(Photoelectric effect) Finding velocity of ejected electron given wavelengths

Aleguan

Hey, I'm in my senior year of highschool and I'm stuck on this problem that deals with things I haven't even learned (Canadian education system for ya). It has to do with the photoelectric effect.

Given the wavelengths of an X-ray before and after it strikes an unknown metal, (initial=1.38e-11, final=1.54e-11), I have to determine the angle at which the X-rays scatter as well as the velocity of the ejected electrons. The electrons start at rest and I am to ignore any relativistic effects.

I can't figure out a way to solve without given the work function of the metal or the threshold frequency. Any help would be much appreciated!

benit13

Hey, I'm in my senior year of highschool and I'm stuck on this problem that deals with things I haven't even learned (Canadian education system for ya).
In my personal experience, this is a problem in every education system, so don't worry about it.

It has to do with the photoelectric effect.
It's probably best to go through a quick and dirty explanation of it then.

The photoelectric effect is the ability for light radiating on a substance (with free electrons) to liberate electrons from the substance. It's a bit like this

Step 1: shine light on a substance
Step 2: ?
Step 3: electrons!

However, there's also an important condition that must be satisfied for the photoelectric effect to occur.

- The frequency of the light must exceed some threshold frequency

This puzzled the hell out of scientists at the time of discovery. Why should the frequency affect it at all? Surely it's the energy of light that should affect the experiment?

So... scientists created experiments where they would vary the frequency of light being shone on the substance and then measure the kinetic energy of the electrons being released from the substance. You can plot the results on a graph (of kinetic energy versus frequency). The result is a straight line curve described by this:

$$\displaystyle KE_{e-} = h f - \phi$$

In this formula, $$\displaystyle KE_{e-}$$ is the kinetic energy of an electron (in Joules, J, or electron-Volts, eV) and f is the frequency of the light (in Hz).

The equation of a straight line is:

$$\displaystyle y = mx + c$$

So... comparing this formula with the straight line one, $$\displaystyle h$$ is the gradient of the line and $$\displaystyle \phi$$ is y-intercept, which is negative (and therefore below the x-axis). The threshold frequency is the x-intercept.

The y-intercept, $$\displaystyle \phi$$, is called the work function and it is the amount of energy required to get the effect to work. It's different for different materials.

The gradient, $$\displaystyle h$$, is actually a very, very interesting quantity...

This experiment meant that a lot of models of radiation and atoms had to be reworked. Old (classical) theories of radiation just couldn't explain this experimental result.

After some time, it was eventually figured out by clever people (including Albert Einstein!) that light can be described as both a particle (photon) and a wave and that the energy of the photon can be described using

$$\displaystyle E = hf$$

There's that h again... it's called Planck's constant (after Max Planck, who also studied this kind of thing) and it relates the energy of a photon to its frequency. This photon energy is a discrete quantity (since you can't have half a photon) and was one of the key pieces of evidence that led to quantum mechanics.

What's really happening in the photoelectric effect is that a photon is absorbed by a free electron in the substance. The photon energy is converted to kinetic energy and the electron can escape. However, if the kinetic energy is too low, it will remain confined. The kinetic energy must be high enough to allow the electron to escape, hence the work function.

The photoelectric effect also applies to free electrons in conducting metals (causing a current) and to valence band electrons in semiconductors (again, causing a current). In those cases, the work function also relates to the ability of the photon to be absorbed in the first place since electrons will only absorb photons that correspond to particular energy level transitions. I wouldn't worry about that though (for now).

Now... as for the question:

Given the wavelengths of an X-ray before and after it strikes an unknown metal, (initial=1.38e-11, final=1.54e-11), I have to determine the angle at which the X-rays scatter as well as the velocity of the ejected electrons. The electrons start at rest and I am to ignore any relativistic effects.

I can't figure out a way to solve without given the work function of the metal or the threshold frequency. Any help would be much appreciated!
In your case the energy is transferred to the electron though scattering, so the formulae probably differ from the usual photoelectric effect ones.

If you look up Compton scattering, you can probably get the equations you need to solve the problem.

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1 person

Woody

The total incoming energy must equal the total outgoing energy.

The change in energy of the X-Rays can be seen in the change of wavelength.
(see:<Wikipediahoton Energy>
The change in energy of the X-Rays must equal the change in energy of the Electrons.

This is where the change from "rays" to "photons" helps.
Each X-Ray photon interacts with just one Electron.
So each X-Ray photon transfers hc/(1.54e-11-1.38e-11) joules to its corresponding electron.

I am guessing, from the amount of information (not included) in the question,
that it may be assumed that all the transferred energy is manifested as increased kinetic energy in the electron.

topsquark

Forum Staff
Since it was mentioned a quick historical note.

When Planck was working on the solution of the blackbody problem he noted that the radiation energy was coming off in little "bits" which we now call quanta or photons. He presumed it had something to do with the nature of the cavity and not the radiation itself. Einstein was the first to (seriously anyway) consider that it was the light energy that was quantized in his landmark paper on the photoelectric effect.

-Dan