Photoelectric effect and determining the planckconstant

In summary, the experiment found that the kinetic energy of expelled electrons is inversely proportional to the frequency of the light used to liberate them. This was determined using the equation Ek=h(f-fw). The relationship between frequency and energy was then explained using the equation h=mx+q.
  • #1
Nikitin
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Homework Statement


In an experiment with the photoelectric effect, the kinetic energy of the expelled electrons was measured. Draw a graph using this data and find the Planckconstant.

Data:
f(1014)hz : 5, 7, 9, 11
Ek(10-18)J: 0.07,0.19,0.3,0.43

Homework Equations


well, Ek + W = hf where Ek is the kinetic energy of the electron, W= the energy needed to liberate it and hf the energy of the photon hitting an atom.

The Attempt at a Solution



As far as I know, one can simplify the above formula to Ek=h(f-fw) where fw is the frequency needed to liberate the electron in the first place. To determine Planck's constant I thought about ΔEk/Δ(f-fw) but I'm a bit confused since Ek isn't proportional to f, but to (f-fw) I'm not sure if that's going to make any difference tho..

So, am I on the right track?

thx 4 all help in advance.
 
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  • #2
Did you draw the graph? That could be helpful.

Anyway, look at the equation you wrote:
[itex]K=h\nu-\phi=h\nu-h\nu_f[/itex]
What's the relationship between the frequency [itex]\nu[/itex] and the energy?
 
  • #3
v=K/h + vf

vf remains constant while K increases, but when v increases by a factor of 10, this doesn't mean that K will increase by a factor of 10...
 
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  • #4
Electron/Positron Impact & Photon Energies

BTW, can you help me with another thing? "An electron and a positron are moving against each other. at the point of impact both have the velocity 0.6c

Two gamma photons are sent out. What is each of theirs energy?"

Well, the sum of momentum equals zero in both cases and thus the momentum of each of the photons must equal [lorentzfactor]*9.11*(10^-31)*1.8*10^8 = 0.615*10^-13. But the book says I'm wrong with the momentum. Please help with this...

You'd better move the second question and write it in a new topic.

Returning to the photoelectric problem,
[itex]K=h\nu-h\nu_f[/itex] is in the same form as [itex]y=mx+q[/itex], where m=h.

What does this suggest you?
 
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  • #5
ah right. thanks, i forgot about that. so h equals the growth rate
 

FAQ: Photoelectric effect and determining the planckconstant

1. What is the photoelectric effect?

The photoelectric effect is the phenomenon where electrons are emitted from a metal surface when it is exposed to light of a certain frequency. This effect was first observed by Heinrich Hertz in 1887.

2. How is the photoelectric effect related to the Planck constant?

The photoelectric effect is directly related to the Planck constant, as it was first explained by Albert Einstein in 1905. Einstein's theory stated that the energy of a single photon of light is directly proportional to its frequency, and the proportionality constant is the Planck constant.

3. How is the Planck constant determined from the photoelectric effect?

The Planck constant can be determined by plotting the stopping voltage of electrons emitted in the photoelectric effect against the frequency of the incident light. The slope of the line on the graph will be equal to the value of the Planck constant.

4. Why is the photoelectric effect important?

The photoelectric effect is important because it provided the first experimental evidence for the existence of photons and the particle-like nature of light. It also led to the development of quantum mechanics and our understanding of the dual nature of light as both a wave and a particle.

5. How does the photoelectric effect impact modern technology?

The photoelectric effect is the basis for many modern technologies, such as solar panels, photodiodes, and digital cameras. It is also used in various scientific instruments, such as spectrometers and electron microscopes. Without our understanding of the photoelectric effect and the Planck constant, many of these technologies would not be possible.

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