Question

We wish to see inside an atom. Assuming the atom to have a diameter of \[100pm\] , this means that one must be able to resolve a width of say \[10pm\] . If an electron microscope is used, the minimum electron energy required is about:
(A) \[15keV\]
(B) \[1.5keV\]
(C) \[150keV\]
(D) \[1.5MeV\]

Answer 1

Hint: An atom is considered or assumed to be spherical here for easier understanding of the structure. Since the diameter of the assumed spherical atom is given as \[100pm\] , it can only be resolved by a microscope that is at least capable of measuring ten times smaller than this size, which is given already in the question. This question is pretty much a direct substitution once you establish the relationship between the wavelength and the resolving power of the electron microscope.
Formulas used: The relationship between the wavelength and the energy is given by the formula, \[\lambda = \dfrac{h}{{\sqrt {2mE} }}\]
Where \[E\] is the energy of the electron
\[m\] is the mass of the electron
\[h\] is Planck's constant.

Complete answer:
We can start by stating that the wavelength of light used in the electron microscope is nearly equal to the resolving\[\lambda = \dfrac{h}{{\sqrt {2mE} }}\] power of electron microscope.
That is, \[\lambda = {10^{ - 11}}m\]
Now that we have the value of wavelength, we can find the energy by substituting in the formula,
We take the square of both the sides and get the value of energy as,
\[E = \dfrac{{{h^2}}}{{2m{\lambda ^2}}} \Rightarrow \dfrac{{{{\left( {6.64 \times {{10}^{ - 34}}} \right)}^2}}}{{2 \times 9.1 \times {{10}^{ - 31}} \times {{\left( {{{10}^{ - 11}}} \right)}^2}}} = 15keV\]
In conclusion the right answer to the question is, option (A) \[15keV\].

Note:
The original form of the electron microscope, the transmission electron microscope (TEM), uses a high voltage electron beam to illuminate the specimen and create an image. The electron beam is produced by an electron gun, commonly fitted with a tungsten filament cathode as the electron source.