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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: 位 = h / p Where 位 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 鈮� 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: 位 = (6.626 x 10^{-34} Js) / ((9.109 x 10^{-31} kg) * (1.875 x 10^8 m/s)) 位 鈮� 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 鈮� 1.22 * 位 / sin(伪) Assuming that 伪 is small and sin(伪) 鈮� 伪, we can simplify the equation to: d 鈮� 1.22 * 位 / 伪 Since the objective lens of an electron microscope has a small numerical aperture, the angle 伪 is very small, and the minimum resolvable distance will be similar to the wavelength of the electrons: d 鈮� 位
05

Determine the smallest resolvable structure

Given the de Broglie wavelength from step 3, the smallest resolvable structure can be approximated as: d 鈮� 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

The particle in a box model is often used to make rough estimates of energy level spacings. For a metal wire \(10 \mathrm{cm}\) long, treat a conduction electron as a particle confined to a one-dimensional box of length \(10 \mathrm{cm} .\) (a) Sketch the wave function \(\psi\) as a function of position for the electron in this box for the ground state and each of the first three excited states. (b) Estimate the spacing between energy levels of the conduction electrons by finding the energy spacing between the ground state and the first excited state.
A double-slit interference experiment is performed with 2.0-ev photons. The same pair of slits is then used for an experiment with electrons. What is the kinetic energy of the electrons if the interference pattern is the same as for the photons (i.e., the spacing between maxima is the same)?
A beam of neutrons is used to study molecular structure through a series of diffraction experiments. A beam of neutrons with a wide range of de Broglie wavelengths comes from the core of a nuclear reactor. In a time-offlight technique, used to select neutrons with a small range of de Broglie wavelengths, a pulse of neutrons is allowed to escape from the reactor by opening a shutter very briefly. At a distance of \(16.4 \mathrm{m}\) downstream, a second shutter is opened very briefly 13.0 ms after the first shutter. (a) What is the speed of the neutrons selected? (b) What is the de Broglie wavelength of the neutrons? (c) If each shutter is open for 0.45 ms, estimate the range of de Broglie wavelengths selected.

An electron passes through a slit of width \(1.0 \times 10^{-8} \mathrm{m}\) What is the uncertainty in the electron's momentum component in the direction perpendicular to the slit but in the plane containing the slit?
What is the ground-state electron configuration of bromine (Br, atomic number 35)?
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