Particle in a Box Questions and Answers

Answer: A

The order is \(E_{2} \gt E_{1} \gt E_{3}\).

Because \(V_{0}\) is small compared with the ground-state energy, first-order perturbation theory applies: each energy rises by \(V_{0}\) times the probability of finding the particle in the raised region. The ground state \(\psi = \sqrt{2/L}\sin(\pi x/L)\) is largest at the centre of the well and small near the walls.

• Box (ii) raises the middle third, where the particle is most likely to be: probability about 0.61.
• Box (i) raises one outer third: \(\frac{1}{3} - \frac{\sin(2\pi/3)}{2\pi} \approx 0.20\).
• Box (iii) raises the outer sixth at each end, where the wavefunction is smallest: \(2\left(\frac{1}{6} - \frac{\sin(\pi/3)}{2\pi}\right) \approx 0.06\).

So \(E_{2} \gt E_{1} \gt E_{3}\). Boxes (i) and (iii) raise the same total width, \(L/3\), so a reading that ignores where the step sits would make \(E_{1} = E_{3}\); the other orders get the position dependence backwards.

Answer: B

The ground-state energy is \(\frac{2}{5}E\).

In a square well of side \(L\), \(\psi = \sin\frac{n_{x}\pi x}{L}\sin\frac{n_{y}\pi y}{L}\) and \(E = E_{0}(n_{x}^{2} + n_{y}^{2})\) with \(E_{0} = \pi^{2}\hbar^{2}/2mL^{2}\). A node only along \(y = L/2\), and none along any line of constant \(x\), means \(n_{x} = 1\) and \(n_{y} = 2\):

\[ \begin{aligned} E &= E_{0}(1 + 4) = 5E_{0} \\ E_{ground} &= E_{0}(1 + 1) = 2E_{0} = \frac{2}{5}E \end{aligned} \]

1/4 comes from comparing only the \(y\) parts (1 against 4), and 1/2 and 3/8 match no pair of states.

Answer: A

The expectation value of energy is \(\dfrac{17}{6}E_{1}\).

For a stationary-state expansion, ⟨E⟩ = Σ |c_n|²E_n; the phase factors drop out of the squared moduli. For an infinite well E_n = n²E₁, so E₂ = 4E₁ and E₃ = 9E₁:

\[ \begin{aligned} \langle E \rangle &= \tfrac{2}{3}E_{1} + \tfrac{1}{6}(4E_{1}) + \tfrac{1}{6}(9E_{1}) \\ &= \tfrac{4 + 4 + 9}{6}E_{1} = \tfrac{17}{6}E_{1} \end{aligned} \]

The probabilities 2/3 + 1/6 + 1/6 add to 1, so the state is normalized. 2/3 E₁ keeps only the ground-state term, 3/2 E₁ treats the levels as E_n = nE₁, and 14E₁ adds the energies without weighting them.

Answer: A

The answer is \(\langle x \rangle = \frac{L}{2}\); \(\langle E \rangle = \frac{\hbar^{2}}{2m}\frac{3\pi^{2}}{L^{2}}\).

Energy: the states have \(E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}\), and the probabilities are \(\frac{1}{3}\) and \(\frac{2}{3}\):

\[ \begin{aligned} \langle E \rangle &= \tfrac{1}{3}E_{1} + \tfrac{2}{3}(4E_{1}) = 3E_{1} \\ &= \frac{\hbar^{2}}{2m}\frac{3\pi^{2}}{L^{2}} \end{aligned} \]

Position: each stationary state alone has \(\langle x \rangle = L/2\). The cross term is proportional to \(2\,\text{Re}(c_{1}^{*}c_{2})\int \psi_{1}x\psi_{2}\,dx\), and since \(c_{1}^{*}c_{2} = i\sqrt{2}/3\) is purely imaginary, it vanishes at this instant. So \(\langle x \rangle = L/2\).

The \(\frac{2L}{3}\) options mistake the probability weights for positions, and \(8\pi^{2}\) is \(2E_{2}\), double counting the excited state.

Answer: B

The fourth excited state has four nodes.

In an infinite square well the states are \( \psi_{n} = \sqrt{2/L}\,\sin(n\pi x/L) \), with \( n = 1 \) the ground state. The \( n \)th state has \( n - 1 \) nodes inside the well, not counting the walls where every state vanishes.

The fourth excited state is \( n = 5 \), so it has \( 5 - 1 = 4 \) nodes. Three nodes belong to the third excited state (\( n = 4 \)), and five or six would count the walls or skip a level.

Answer: B

The minimum depth is \(\frac{\hbar^{2}\pi^{2}}{8ma^{2}}\).

