For electron microscopes:
The de Broglie wavelength associated with a particle having momentum p is given as follows.
λ = h p λ = \frac{h}{p} λ = p h
Substitute K 2 + 2 K E 0 c \frac{\sqrt{K^2+2KE_0}}{c} c K 2 + 2 K E 0 for p.
λ = h K 2 + 2 K E 0 c = h c K 2 + 2 K E 0 λ = \frac{h}{\frac{\sqrt{K^2+2KE_0}}{c}} \\
= \frac{hc}{\sqrt{K^2+2KE_0}} λ = c K 2 + 2 K E 0 h = K 2 + 2 K E 0 h c
For a 3.0-MV transmission electron microscope:
K = 3.0 × 1 0 6 e V E 0 = 0.511 × 1 0 6 e V λ = 1240 e V × n m ( 3.0 × 1 0 6 e V ) 2 + 2 ( 3.0 × 1 0 6 e V ) ( 0.511 × 1 0 6 e V ) = 3.57 × 1 0 − 12 m = 3.57 p m 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 K = 3.0 × 1 0 6 e V E 0 = 0.511 × 1 0 6 e V λ = ( 3.0 × 1 0 6 e V ) 2 + 2 ( 3.0 × 1 0 6 e V ) ( 0.511 × 1 0 6 e V ) 1240 e V × nm = 3.57 × 1 0 − 12 m = 3.57 p m
For optical microscopes:
d = λ 2 N A d= \frac{λ}{2NA} d = 2 N A λ
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 = 514 2 × 1.45 = 177 n m d= \frac{514}{2 \times 1.45} = 177 \;nm d = 2 × 1.45 514 = 177 nm
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