# 7.4 The quantum particle in a box  (Page 3/12)

 Page 3 / 12

## A simple model of the nucleus

Suppose a proton is confined to a box of width $L=1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\text{m}$ (a typical nuclear radius). What are the energies of the ground and the first excited states? If the proton makes a transition from the first excited state to the ground state, what are the energy and the frequency of the emitted photon?

## Strategy

If we assume that the proton confined in the nucleus can be modeled as a quantum particle in a box, all we need to do is to use [link] to find its energies ${E}_{1}$ and ${E}_{2}$ . The mass of a proton is $m=1.76\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\phantom{\rule{0.2em}{0ex}}\text{kg.}$ The emitted photon carries away the energy difference $\text{Δ}E={E}_{2}-{E}_{1}.$ We can use the relation ${E}_{f}=hf$ to find its frequency f .

## Solution

The ground state:

${E}_{1}=\frac{{\pi }^{2}{\hslash }^{2}}{2m\phantom{\rule{0.2em}{0ex}}{L}^{2}}=\frac{{\pi }^{2}{\left(1.05\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\text{J}·\text{s}\right)}^{2}}{2\left(1.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\phantom{\rule{0.2em}{0ex}}{\left(1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\text{m}\right)}^{2}}=3.28\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}\text{J}=2.05\phantom{\rule{0.2em}{0ex}}\text{MeV}.$

The first excited state: ${E}_{2}={2}^{2}{E}_{1}=4\left(2.05\phantom{\rule{0.2em}{0ex}}\text{MeV}\right)=8.20\phantom{\rule{0.2em}{0ex}}\text{MeV}$ .

The energy of the emitted photon is ${E}_{f}=\text{Δ}E={E}_{2}-{E}_{1}=8.20\phantom{\rule{0.2em}{0ex}}\text{MeV}-2.05\phantom{\rule{0.2em}{0ex}}\text{MeV}=6.15\phantom{\rule{0.2em}{0ex}}\text{MeV}$ .

The frequency of the emitted photon is

$f=\frac{{E}_{f}}{h}=\frac{6.15\phantom{\rule{0.2em}{0ex}}\text{MeV}}{4.14\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-21}\text{MeV}·\text{s}}=1.49\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{21}\phantom{\rule{0.2em}{0ex}}\text{Hz}.$

## Significance

This is the typical frequency of a gamma ray emitted by a nucleus. The energy of this photon is about 10 million times greater than that of a visible light photon.

The expectation value of the position for a particle in a box is given by

$⟨x⟩=\underset{0}{\overset{L}{\int }}dx{\psi }_{n}^{*}\left(x\right)x{\psi }_{n}\left(x\right)=\underset{0}{\overset{L}{\int }}dxx|{\psi }_{n}^{*}\left(x\right){|}^{2}=\underset{0}{\overset{L}{\int }}dxx\frac{2}{L}{\text{sin}}^{2}\frac{n\pi x}{L}=\frac{L}{2}.$

We can also find the expectation value of the momentum or average momentum of a large number of particles in a given state:

$\begin{array}{cc}\hfill ⟨p⟩& =\underset{0}{\overset{L}{\int }}dx{\psi }_{n}^{*}\left(x\right)\left[\text{−}i\hslash \frac{d}{dx}{\psi }_{n}\left(x\right)\right]\hfill \\ & =\text{−}i\hslash \underset{0}{\overset{L}{\int }}dx\sqrt{\frac{2}{L}}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{n\pi x}{L}\left[\frac{d}{dx}\sqrt{\frac{2}{L}}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{n\pi x}{L}\right]=\text{−}i\frac{2\hslash }{L}\underset{0}{\overset{L}{\int }}dx\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{n\pi x}{L}\left[\frac{n\pi }{L}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\frac{n\pi x}{L}\right]\hfill \\ & =\text{−}i\frac{2n\pi \hslash }{{L}^{2}}\underset{0}{\overset{L}{\int }}dx\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{2n\pi x}{L}=\text{−}i\frac{n\pi \hslash }{{L}^{2}}\phantom{\rule{0.2em}{0ex}}\frac{L}{2n\pi }\underset{0}{\overset{2\text{π}n}{\int }}d\phi \phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\phi =\text{−}i\frac{\hslash }{2L}·0=0.\hfill \end{array}$

Thus, for a particle in a state of definite energy, the average position is in the middle of the box and the average momentum of the particle is zero—as it would also be for a classical particle. Note that while the minimum energy of a classical particle can be zero (the particle can be at rest in the middle of the box), the minimum energy of a quantum particle is nonzero and given by [link] . The average particle energy in the nth quantum state—its expectation value of energy—is

${E}_{n}=⟨E⟩={n}^{2}\frac{{\pi }^{2}{\hslash }^{2}}{2m}.$

The result is not surprising because the standing wave state is a state of definite energy. Any energy measurement of this system must return a value equal to one of these allowed energies.

Our analysis of the quantum particle in a box would not be complete without discussing Bohr’s correspondence principle. This principle states that for large quantum numbers, the laws of quantum physics must give identical results as the laws of classical physics. To illustrate how this principle works for a quantum particle in a box, we plot the probability density distribution

$|{\psi }_{n}\left(x\right){|}^{2}=\frac{2}{L}{\text{sin}}^{2}\left(n\text{π}x\text{/}L\right)$

for finding the particle around location x between the walls when the particle is in quantum state ${\psi }_{n}$ . [link] shows these probability distributions for the ground state, for the first excited state, and for a highly excited state that corresponds to a large quantum number. We see from these plots that when a quantum particle is in the ground state, it is most likely to be found around the middle of the box, where the probability distribution has the largest value. This is not so when the particle is in the first excited state because now the probability distribution has the zero value in the middle of the box, so there is no chance of finding the particle there. When a quantum particle is in the first excited state, the probability distribution has two maxima, and the best chance of finding the particle is at positions close to the locations of these maxima. This quantum picture is unlike the classical picture.

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