Answer to Question #164692 in Quantum Mechanics for Hima bindu

Question

caluculate the Resolution of electron microscopes and optical microscopes

Answer 1

For electron microscopes:

The de Broglie wavelength associated with a particle having momentum p is given as follows.

$λ = \frac{h}{p}$

Substitute $\frac{\sqrt{K^2+2KE_0}}{c}$ for p.

$λ = \frac{h}{\frac{\sqrt{K^2+2KE_0}}{c}} \\ = \frac{hc}{\sqrt{K^2+2KE_0}}$

For a 3.0-MV transmission electron microscope:

$K = 3.0 \times 10^6 \;eV \\ E_0 = 0.511 \times 10^6 \;eV \\ λ = \frac{1240 \;eV \times nm}{\sqrt{(3.0 \times 10^6 \;eV)^2+2(3.0 \times 10^6 \;eV)(0.511 \times 10^6 \;eV)}} \\ = 3.57 \times 10^{-12} \;m \\ = 3.57 \;pm$

For optical microscopes:

$d= \frac{λ}{2NA}$

d is the resolution

λ is the wavelength

NA is the numerical aperture, which gathers light and resolve the specimen.

Where λ is the wavelength of light used to image a specimen. If using a green light of 514 nm and an oil immersion objective with an NA of 1.45, then the (theoretical) limit of resolution will be 177 nm.

$d= \frac{514}{2 \times 1.45} = 177 \;nm$