I found an answer from www.khanacademy.org

**De Broglie wavelength** (video) | Khan Academy

Bohr model **energy** levels (derivation using **physics**) ... Since visible **light** has a
wavelength of about 500 nanometers, this means that visible **light** ... Since
**electrons** have a rest mass, unlike **photons**, they have a **de Broglie wavelength** ...
For example in the **double** slit experiment.... **electron waves** diffract and then '
condense' ...

For more information, see **De Broglie wavelength** (video) | Khan Academy

I found an answer from www.quora.com

Why **does** TEM have higher resolution than SEM? - Quora

In general, TEM has a higher resolution than SEM by a **factor** of 10 or more. ... In
a TEM, a **nearly** parallel beam of **electrons** travels through a thin specimen, ...
Why don't they **use** an **electron microscope** to test for covid19? ... SEM is
operating at acceleration **voltage** up to 30 **kV** while TEM have **electrons**
**accelerated** up to ...

For more information, see Why **does** TEM have higher resolution than SEM? - Quora

I found an answer from www-ssrl.slac.stanford.edu

Exchange Coupling of Co and Fe on Antiferromagnetic NiO ...

2.3 Resonant X-ray Absorption Process and **Electron** Yield . ... The **microscopic**
origin of the magnetic anisotropy is the spin-orbit coupling which couples the spin
... By making **use** of the magnetic moment of the neutron and the fact that its **de**-
**Broglie** ... **Such** a **microscope will** allow to achieve a lateral resolution of 20-**50nm**
.

For more information, see Exchange Coupling of Co and Fe on Antiferromagnetic NiO ...

Given that

Accelerating voltage E = 50 kV = 50*10^3 V

E = 50 * 10^3 * 1.6 * 10^{-19} Joules

Step 1: Relation between the de Broglie wavelength and kinetic energy

Recall the formula of momentum and kinetic energy

Momentum p = mv

Kinetic energy K.E = \frac{1}{2} mv^2

Where, p - momentum, m - mass and v - velocity

K.E = \frac{p^2}{2m} \because v = \frac{p}{m}

p = \sqrt{2m K.E} ...........................(1)

de Broglie wavelength \lambda = \frac{h}{p}

\lambda = \frac{h}{ \sqrt{2m K.E}} \because \text{ Equation (1)}

Step 2: Substituting known values in the above de Broglie wavelength wave length

Mass of an electron m = 9.1 * 10^{-31} kg.

de Broglie wavelength \lambda = \frac{6.63 * 10^{-31}}{\sqrt{2 * 9.1 * 10^{-31} * 50 * 10^3 * 1.6 * 10^{-19} }}

\lambda = \frac{6.63 * 10^{-31}}{1.21 * 10^{-22}}

\lambda = 5.47 * 10^{-12} m

Hence, de Broglie wavelength of electron \lambda = 5.47 * 10^{-12} m

Step 3: Resolving power of an electron microscope

Wavelength of yellow light \lambda_y=5.99*10^{-7} m

Wavelength of electron \lambda_e=5.47*10^{-12} m

The microscope's resolving power is inversely proportional to the radiation's wavelength.

\frac{ \text{ electron microscope's resolving power }}{ \text{ optical microscope's resolving power}} = N * \frac{5.99 * 10^{-7}}{5.47 * 10^{-12}} \because \text{ N - constant }

Therefore, electron microscope's resolving power > 10^{5} * optical microscope's resolving power