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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

(a) Make a qualitative sketch of the wave function for the \(n=5\) state of an electron in a finite box \([U(x)=0\) for \(00\) elsewherel. (b) If \(L=1.0 \mathrm{nm}\) and \(U_{0}=1.0\) keV, estimate the number of bound states that exist.
What is the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus \((1.0 \mathrm{fm}) ?\)
An electron moving in the positive \(x\) -direction passes through a slit of width \(\Delta y=85 \mathrm{nm} .\) What is the minimum uncertainty in the electron's velocity in the \(y\) -direction?
An 81 -kg student who has just studied matter waves is concerned that he may be diffracted as he walks through a doorway that is \(81 \mathrm{cm}\) across and \(12 \mathrm{cm}\) thick. (a) If the wavelength of the student must be about the same size as the doorway to exhibit diffraction, what is the fastest the student can walk through the doorway to exhibit diffraction? (b) At this speed, how long would it take the student to walk through the doorway?
An electron is confined to a one-dimensional box of length \(L\) (a) Sketch the wave function for the third excited state. (b) What is the energy of the third excited state? (c) The potential energy can't really be infinite outside of the box. Suppose that \(U(x)=+U_{0}\) outside the box, where \(U_{0}\) is large but finite. Sketch the wave function for the third excited state of the electron in the finite box. (d) Is the energy of the third excited state for the finite box less than, greater than, or equal to the value calculated in part (b)? Explain your reasoning. [Hint: Compare the wavelengths inside the box.] (e) Give a rough estimate of the number of bound states for the electron in the finite box in terms of \(L\) and \(U_{0}\).
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