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Chapter 6: Problem 50

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of $9.47 \times 10^{6} \mathrm{~m} / \mathrm{s}$ What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

Short Answer

Expert verified
The characteristic wavelength of the electron with a speed of \(9.47 \times 10^{6} \mathrm{~m} / \mathrm{s}\) is approximately \(7.68 \times 10^{-11}\mathrm{m}\). Since the size of atoms typically ranges from \(1 \times 10^{-10}\mathrm{m}\) to \(5 \times 10^{-10}\mathrm{m}\), the electron's wavelength is indeed comparable to the size of atoms. Consequently, electron microscopes can generate highly magnified images of biological and other types of materials.

Step by step solution

01

Recall the de Broglie wavelength formula

In order to find the characteristic wavelength of the electron, we need to utilize the de Broglie wavelength formula, which is given by: \[\lambda = \frac{h}{p}\] Where \(\lambda\) is the characteristic wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the electron.
02

Calculate the momentum of the electron

The momentum of the electron can be found by multiplying its mass by its velocity. The mass of an electron is approximately \(9.11 \times 10^{-31} \mathrm{kg}\), and the velocity is given in the exercise as \(9.47 \times 10^{6} \mathrm{~m} / \mathrm{s}\). To find the electron's momentum, we multiply its mass and velocity: \[p = mv = (9.11 \times 10^{-31} \mathrm{kg})(9.47 \times 10^6 \mathrm{m/s}) = 8.62 \times 10^{-24}\mathrm{kg\ m/s}\]
03

Calculate the characteristic wavelength

Now that we have the momentum of the electron, we can find its characteristic wavelength using the de Broglie formula. The Planck's constant is approximately \(6.626 \times 10^{-34} \mathrm{J\cdot s}\). Therefore, the wavelength is: \[\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}\mathrm{J\cdot s}}{8.62 \times 10^{-24}\mathrm{kg\ m/s}} = 7.68 \times 10^{-11}\mathrm{m}\]
04

Compare the wavelength to the size of atoms

The size of atoms typically ranges from \(1 \times 10^{-10}\mathrm{m}\) to \(5 \times 10^{-10}\mathrm{m}\). Comparing the characteristic wavelength of the electron, which is \(7.68 \times 10^{-11}\mathrm{m}\), to the size of atoms, we find that the wavelength of the electron is in fact smaller than the size of atoms. So, the electron's wavelength is comparable to the size of atoms. This is why electron microscopes can generate highly magnified images of biological and other materials, as their characteristic wavelength is similar or smaller than the size of the object they are observing.

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