Electron-Matter interaction

Question 1

Why electrons allow for better resolution than light?

Answer 1

They can also be used for imaging because of wave particle duality. From Do Broglie relation you can find the wavelength of an accelerated photon. For a 1kV, it is 0.03nm => diffraction limitation is not a problem at all since atoms are like 0.1nm.

Question 2

What is the wavelength/velocity of an electron in a 1 kV electron microscope?

Electron mass: me = 9.10938356 × 10−31 kg
Electron charge: e = 1.6021766208 × 10−19 C
Planck constant: h = 6.626 x 10-24 m2 kg s-1
1J = 1 kg m2 s-2

Answer 2

E = eU = mv^2/2
=> v = sqrt(2eU/m)
=> lambda = h/mv = h/sqrt(2meU) = hc/sqrt(2mc^2[eV]*E) = 0.04 nm

Question 3

Describe the different types of electron scattering possible!

Answer 3

• elastic backscattered => high energy/coherent, with preserved energy and phase of the incident electron beam
  -inelastically backscattered: also referred to as incoherent, they would ave collided with electrons(see the same for the scattered)
• transmission: no interaction
• elastically scattered: they are only deflected but preserve energy and phase
• inelastically scattered: energy is lost/transmitted to a secondary electron which then has low energy compared to the scattered electron. In addition higher shell electron would then fall to the lower orbital state resulting in X-ray photons.

Question 4

Interaction cross section

Answer 4

doesnt really represent area but the probability of scattering event.

𝜎𝑎𝑡𝑜𝑚 = 𝜋𝑟2

where r is the effective radius of the atom/the matter that the electron will interact with.

A higher r would thus mean that a bigger interaction cross section is present meaning more electron scattering events can occur

Question 5

Formula for interaction cross section effective radius
Hint: think of the factors that influence this process

Answer 5

The factors are:
1. The atomic number, heavier atoms generally deflect electrons more strongly since they have bigger atom nucleus charge
2. Incident beam voltage: electrons scatter less strongly for higher voltage
3. Scattering into high-angles is less likely since it requires less momentum/energy transfer.

Thus taking all of this into account we get the following formula.
𝑟𝑒𝑙𝑎𝑠𝑡𝑖𝑐 = 𝑍𝑒/𝑈𝜃

Question 6

Solid angle of scattering

Answer 6

measure of the “spread” of scattering directions

Ω = 2𝜋(1 − 𝑐𝑜𝑠𝜃)

Question 7

Total scattering cross section

Answer 7

𝜎𝑡 = 𝑁𝜎𝑎𝑡𝑜𝑚 = 𝑁𝜋𝑟2, 𝑁 = 𝑁A * 𝜌/𝑚𝑎

Question 8

Why do we observe more scattering at high angles for Au foil compared to Cu one?

Answer 8

higher Z

Question 9

Mean Free path

Answer 9

Mean free path = distance between two scattering events
Λ = 1/𝜎𝑡 = 𝑚𝑎 / (𝑁𝐴𝜎𝑎𝑡𝑜𝑚𝜌)

Question 10

What are the most commons scattering events?

Answer 10

Plasmon and elastic scattering
events account for the majority
of total scattering cross section.

Question 11

Why is backscattering more likely if the electron passes close to the nucleus?

Answer 11

The closer it passes the higher pull by the nucleus => deflection angle becomes close to 180 => backscattering

Question 12

Plasmons and Plasmon scattering

Answer 12

plasmon is a quasiparticle that represents the collective oscillation of free electron density in a material, typically a metal, in response to an external electromagnetic field

When electrons pass through a material, they can excite plasmons, which leads to energy loss. This loss can be measured and analyzed in electron energy loss spectroscopy

Question 13

Auger scattering

Answer 13

Auger scattering refers to a process in which an atom, excited by the removal of an inner-shell electron, undergoes a series of electron transitions that result in the ejection of a secondary electron, known as an Auger electron. This process, named after the French physicist Pierre Auger, occurs without the emission of an X-ray photon, which would be typical in X-ray fluorescence following inner-shell ionization.

Question 14

electron interaction volume

Answer 14

the region within a material where incoming electrons interact with the atoms in the sample, undergoing various scattering and energy-loss processes.

Question 15

When can secondary electrons (SE) be formed?

Answer 15

SE1: SEs directly generated by primary electrons
SE2: SEs generated by BSEs leaving the sample
SE3: SEs generated by BSEs hitting objects in the
vacuum chamber

Question 16

Describe the shape of the electron interation volume dpending on the energy of the incident beam!

Answer 16

The higher the energy the greater the depth but also the lower the width on the surface of the sample. (see figure if you dont remember)

Question 17

What kind of interaction volume would be needed in TEM and SEM, and thus, what energy of beam do these EM techniques need?

Answer 17

SEM generally requires low Energy beams to allow for a bigger width of interaction volume and thus more information of the surface.

TEM on the other hand requires higher energy beams to reduce any inelastic interactions as much as possible and would thus have a thinner but deeper interaction volume.