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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' ...

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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 ...

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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 ...

Veda
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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