## Electron and optical microscope

### Task number: 4378

Reachable resolution power of the microscope is given by the wavelength of the used particles. This means that the smallest detail that we can observe has approximately the size of the wavelength. Let’s assume we want to see inside the atom which has a diameter of approximately 100 pm, this means we want to see details of approximately 10 pm.

a) What energy must the electrons have in the case we want to use electron microscope?

b) What energy must the photons have in the case we want to use optical microscope?

c) Which microscope is more practical and why?

#### Hint

We have the de Broglie formula Broglieho vztahu connecting the wavelength and momentum of the particle given by

\[\lambda=\frac{h}{p} ,\]where

*h*= 6.63·10^{−34}Js is Planck’s constant.Energy of the photon is proportional to its frequency; therefore, not proportionate to its wavelength.

#### Notation

*λ*= 10 pm = 10^{−11}mwavelength *E*_{e}= ? (eV)energy of the electron *E*_{f}= ? (eV)energy of the photon *From tables:**h*= 6.63·10^{−34}JsPlanck’s constant *m*_{e}= 9.11·10^{−31}kgelectron mass *c*= 3.00·10^{8}m s^{−1}speed of light in vacuum #### Solution

a) Momentum

\[\lambda=\frac{h}{p_{\mathrm{e}}} ,\]*p*_{e}and the wavelength λ_{e}of the electron are related by de Broglie formula:where

*h*= 6.63·10^{−34}Js is Planck’s constant.We can determine the kinetic energy of the electron

\[E_{\mathrm{e}}=\frac{p_{\mathrm{e}}^2}{2m_{\mathrm{e}}}=\frac{\left(\frac{h}{\lambda_{\mathrm{e}}}\right)^2}{2m_{\mathrm{e}}}=\frac{h^2}{2m_{\mathrm{e}}\lambda_{\mathrm{e}}^2}=\frac{\left(6.63{\cdot}10^{-34}\right)^2}{2{\cdot}9.11{\cdot}10^{-31}\cdot\left(10^{-11}\right)^2}\,\mathrm{J}=2.4{\cdot}10^{-15}\,\mathrm{J}=15\,\mathrm{keV}.\]*E*_{e}from its momentum*p*_{e}b) Energy of the photon

\[E_f=hf=h\frac{c}{\lambda}=\frac{6.63{\cdot}10^{-34}\cdot3.00{\cdot}10^8}{10^{-11}}\,\mathrm{J}=1.99{\cdot}10^{-14}\,\mathrm{J}=124\,\mathrm{keV} .\]*E*_{f}is equal toc) In terms of more resolution, the electron microscope is more practical – it is sufficient to accelerate the electrons by voltage of 1.5 kV which is not so big. In optical microscope we would need to use gamma radiation (see electromagnetical spectra ) which is hard to obtain and dangerous.

#### Answer

To reach the desired resolution power we would need to use electrons with energy of 15 keV or photons with energy of 124 keV.