Chapter 38, Problem 055 The highest achievable resolving power of a microscope i

Question

Chapter 38, Problem 055

The highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest item that can be distinguished has dimensions about equal to the wavelength. Suppose one wishes to "see" inside an atom. Assuming the atom to have a diameter of 400 pm, this means that one must be able to resolve a width of, say, 40 pm.
 (a) If an electron microscope is used, what minimum electron energy is required?
 (b) If a light microscope is used, what minimum photon energy is required?
 (c) Which microscope seems more practical and why?

Explanation / Answer

(a)To be able to resolve an object the wavelength is required to be of the similar dimensions.

Therefore lambda = 40pm = 40*10^(-12)m

and so the momentum for electrons with this wavelength will be: p = hlambda = (6.626*10^(-34))/(40*10^(-12)) = 1.66*10^(-23) kgm/s

and momentum and kinetic energy are related to each other by expression:

p = (2mE)^(1/2)

So E = (p^(2))/2m

=> E = (1.66*10^(-23))^2)/(2*9.1*10^(-31)) = 1.51*10^(-16) J = 0.95 keV

b) If light of wavelength 40pm is used then the energy of the photons will be:

E = (6.626*10^(-34)*3*10^(8))/(40*10^(-12)) = 4.969*10^(-15) J = 31.0164 keV

c) An Electron microscope requires much less energy than the electromagnetic wave of same wavelength to probe the atom. Hence an Electron microscope is more practical.

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