X-Ray Shielding (Scanning Electron Microscope)

AI Thread Summary
X-ray shielding in scanning electron microscopes (SEM) is crucial due to the characteristic radiation generated when electrons hit the specimen, particularly at energies up to 10 keV. Concerns were raised about the adequacy of aluminum and thin mu-metal coatings for shielding, as aluminum has a half-value thickness of 112 mm for 10 keV radiation, which may not provide sufficient protection. Experts emphasized the importance of conducting a radiation survey before operation, especially after modifications to the SEM, to ensure safety protocols are followed. The discussion highlighted that while 10 keV radiation is manageable, higher energies, such as 30 keV, require careful consideration of shielding effectiveness and potential exposure risks. It was advised to consult health physics professionals for accurate assessments and to use appropriate detection tools for lower energy X-rays.
Karido
Messages
4
Reaction score
0
Hallo everyone,

I have a question about X-Ray shielding in a scanning electron microscope and I hope you can help me!
The topic is that when electrons hit the specimen, characteristic radiation up to 10keV is generated. This radiation is blocked/decelerated my the surrounding tower and chamber that are shielded with mu-metal (how thick is this coating). If I would just use pure Aluminium the X-Ray shielding wouldn't be enough. So far, is this correct?
Because Aluminium has a half-value thickness of about 112mm for 10keV x-radiation.
I am a bit concerned that some new-built parts of our SEM-chamber-cover aren't enough to shield the generated x-radiation. They are made out of alluminium with a thin mu-metal coating. It may be enough for magnetic field but I don't know if it is enough for the x-rays.. Can anyone tell me if my concerns are legitimate or is the X-radiation easily shielded with such layers?

Thank you for your help!
 
Physics news on Phys.org
Karido said:
Hallo everyone,

I have a question about X-Ray shielding in a scanning electron microscope and I hope you can help me!
The topic is that when electrons hit the specimen, characteristic radiation up to 10keV is generated. This radiation is blocked/decelerated my the surrounding tower and chamber that are shielded with mu-metal (how thick is this coating). If I would just use pure Aluminium the X-Ray shielding wouldn't be enough. So far, is this correct?
Because Aluminium has a half-value thickness of about 112mm for 10keV x-radiation.
I am a bit concerned that some new-built parts of our SEM-chamber-cover aren't enough to shield the generated x-radiation. They are made out of alluminium with a thin mu-metal coating. It may be enough for magnetic field but I don't know if it is enough for the x-rays.. Can anyone tell me if my concerns are legitimate or is the X-radiation easily shielded with such layers?

Thank you for your help!

It is most impossible to give you any accurate answer to this, because it also depends on the configuration. You can easily use shielding of an adequate thickness, but if the coverage is like swiss cheese, then it is of no use either. So without actually looking at what you have, it is almost impossible to say.

You have a very legitimate concern, and if your organization is following the proper safety protocol, they should have someone do a radiation survey before operation, especially when an instrument has been modified. Relay your concern to your supervisor, and request a radiation survey be performed. If you have a nagging feeling that something isn't quite right, do not ignore your instinct.

If you are working at a US Nat'l Lab, this is a valid "Stop Work" order before you continue working any further.

Zz.
 
  • Like
Likes Andy Resnick, Karido and M Quack
I agree that this has to be taken seriously. The information you provide is not sufficient to give a yes/no answer.

10 keV is not too hard, but it still has some penetration power. In case of doubt, ask someone from health physics to check for radiation leaks with a hand held detector. In general, shielding should be calculated for the maximum voltage/electron energy in the system, not for characteristic lines.

To get an estimate of the absorption of different materials you can look at this database:

http://henke.lbl.gov/optical_constants/filter2.html
 
  • Like
Likes Karido
Thanks for your answers! You helped me a lot!

I will check this out. But I have another theoretical question:

So in my SEM x-radiation up to 30keV can be generated because I accelerate my electrons with a voltage up to 30 keV.
If I have a aluminium plate that is 10mm thick, I have a transmission of 5% of the radiation. (http://web-docs.gsi.de/~stoe_exp/web_programs/x_ray_absorption/index.php)
Now I ask myself, how dangerous is x-radiation that has an energy of 30keV?
I know that it depends on the amount of the x-rays and the time you are exposed to them!
But can you tell me a quick overview about how to calculate an amount in sievert?
Because most of hand-held detectors start at a detectable energy level of 100keV.
Nevertheless I will check this with a professional tool that is able to detect x-rays at 10keV and higher energy levels if possible!
 
This is from Griffiths' Electrodynamics, 3rd edition, page 352. I am trying to calculate the divergence of the Maxwell stress tensor. The tensor is given as ##T_{ij} =\epsilon_0 (E_iE_j-\frac 1 2 \delta_{ij} E^2)+\frac 1 {\mu_0}(B_iB_j-\frac 1 2 \delta_{ij} B^2)##. To make things easier, I just want to focus on the part with the electrical field, i.e. I want to find the divergence of ##E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2##. In matrix form, this tensor should look like this...
Thread 'Applying the Gauss (1835) formula for force between 2 parallel DC currents'
Please can anyone either:- (1) point me to a derivation of the perpendicular force (Fy) between two very long parallel wires carrying steady currents utilising the formula of Gauss for the force F along the line r between 2 charges? Or alternatively (2) point out where I have gone wrong in my method? I am having problems with calculating the direction and magnitude of the force as expected from modern (Biot-Savart-Maxwell-Lorentz) formula. Here is my method and results so far:- This...
Back
Top