Photoelectric effect and kinetic energy

In summary, the maximum kinetic energy of the emitted electrons is given by the equation K_{electron} = E_{photon} - W_{metal}, where W_{metal} is the work function of the metal. In the case of aluminum, 4.2 eV is required to move an electron, so the maximum kinetic energy of the emitted electrons will be equal to the energy of the photon (in this case, 2000 Å) minus 4.2 eV. The minimum energy needed to free an electron from aluminum is also known as the work function, and it can be converted to joules by multiplying by 1.602x10^{-19} J/eV.
  • #1
tony873004
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Homework Statement


Light of a wavelength 2000 Å falls on an aluminum surface. In aluminum, 4.2 eV are required to move an electron. What is the kinetic energy of

(a) the fastest, and
(b) the slowest emitted photoelectrons?



Homework Equations



[tex]K_{{\rm{max}}} = eV_0 [/tex]

The Attempt at a Solution


[tex]K_{{\rm{max}}} = eV_0 = - {\rm{1}}{\rm{.602}} \times 10^{ - 19} {\rm{C}} \times 4.2{\rm{eV}} \times {\rm{1}}{\rm{.602}} \times 10^{ - 19} \frac{{\rm{J}}}{{{\rm{eV}}}} = - 1.078 \times 10^{ - 37} {\rm{JC}}[/tex]
I doubt I'm doing this right, by using the required energy as the stopping potential. The units don't work out. But there's nothing in the book, except the formula I listed, that tells me how to do such a problem. The book gives me the impression that the stopping potential is experimentally derived. They give a graph for sodium, with points and a trendline, but it doesn't tell me what it is for aluminum (unless they buried it in an appendix). How do I do this problem?
 
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  • #2
It looks like you need the work function:

[tex] K_{electron} = E_{photon} - W_{metal} [/tex]

where [tex] W_{metal} [/tex] is the minimum energy needed to free an electron from some particular metal.

The joules-electronvolt conversion would be helpful as well:

1 eV = 1.6022 X [tex] 10^{-19} [/tex] J
 
  • #3
wow..i have this same exact problem.
 
  • #4
@ buffordboy; the Work function is [itex]eV_0=4.2 eV[/itex]

In aluminum, 4.2 eV are required to move an electron

@tony; why have you multiplied by the fundamental charge? you are given an energy (4.2eV) not a voltage; to convert it to Joules you only need to multiply it by [itex]1.602x10^{-19} \frac{J}{eV}[/itex]

also, [itex]eV_0[/itex] gives you the work function not the maximum kinetic energy of the photo emitted electrons. The energy of these electrons is coming from the light that is bombarding the aluminum, not from a voltage [itex]V_o[/itex]. What is the maximum amount of energy that an electron will get from 2000 angstrom light? How much of that energy will be leftover kinetic energy after it escapes the aluminum?
 
  • #5

1. What is the photoelectric effect?

The photoelectric effect is the phenomenon in which electrons are emitted from a material when it is exposed to light of a certain frequency.

2. How does the photoelectric effect relate to kinetic energy?

When electrons are emitted from a material due to the photoelectric effect, they gain kinetic energy. This energy is equal to the difference between the energy of the incident photons and the work function of the material.

3. What is the work function?

The work function is the minimum amount of energy needed for an electron to escape from the surface of a material. It is different for each material and is dependent on the properties of the material.

4. What is the significance of the threshold frequency in the photoelectric effect?

The threshold frequency is the minimum frequency of light that can cause the photoelectric effect in a specific material. If the frequency of the incident light is below the threshold frequency, no electrons will be emitted regardless of the intensity of the light.

5. How does the kinetic energy of the emitted electrons change with increasing frequency of incident light?

According to the photoelectric effect equation, the kinetic energy of the emitted electrons increases with increasing frequency of incident light. This is because higher frequency light carries more energy per photon, so each photon can transfer more energy to the electrons in the material.

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