# Can an Electron Be Observed in a Light Microscope?

• Juli
Juli
Homework Statement
Electrons cannot move in fixed orbits in an atom. We want to “observe” such an electron with a light microscope. The accuracy with which we want to observe the position of the electron on its orbit is 10 pm. So the wavelength of the light in this microscope must also be around 10 pm.
1. What would be the energy of a photon of this light?
2. How much energy would such a photon be transferred to the electron in a head-on collision?
3. What do these results say about the possibility of being able to “observe” an electron’s presumed orbit at two or more points?
Relevant Equations
##E_e = E_{Ph} - E_{Ph'}##
$$\lambda' = \lambda + \lambda_c * (1-cos\phi)$$
Hello everyone,
I solved the problem above in the following way:

1. ##E_{Ph} = 124##keV
2. ##E_e = E_{Ph} - E_{Ph'}##
##E_{Ph}## is the energy of the incoming photon, ##E_{Ph'}## is the energy of the photon, after the scattering with the electron (I am using the formulas assuming there is Compton scattering happening).
I got the wavelength for ## E_{Ph'}## this way: $$\lambda' = \lambda + \lambda_c * (1-cos\phi) = 14.84\text{pm}$$.
For ##\phi## I chose ##\pi## because of the head-on collision. ## \lambda_c = 2.42\cdot 10^{-12}##m.
I got $$E_e = 124 \text{keV} - 84\text{keV} = 40 \text{keV}$$

3. Now we get to the part where I need advice on.
My answer to the question is, that the probability to dislodge the electron in the process of observing and thereby irradiate, is quite high. Since the binding energy of an electron in the outer shell of the atom is a few eV and 40keV is a lot in comparison. Therefore the I predict a bad possibility to observe an electron's presumed orbit at two or more points. (It would be possible on one point and then the electron would be gone (completely or in a higher energy state)).
A friend made me doubt my theory, he said the answer might me related with Heisenbergy uncertanty relation.
Do you have thoughts on this?
I am always thankful for answers :)

Your conclusion is that in order to determine an electrons position to the required accuracy, you need a relatively high energy photon. And that this interaction would inevitably ionise the atom?

PS I suspect your friend may be wrong about the relevance of the HUP (Heisenberg Uncertainty Principle).

Juli
Why do you think the incoming Xray will interact with only an individual electron? What is the target? Shooting beams of Xrays and γrays at targets has been done for more than a century. I suggest a look at the rich history of this very useful technique. It is inherently messy but lovely Physics.

PeroK said:
Your conclusion is that in order to determine an electrons position to the required accuracy, you need a relatively high energy photon. And that this interaction would inevitably ionise the atom?
So the first part is not my conclusion but the problem statement. The second part is my conclusion. The homework question is "What do these results (from 1. and 2.) say about the possibility of being able to “observe” an electron’s presumed orbit at two or more points?" And that was my suggestion.

hutchphd said:
Why do you think the incoming Xray will interact with only an individual electron? What is the target? Shooting beams of Xrays and γrays at targets has been done for more than a century. I suggest a look at the rich history of this very useful technique. It is inherently messy but lovely Physics.
The question is stating that we just take a look at one electron. So the target is the electron. There isn't more information in the question than I gave you. So I thought that in the end it would lead to the Compton- effect: A photon that hits an electron.

Juli said:
So the first part is not my conclusion but the problem statement. The second part is my conclusion. The homework question is "What do these results (from 1. and 2.) say about the possibility of being able to “observe” an electron’s presumed orbit at two or more points?" And that was my suggestion.
Sounds right to me. You don't get the chance of a second orbital position measurement, because the atom has been ionised.

Note that this is not the HUP. Although many sources online may claim it is! The HUP relates to the spread of position measurements made on a large number (ensemble) of identical atoms. You find the electron across a range of possible positions. That's the ##\Delta x## in this case. It's definitely not the displacement of the electron as a result of the interaction with the photon.

Juli
Thank you :)
Also for making it clear with th HUP!

PeroK
I would argue that in any real situation, the scattering you observe will of necessity be from scattering many events. The details of how they add will depend upon geometrical issues of coherence. These will produce an ensemble average and therefore be subject to the limitations of Heisenberg uncertainty, which at heart is a fundamental statement of information theory. So I disagree with @PeroK on this and think the fundamental limitation in resolution is complicated, but essentially Heisenberg. It is the usual (spatial) frequency-bandwidth problem which cannot, to my knowledge, be superseded..

hutchphd said:
I would argue that in any real situation, the scattering you observe will of necessity be from scattering many events. The details of how they add will depend upon geometrical issues of coherence. These will produce an ensemble average and therefore be subject to the limitations of Heisenberg uncertainty, which at heart is a fundamental statement of information theory. So I disagree with @PeroK on this and think the fundamental limitation in resolution is complicated, but essentially Heisenberg. It is the usual (spatial) frequency-bandwidth problem which cannot, to my knowledge, be superseded..
The question was whether you could do two measurements of an electron in an orbital. The answer to that is no, but not because of the HUP.

hutchphd
I guess misinterpretted a rather convoluted question/answer pair. Indeed I don't even know what "two measurements in an orbital" means. But I also do not know what "because of" HUP means. (The Uncertainty Principle follows fundamentally from the structure of the physics ...the question is then tautological)
That was what I wished to emphasize.

## Can an electron be seen with a light microscope?

No, an electron cannot be seen with a light microscope. Light microscopes use visible light to magnify objects, but electrons are much smaller than the wavelength of visible light, making them impossible to observe directly with this type of microscope.

## Why can't light microscopes observe electrons?

Light microscopes are limited by the wavelength of visible light, which ranges from about 400 to 700 nanometers. Electrons are subatomic particles with sizes on the scale of femtometers (10^-15 meters), far smaller than the wavelength of visible light. Therefore, they cannot be resolved or observed using light microscopy.

## What type of microscope is used to observe electrons?

To observe electrons, scientists use electron microscopes, such as transmission electron microscopes (TEM) and scanning electron microscopes (SEM). These microscopes use beams of electrons instead of light to achieve much higher resolution and magnification.

## How do electron microscopes work?

Electron microscopes work by directing a beam of electrons at a sample. The interaction between the electrons and the sample produces various signals that can be detected and used to create an image. Because electrons have much shorter wavelengths than visible light, electron microscopes can resolve much smaller structures.

## What are the limitations of electron microscopes compared to light microscopes?

While electron microscopes offer higher resolution and magnification, they also have several limitations. They require a vacuum environment, which means samples must be specially prepared and cannot be living. Additionally, electron microscopes are more expensive, complex, and require more expertise to operate compared to light microscopes.

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