Chapter 38, Problem 055 The highest achievable resolving power of a microscope i
ID: 1655114 • Letter: C
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.