The infinite wall at \(x = 0\) forces \(\psi(0) = 0\), so inside the well \(\psi = A\sin kx\) with \(k = \sqrt{2m(V_{0} + E)}/\hbar\), and outside it decays as \(e^{-\kappa x}\). A bound state just appears when E reaches zero from below and the inside wave has turned through a quarter wavelength by \(x = a\), so that it can join a flat exterior solution:

\[ \begin{aligned} ka &= \frac{\pi}{2},\quad k^{2} = \frac{2mV_{0}}{\hbar^{2}} \\ V_{0} &= \frac{\pi^{2}\hbar^{2}}{8ma^{2}} \end{aligned} \]

This is the odd-parity condition of a finite well of width 2a, so unlike a symmetric well it needs a minimum depth. \(\frac{\hbar^{2}\pi^{2}}{2ma^{2}}\) is the ground-state energy of an infinite well of width a, a different quantity.

Answer: D

There are 9 eigenstates with this energy.

The levels of a cubical box are

\[ \begin{aligned} E &= \frac{\pi^{2}\hbar^{2}}{2mL^{2}} \\ &\quad\times(n_{x}^{2} + n_{y}^{2} + n_{z}^{2}) \end{aligned} \]

so \(19\hbar^{2}\pi^{2}/mL^{2}\) needs \(n_{x}^{2} + n_{y}^{2} + n_{z}^{2} = 38\) with positive integers. The only ways to write 38 as a sum of three positive squares are

\[ \begin{aligned} 38 &= 36 + 1 + 1 \quad (6, 1, 1) \\ 38 &= 25 + 9 + 4 \quad (5, 3, 2) \end{aligned} \]

(6, 1, 1) can be arranged in 3 ways and (5, 3, 2) in 3! = 6 ways, giving 3 + 6 = 9 states. Stopping at one of the two sets gives 3 or 6, not an offered option; 10 and 4 come from miscounting the arrangements.

Answer: D

The WKB energies are \(E_{n}^{0} + \frac{V_{0}}{2}\left(1 + \frac{V_{0}}{8E_{n}^{0}}\right)\).

With infinite walls the WKB condition is \(\int_{0}^{a}p\,dx = n\pi\hbar\). The momentum is \(\sqrt{2m(E - V_{0})}\) over the shelf and \(\sqrt{2mE}\) beyond it, each for a/2. For \(E \gg V_{0}\) expand \(\sqrt{E - V_{0}}\) to second order in \(x = V_{0}/E\):

\[ \begin{aligned} \tfrac{a}{2}\sqrt{2mE}\left(2 - \tfrac{x}{2} - \tfrac{x^{2}}{8}\right) &= n\pi\hbar \\ 2mEa^{2}\left(1 - \tfrac{x}{4} - \tfrac{x^{2}}{16}\right)^{2} &= n^{2}\pi^{2}\hbar^{2} \\ E\left(1 - \tfrac{x}{2} - \tfrac{x^{2}}{16}\right) &= E_{n}^{0} \end{aligned} \]

\[ \begin{aligned} E &= E_{n}^{0} + \frac{V_{0}}{2} + \frac{V_{0}^{2}}{16E_{n}^{0}} \\ &= E_{n}^{0} + \frac{V_{0}}{2}\left(1 + \frac{V_{0}}{8E_{n}^{0}}\right) \end{aligned} \]

\(E_{n}^{0} + V_{0}/2\), the average of the shelf over the well, is only the first-order result, \(E_{n}^{0} + V_{0}\) would raise the whole floor, and the factor \(V_{0}/4E_{n}^{0}\) doubles the second-order term.

Answer: B

The keyed pressure is \(\frac{3\pi\hbar^{2}}{2mR^{5}}\).

The key treats the electrons in the s-wave levels of the spherical well, \(E_{n} = n^{2}\pi^{2}\hbar^{2}/2mR^{2}\). Two electrons (opposite spins) fill \(n = 1\) and the third goes to \(n = 2\):

\[ \begin{aligned} E &= (1 + 1 + 4)\frac{\pi^{2}\hbar^{2}}{2mR^{2}} \\ &= \frac{3\pi^{2}\hbar^{2}}{mR^{2}} \\ P &= -\frac{dE}{dV} = -\frac{dE/dR}{4\pi R^{2}} \\ &= \frac{6\pi^{2}\hbar^{2}}{mR^{3}}\cdot\frac{1}{4\pi R^{2}} \\ &= \frac{3\pi\hbar^{2}}{2mR^{5}} \end{aligned} \]

The pressure scales as \(R^{-5}\) because the energy goes as \(R^{-2}\) and the volume as \(R^{3}\), which rules out the \(R^{3}\) option; the factor 3/4 is half the correct coefficient and 7/2 does not come from these levels.

Note that in a full treatment the third electron would occupy the lowest \(l = 1\) level, whose energy (from the first zero of the spherical Bessel function \(j_{1}\), at 4.49) lies below the \(n = 2\), \(l = 0\) level; that gives a pressure not among the options, so the key rests on the s-wave picture.