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Chapter 28: Problem 4

What is the magnitude of the momentum of an electron with a de Broglie wavelength of \(0.40 \mathrm{nm} ?\)

Short Answer

Expert verified
Answer: The magnitude of the electron's momentum is approximately \(1.6565 \times 10^{-24} \mathrm{kg \cdot m/s}\).

Step by step solution

01

Understand the de Broglie wavelength formula

The de Broglie wavelength formula relates the wavelength (\(λ\)) of a particle to its momentum (\(p\)). The formula is given by: \(\lambda = \frac{h}{p}\), where \(h\) is the Planck constant.
02

Recall the Planck constant

The Planck constant (\(h\)) is a fundamental constant in quantum mechanics, with the approximate value of \(6.626 \times 10^{-34} \mathrm{Js}\).
03

Express the given wavelength in meters

The wavelength provided in the exercise is \(0.40 \mathrm{nm}\). Before plugging it into the de Broglie formula, we need to convert it to meters. To convert nanometers to meters, we use the following conversion factor: \(1 \mathrm{nm} = 10^{-9} \mathrm{m}\). So, \(0.40 \mathrm{nm} = 0.40 \times 10^{-9} \mathrm{m}\).
04

Calculate the momentum of the electron

Now, we can plug the given wavelength and the Planck constant into the de Broglie formula to find the momentum of the electron: $$\lambda = \frac{h}{p}$$ $$p = \frac{h}{\lambda}$$ $$p = \frac{6.626 \times 10^{-34} \mathrm{Js}}{0.40 \times 10^{-9} \mathrm{m}}$$ $$p = \frac{6.626 \times 10^{-34}}{0.40 \times 10^{-9}}$$ Evaluating the expression gives us the momentum of the electron: $$p \approx 1.6565 \times 10^{-24} \mathrm{kg \cdot m/s}$$ Thus, the magnitude of the electron's momentum is approximately \(1.6565 \times 10^{-24} \mathrm{kg \cdot m/s}\).

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Most popular questions from this chapter

What is the electronic configuration of the ground state of the carbon atom? Write it in the following ways: (a) using spectroscopic notation \(\left(1 s^{2}, \ldots\right) ;\) (b) listing the four quantum numbers for each of the electrons. Note that there may be more than one possibility in (b).
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 $\left.1 \mathrm{fm}, 1 \mathrm{fm}=10^{-15} \mathrm{m}\right)$.
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.)
How many electron states of the H atom have the quantum numbers \(n=3\) and \(\ell=1 ?\) Identify each state by listing its quantum numbers.
A magnesium ion \(\mathrm{Mg}^{2+}\) is accelerated through a potential difference of \(22 \mathrm{kV}\). What is the de Broglie wavelength of this ion?
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