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Chapter 6: Problem 44

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 \(9.38 \times 10^{6} \mathrm{~m} / \mathrm{s}\). 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 is approximately \(7.75 \times 10^{-11} \mathrm{~m}\) and is comparable to the size of atoms, which typically range from \(1 \times 10^{-10} \mathrm{~m}\) to \(5 \times 10^{-10} \mathrm{~m}\).

Step by step solution

01

Write down the given information

The electron's speed is given as \(9.38 \times 10^{6} \mathrm{~m} / \mathrm{s}\).
02

Recall the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \[\lambda = \frac{h}{p}\] where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle.
03

Find the momentum of the electron

The momentum of the electron can be calculated using the formula: \[p = mv\] where \(m\) is the mass of the electron and \(v\) is its speed. Given that the speed of the electron is \(9.38 \times 10^{6} \mathrm{~m} / \mathrm{s}\), and using the constant mass of the electron as \(9.11 \times 10^{-31} \mathrm{~kg}\), we can calculate the momentum as follows: \[p = (9.11 \times 10^{-31} \mathrm{~kg})(9.38 \times 10^{6} \mathrm{~m} / \mathrm{s})\] \[p = 8.545 \times 10^{-24} \mathrm{~kg~m/s}\]
04

Calculate the wavelength of the electron

Using the de Broglie wavelength formula and the given value of Planck's constant, \(h = 6.626 \times 10^{-34} \mathrm{~Js}\), we can calculate the wavelength of the electron: \[\lambda = \frac{6.626 \times 10^{-34} \mathrm{~Js}}{8.545 \times 10^{-24} \mathrm{~kg~m/s}}\] \[\lambda = 7.75 \times 10^{-11} \mathrm{~m}\]
05

Compare the wavelength with the size of atoms

The typical size of an atom is about \(1 \times 10^{-10} \mathrm{~m}\) to \(5 \times 10^{-10} \mathrm{~m}\). Comparing this size with the calculated wavelength of the electron, \(7.75 \times 10^{-11} \mathrm{~m}\), we can see that the electron's wavelength is comparable to the size of atoms. In conclusion, the characteristic wavelength of the electron is around \(7.75 \times 10^{-11} \mathrm{~m}\), and it is comparable to the size of atoms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

de Broglie Wavelength
The de Broglie wavelength is a core concept in quantum mechanics that describes the wave-like nature of particles, such as electrons. This concept was introduced by French physicist Louis de Broglie in 1924.
The de Broglie wavelength (\(\lambda\)) of any particle is calculated using the formula:
  • \(\lambda = \frac{h}{p}\)
Here, \(h\) is Planck's constant, and \(p\) is the momentum of the particle. The de Broglie wavelength provides insight into the quantum behavior of particles, especially when their sizes are comparable to their wavelengths.
When an electron is accelerated to high speeds, like in an electron microscope, its de Broglie wavelength becomes small enough to be comparable to atomic sizes, allowing it to resolve features smaller than those visible with a light microscope.
Momentum of Electron
The momentum of an electron is a measure of its motion, calculated using its mass and velocity. The momentum \(p\) of a particle such as an electron is given by the formula:
  • \(p = mv\)
where \(m\) is the mass of the electron, and \(v\) is its velocity. Electrons have a constant mass of approximately \(9.11 \times 10^{-31} \text{ kg}\).
For example, an electron moving at a speed of \(9.38 \times 10^6 \text{ m/s}\) would have a momentum:
  • \(p = (9.11 \times 10^{-31} \text{ kg})(9.38 \times 10^6 \text{ m/s}) = 8.545 \times 10^{-24} \text{ kg m/s}\)
This calculated momentum is crucial for determining the de Broglie wavelength and hence the resolving power of electron microscopes.
Planck's Constant
Planck's constant is a fundamental constant in physics, represented by the symbol \(h\). It is crucial in quantum mechanics and aids in the calculation of wavelengths and frequencies.
Planck's constant has a fixed value of approximately \(6.626 \times 10^{-34} \text{ Js}\). This constant bridges the physics of the macroscopic world with the quantum world by quantifying the energy of particles oscillating at particular frequencies or wavelengths.
In the context of de Broglie's equation, Planck's constant helps determine the wave nature of particles like electrons. Through the formula \(\lambda = \frac{h}{p}\), Planck's constant is key to finding the de Broglie wavelength, which in turn allows us to understand quantum behaviors such as electron diffraction.
Atomic Size Comparison
When discussing the de Broglie wavelength of electrons, it's important to understand how these wavelengths compare to the size of atoms. Atoms typically have sizes ranging between \(1 \times 10^{-10} \text{ m}\) and \(5 \times 10^{-10} \text{ m}\).
The calculated wavelength of an electron moving at a high speed is around \(7.75 \times 10^{-11} \text{ m}\). This wavelength is on the scale of atomic dimensions, making it suitable for imaging at the atomic level with an electron microscope.
Thus, understanding this comparison is vital since it explains how electron microscopes can visualize structures much smaller than those visible with traditional optical microscopes, which rely on visible light with longer wavelengths.

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Most popular questions from this chapter

Neutron diffraction is an important technique for determining the structures of molecules. Calculate the velocity of a neutron needed to achieve a wavelength of \(0.955 \AA\). (Refer to the inside cover for the mass of the neutron).

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{~nm}\). (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? (d) These UV photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the 325-nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

Consider a fictitious one-dimensional system with one electron. The wave function for the electron, drawn at the top of the next column, is \(\psi(x)=\sin x\) from \(x=0\) to \(x=2 \pi .\) (a) Sketch the probability density, \(\psi^{2}(x)\), from \(x=0\) to \(x=2 \pi\) (b) At what value or values of \(x\) will there be the greatest probability of finding the electron? (c) What is the probability that the electron will be found at \(x=\pi ?\) What is such a point in a wave function called? \([\) Section \(6.5]\)

In the transition metals (the \(d\) -block), the electron configuration of cations is different than what you might expect. Instead of the \(\mathrm{d}\) electrons being lost first, \(s\) electrons are lost first. For example, the electron configuration of iron, \(\mathrm{Fe}\), is \([\mathrm{Ar}] 4 s^{2} 3 d^{6}\); but the electron configuration of \(\mathrm{Fe}^{2+}\) is \([\mathrm{Ar}] 3 d^{6} ;\) the \(4 \mathrm{~s}\) electrons are eliminated to make the cation. Write out the electron configurations of (a) \(\mathrm{Zn}^{2+}\) (b) \(\mathrm{Pt}^{2+}\) (c) \(\mathrm{Cr}^{3+}\) (d) \(\mathrm{Ti}^{4+}\).

Using Heisenberg's uncertainty principle, calculate the uncertainty in the position of (a) a \(1.50-\mathrm{mg}\) mosquito moving at a speed of \(1.40 \mathrm{~m} / \mathrm{s}\) if the speed is known to within \(\pm 0.01 \mathrm{~m} / \mathrm{s} ;\) (b) a proton moving at a speed of \((5.00 \pm 0.01) \times 10^{4} \mathrm{~m} / \mathrm{s}\). (The mass of a proton is given in the table of fundamental constants in the inside cover of the text.)
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