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Chapter 28: Problem 29

A marble of mass \(10 \mathrm{g}\) is confined to a box \(10 \mathrm{cm}\) long and moves at a speed of \(2 \mathrm{cm} / \mathrm{s} .\) (a) What is the marble's quantum number \(n ?\) (b) Why can we not observe the quantization of the marble's energy? [Hint: Calculate the energy difference between states \(n\) and \(n+1 .\) How much does the marble's speed change?]

Short Answer

Expert verified
Answer: The quantum number of the marble is approximately \(6 \times 10^{30}\). We cannot observe the quantization of its energy because the energy difference between the quantum energy levels is extremely small (\(\approx 7.42\times 10^{-62} \mathrm{J}\)), making it impossible to detect the changes in the marble's speed with the naked eye or even with standard measurement tools.

Step by step solution

01

Convert the marble's mass and speed to SI units

The marble's mass is given in grams, and the speed is given in cm/s. We need to convert these to SI units (kilograms and m/s) for our calculations. Mass conversion: 1 g = 0.001 kg Marble's mass = 10 g = (10)(0.001) kg = 0.01 kg Length conversion: 1 cm = 0.01 m Box length = 10 cm = (10)(0.01) m = 0.1 m Speed conversion: 2 cm/s = 0.02 m/s Marble's speed = 2 cm/s = 0.02 m/s
02

Find the de Broglie wavelength

We will use the de Broglie wavelength formula to find the marble's wavelength: \(\lambda = \frac{h}{p} = \frac{h}{mv}\) where \(\lambda\) is the wavelength, h is Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), m is the marble's mass, and v is the marble's speed. \(\lambda = \frac{6.626 \times 10^{-34}}{(0.01)(0.02)} = 3.313\times 10^{-32}\) m
03

Find the quantum number n

The mathematical relationships between wavelength, box length, and quantum number n for a particle in a 1D box are given as: \(\lambda = \frac{2L}{n} \) We will rearrange the formula to solve for n: \(n = \frac{2L}{\lambda}\) \(n = \frac{2(0.1)}{3.313\times 10^{-32}} = 6.039\times 10^{30}\) As n must be an integer value, rounding off, we get: \(n \approx 6 \times 10^{30}\)
04

Calculate the energy difference between states n and n+1

We will use the energy formula for a particle in a 1D box to find the energy difference: \(E_n = \frac{n^2h^2}{8mL^2}\) The energy difference between states n and n+1 is: \(\Delta E = E_{(n+1)} - E_n\) \(\Delta E = \frac{(n+1)^2h^2}{8mL^2} - \frac{n^2h^2}{8mL^2}\) Now plugging in the values: \(\Delta E \approx \frac{[(6\times 10^{30}+1)^2-(6\times 10^{30})^2](6.626 \times 10^{-34})^2}{8(0.01)(0.1)^2} = 7.421 \times 10^{-62}\) J
05

Discuss the observability of energy quantization

The energy difference between the quantum energy levels, as calculated in the previous step, is extremely small (\(\approx 7.42\times 10^{-62} \mathrm{J}\)); thus, the marble's speed changes very little and it is impossible to observe quantization with the naked eye or even with standard measurement tools. This is why we cannot observe the quantization of the marble's energy in this case.

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Most popular questions from this chapter

An electron is confined to a one-dimensional box of length \(L\). When the electron makes a transition from its first excited state to the ground state, it emits a photon of energy 0.20 eV. (a) What is the ground-state energy (in electron-volts) of the electron in this box? (b) What are the energies (in electron-volts) of the photons that can be emitted when the electron starts in its third excited state and makes transitions downwards to the ground state (either directly or through intervening states)? (c) Sketch the wave function of the electron in the third excited state. (d) If the box were somehow made longer, how would the electron's new energy level spacings compare with its old ones? (Would they be greater, smaller, or the same? Or is more information needed to answer this question? Explain.)
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.)
An electron in an atom has an angular momentum quantum number of \(2 .\) (a) What is the magnitude of the angular momentum of this electron in terms of \(\hbar ?\) (b) What are the possible values for the \(z\) -components of this electron's angular momentum? (c) Draw a diagram showing possible orientations of the angular momentum vector \(\overrightarrow{\mathbf{L}}\) relative to the z-axis. Indicate the angles with respect to the z-axis.
An electron is confined in a one-dimensional box of length \(L\). Another electron is confined in a box of length 2 \(L\). Both are in the ground state. What is the ratio of their energies $E_{2 l} / E_{L} ?$
(a) Show that the number of electron states in a subshell is \(4 \ell+2 .\) (b) By summing the number of states in each of the subshells, show that the number of states in a shell is \(2 n^{2} .\) [Hint: The sum of the first \(n\) odd integers, from 1 to \(2 n-1,\) is \(n^{2} .\) That comes from regrouping the sum in pairs, starting by adding the largest to the smallest: \(1+3+5+\dots+(2 n-5)+(2 n-3)+(2 n-1)\) \(=[1+(2 n-1)]+[3+(2 n-3)]+[5+(2 n-5)]+\cdots\) \(=2 n+2 n+2 n+\cdots=2 n \times \frac{n}{2}=n^{2}\)
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