Photoeletric threshold

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In summary: All you need to do is substitute the electron volt for the energy in Joules and you're good to go.In summary, the photoelectric threshold wavelength of a tungsten surface is 272nm, the work function in ElectonVolts of this tungsten is \frac{hc}{\lambda}=5.7 eV, and the maximum kinetic energy in eV of the electrons ejected from this tungsten by ultraviolet surface is Kmax=5.7eV.
  • #1
shakejuhn
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Homework Statement


The photoelectric threshold wavelength of a tungsten surface is 272nm

A) what is the threshold frequency of this tungsten
b) what is the work function in electonVolts of this tungsten
C) Calculate the maximum kinetic energy in eV of the electrons ejected from this tungsten by ultraviolet surface.


Homework Equations


A) f= W/h
B) ?
C) Kmax = hf-W


The Attempt at a Solution



i have no idea how to implement these equations, I am only given the wavelegnth can some one point me in the correct direction and also if anyone know a formula for part B of this problem
 
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  • #2
a) for any wave phenomenon: wavelength = wave speed * frequency
b) I'm not sure what you mean by a "work function".
c) should be easy if you have a)
 
  • #3
For part a, kamerling is right. You can calculate the frequency of the photon directly from its wavelength.

For part b, the work function (W) is a term specific to the photoelectric effect. W has units of energy and represents the amount of "work" required to completely liberate an electron from the metal (in this case tungsten). Since you know the frequency of a photon that *will* accomplish this, you know its energy, therefore you know the work function.
 
  • #4
wave speed = wavelength*frequency
 
  • #5
yeah...good catch.
 
  • #6
(Sorry about the long-winded intro, but you said you had some trouble manipulating the equations; I myself am not good at remembering equations so I find it helpful to keep the main (applicable) concepts in mind and more often than not, that helps me to come to terms with a problem much more easily than digging out seemingly meaningless equations from a book.)

The key thing to remember with the Photoelectric Effect is that one photon gives all of its energy to one electron.

The energy of the photon is given by:

[tex]E = hf = \frac{hc}{\lambda}[/tex]

Considering conservation of energy for the emission:

[tex]E = W + K[/tex]

Where, W is the work done in liberating the electron from the surface and K is the kinetic energy it leaves the surface of the metal with.

The work function (a property of the metal in question) is defined as being the very least energy (of an incoming photon) that will stimulate electron emission; thus, the emitted electron has a kinetic energy of approximately zero. Such a photon is said to be at a threshold frequency, [tex]f_{0}[/tex] (or wavelength [tex]\lambda_{0}[/tex]) and hence it's energy is given by:

[tex]E_{0} = hf_{0} = \frac{hc}{\lambda_{0}}[/tex]

But, since the kinetic energy of the electron here is zero this energy is equal to the work function:

[tex]W = hf_{0} = \frac{hc}{\lambda_{0}}[/tex]

Putting this into the conservation of energy equation above:

[tex]\frac{hc}{\lambda} = \frac{hc}{\lambda_{0}} + K[/tex]

[tex]K = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}}[/tex]




Now, for part (a) you are right; the equation you are using will give you the work function of the metal in units of Joules.

For part (b) all you need to know is the value of the electron volt. The electron volt is defined as the energy gained/lost by an electron when it is accelerated/decelerated by a potential of 1 volt:

[tex]1eV = 1.6 x 10^{-19}J.[/tex]

So, in order to get the work function of the metal in eV, all you need to do is multiply your answer from part (a) by that number.

You are also correct about part (c); your equation is the same as my expression for K.
 
Last edited:

1. What is the photoelectric threshold?

The photoelectric threshold is the minimum amount of energy required to cause an electron to be emitted from a metal surface when it is exposed to light.

2. How is the photoelectric threshold determined?

The photoelectric threshold is determined by measuring the voltage at which electrons begin to be emitted from a metal surface when it is exposed to light of a specific wavelength. This voltage is known as the stopping potential.

3. What is the significance of the photoelectric threshold?

The photoelectric threshold is significant because it provides evidence for the particle nature of light. It also led to the development of the quantum theory and helped to explain the photoelectric effect.

4. How does the photoelectric threshold vary for different metals?

The photoelectric threshold can vary for different metals depending on their physical and chemical properties. Metals with lower work functions (the amount of energy needed to remove an electron from the surface) will have a lower photoelectric threshold.

5. What applications does the photoelectric threshold have?

The photoelectric threshold has a wide range of applications, including solar cells, photodiodes, and photoelectric sensors. It is also used in spectroscopy to determine the energy levels of atoms and molecules.

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