #11), then square the result and divide by 2 you do get somewhere near the numerical value of $c$ in m/s. But that's just because OP has calculated the constant $(hc)^2/2$ in some funky units and it happens to have a numerical value near to the numerical value of $c$ in SI.

 

[in2infinity]

This is just numerology, isn't it? If you take the quoted energies in eV and multiply them by the wavelengths in nm (post #8), divide by 1.6 because reasons (post #11), then square the result and divide by 2 you do get somewhere near the numerical value of $c$ in m/s. But that's just because OP has calculated the constant $(hc)^2/2$ in some funky units and it happens to have a numerical value near to the numerical value of $c$ in SI.

I am sorry, by 1.6, I mean the electrical charge constant e. Those are the values I am showing in the data set. When you multiply the threshold wavelength, by the work function in EVs and square the result is very nearly is equal to half the speed of light. I don't think it's numerology.

It seems like a fairly straight forwards calculation. Although, I am not exactly sure what you mean by numerology?

 

[Ibix]

I am sorry, by 1.6, I mean the electrical charge constant e.

In what units? And why are you dividing by that anyway?

 

[Ibix]

And can you write down your exact process? Taking your first line of data for Cs you take W=212eV and $\lambda$=580nm (I guess) how do you get the 122.9 in the last column in your table? What do you get when you square that value? When do you divide by 1.6 and when do you divide by 2? In short, please show every step of your calculation for one example. And please state in what units is your final answer.

 

[in2infinity]

Sure

2.12 EV * 580nm = 1229.60

122960^2 = 1511916

1511916 x 2 = 3023832

3023832 / 299792 = 10.086434

I scaled the answer as I originally used it to make a graph.

This concept of 1.6 is because in the equation W= c*h / f provided previously need to be divided by e (roughly 1.6) to produce the work function in EVs not E.

Therefore, W = (c*h/f)/e

I wasn't expecting that this was going to be such a big deal, by the way. I just assumed this was in the litterature somewhere.

And can you write down your exact process? Taking your first line of data for Cs you take W=212eV and $\lambda$=580nm (I guess) how do you get the 122.9 in the last column in your table? What do you get when you square that value? When do you divide by 1.6 and when do you divide by 2? In short, please show every step of your calculation for one example. And please state in what units is your final answer.

 

[topsquark]

Sure

2.12 EV * 580nm = 1229.60

122960^2 = 1511916

1511916 x 2 = 3023832

3023832 / 299792 = 10.086434

I scaled the answer as I originally used it to make a graph.

This concept of 1.6 is because in the equation W= c*h / f provided previously need to be divided by e (roughly 1.6) to produce the work function in EVs not E.

Therefore, W = (c*h/f)/e

I wasn't expecting that this was going to be such a big deal, by the way. I just assumed this was in the litterature somewhere.

Check back on PHF, here. Post #4.

-Dan

 

[Ibix]

So the process you've followed calculates $2(hc)^2$ in funky units. The numerical value in those funky units happens to be in the ballpark of the numerical value of $c$ in SI units (or ten times it - you don't seem to be clear). This is a meaningless coincidence (and dimensionally inconsistent as already pointed out), which is what I mean by "this is just numerology".

 

[in2infinity]

So the process you've followed calculates $2(hc)^2$ in funky units. The numerical value in those funky units happens to be in the ballpark of the numerical value of $c$ in SI units (or ten times it - you don't seem to be clear). This is a meaningless coincidence (and dimensionally inconsistent as already pointed out), which is what I mean by "this is just numerology".

By your response, I assume then that this appears not to be in the literature. Which was actually my original question. And no, I did not use h in the calculation. I was only explaining that Doc Al's difference in expected value of E from eV.

Although that does make me think. Doesn't the value for h have no mathematical explanation. I mean, isn't it just derived from experimentation?

btw: Does anyone actually know of a proper reference for the experimental values of work function and elements?

I really appreaciate all of the view that are being presented here. Thanks

 

[berkeman]

I don't think it's numerology.

Thread closed for Moderation...

 

[DrClaude]

This is numerology: playing around with numbers until an "interesting" result is obtained. The problem is that this can be done with any set of numbers so generally doesn't lead to any real insight.

Note also that some numbers in the OP may be incorrect (see http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/photoelec.html), leading to even more spurious coincidences.

This thread will remained closed.