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Chapter 28: Problem 28

An electron is confined to a box of length \(1.0 \mathrm{nm} .\) What is the magnitude of its momentum in the \(n=4\) state?

Short Answer

Expert verified
Answer: The magnitude of the momentum of the electron in the n=4 state is \(1.33 \times 10^{-24} \ \mathrm{kg\: m/s}\).

Step by step solution

01

Find the wavelength

Using the quantum number concept, if n=4, there will be four half wavelengths within the box. Therefore, the wavelength of the electron can be determined by the length of the box and the number of half wavelengths: \(\lambda = \frac{2L}{n}\) where n = 4 and L is the length of the box, which is 1.0 nm. Plug in the values and calculate the wavelength: \(\lambda = \frac{2 * 1.0}{4} = 0.5 \ \mathrm{nm}\)
02

Convert wavelength to meters

Since we need to find the momentum in SI units, we need to convert the wavelength from nm to meters: \(\lambda = 0.5 \ \mathrm{nm} * 10^{-9} \ \frac{\mathrm{m}}{\mathrm{nm}} = 0.5 * 10^{-9} \ \mathrm{m}\)
03

Calculate the momentum

Now, we can use the de Broglie wavelength formula to find the momentum: \(p = \frac{h}{\lambda}\) where \(h\) is the Planck's constant, \(6.63 \times 10^{-34} \ \mathrm{Js}\), and \(\lambda\) is the wavelength in meters. Plug in the values and calculate the momentum: \(p = \frac{6.63 \times 10^{-34}}{0.5 * 10^{-9}} = 1.33 \times 10^{-24} \ \mathrm{kg\: m/s}\) So, the magnitude of the momentum of the electron in the n=4 state is \(1.33 \times 10^{-24} \ \mathrm{kg\: m/s}\).

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Most popular questions from this chapter

What are the de Broglie wavelengths of electrons with the following values of kinetic energy? (a) \(1.0 \mathrm{eV}\) (b) $1.0 \mathrm{keV} .
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