Question

A scanning electron microscope is used to look at cell structure with 10 -nm resolution. A beam of electrons from a hot filament is accelerated with a voltage of \(12 \mathrm{kV}\) and then focused to a small spot on the specimen. (a) What is the wavelength in nanometers of the beam of incoming electrons? (b) If the size of the focal spot were determined only by diffraction, and if the diameter of the electron lens is one fifth the distance from the lens to the specimen, what would be the minimum separation resolvable on the specimen? (In practice, the resolution is limited much more by aberrations in the magnetic lenses and other factors.)

Short answer

Answer: To find the minimum resolvable separation on the specimen, we need to first find the wavelength of the electron beam using the de Broglie wavelength equation and the given voltage. Next, we use the diffraction equation to calculate the minimum separation resolvable on the specimen based on this wavelength.

Step-by-step solution

Find the wavelength of the electron beam

To find the wavelength of the electron beam, we will use the de Broglie wavelength equation: \(λ = \dfrac{h}{p}\) where λ is the wavelength, h is the Planck's constant (\(6.626 \times 10^{-34} \ \text{J} \cdot \text{s}\)), and p is the momentum of the electron. To find the momentum of the electron, we need to find its kinetic energy and then convert it to momentum using the formula: \(K = \dfrac{p^2}{2m}\) \(p = \sqrt{2mK}\) We can find the kinetic energy of the electron using the given voltage: \(K = eV\) where e is the electron charge (\(1.602 \times 10^{-19} \ \text{C}\)) and V is the voltage (12 kV). Now, we can plug these values into the equation and find the wavelength.

Find the minimum separation on the specimen

Once we have the wavelength of the electron beam, we can use the diffraction equation to find the minimum separation resolvable on the specimen: \(θ = \dfrac{1.22 λ}{D}\) where \(θ\) is the angle of the electron beam, λ is the wavelength of the electron beam, and D is the diameter of the electron lens. We know that the diameter of the electron lens is one fifth of the distance from the lens to the specimen, so we can write: \(D = \dfrac{l}{5}\) where l is the distance from the lens to the specimen. Now, we can rewrite the diffraction equation in terms of the minimum separation resolvable on the specimen: \(s = l \cdot θ\) Substituting the values and solving for the minimum separation, we can find the final answer.