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Chapter 28: Problem 12

If diffraction were the only limitation on resolution, what would be the smallest structure that could be resolved in an electron microscope using 10-keV electrons?

Short Answer

Expert verified
Answer: The smallest structure that can be resolved is approximately 4.016 x 10^{-11} m, or 40.16 pm (picometers).

Step by step solution

01

Calculate the de Broglie wavelength of the electrons

To determine the smallest resolvable structure, we first need to find the de Broglie wavelength of the electrons using the formula: λ = h / p Where λ is the de Broglie wavelength, h is the Planck constant (approximately 6.626 x 10^{-34} Js), and p is the momentum of the electron. Since p = mv, where m is the mass of the electron and v is its velocity, we can find the velocity by using the electron's kinetic energy: KE = (1/2)mv^2 v = sqrt(2 * KE / m) The mass of an electron is about 9.109 x 10^{-31} kg, and the kinetic energy (KE) is given as 10 keV, which is equivalent to 1.602 x 10^{-16} J. Using these values, we can find the de Broglie wavelength of the electron.
02

Calculate the electron velocity

Using the kinetic energy and electron mass, we can calculate the velocity: v = sqrt(2 * (1.602 x 10^{-16} J) / (9.109 x 10^{-31} kg)) v ≈ 1.875 x 10^8 m/s
03

Calculate the de Broglie wavelength

Now, we can apply the velocity to the de Broglie wavelength formula: λ = (6.626 x 10^{-34} Js) / ((9.109 x 10^{-31} kg) * (1.875 x 10^8 m/s)) λ ≈ 4.016 x 10^{-11} m
04

Use the Rayleigh criterion to find the minimum resolvable distance

Now that we have the de Broglie wavelength, we can use the Rayleigh criterion to find the minimum resolvable distance (d). The Rayleigh criterion is given by: d ≈ 1.22 * λ / sin(α) Assuming that α is small and sin(α) ≈ α, we can simplify the equation to: d ≈ 1.22 * λ / α Since the objective lens of an electron microscope has a small numerical aperture, the angle α is very small, and the minimum resolvable distance will be similar to the wavelength of the electrons: d ≈ λ
05

Determine the smallest resolvable structure

Given the de Broglie wavelength from step 3, the smallest resolvable structure can be approximated as: d ≈ 4.016 x 10^{-11} m Therefore, the smallest structure that can be resolved in an electron microscope using 10-keV electrons, considering diffraction as the only limitation, is approximately 4.016 x 10^{-11} m, or 40.16 pm (picometers).

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Most popular questions from this chapter

A fly with a mass of \(1.0 \times 10^{-4} \mathrm{kg}\) crawls across a table at a speed of \(2 \mathrm{mm} / \mathrm{s} .\) Compute the de Broglie wavelength of the fly and compare it with the size of a proton (about $\left.1 \mathrm{fm}, 1 \mathrm{fm}=10^{-15} \mathrm{m}\right)$.
The particle in a box model is often used to make rough estimates of energy level spacings. Suppose that you have a proton confined to a one-dimensional box of length equal to a nuclear diameter (about \(10^{-14} \mathrm{m}\) ). (a) What is the energy difference between the first excited state and the ground state of this proton in the box? (b) If this energy is emitted as a photon as the excited proton falls back to the ground state, what is the wavelength and frequency of the electromagnetic wave emitted? In what part of the spectrum does it lie? (c) Sketch the wave function \(\psi\) as a function of position for the proton in this box for the ground state and each of the first three excited states.
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