- #1

Juli

- 21

- 5

- 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 :)

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 :)