Question bank — Quantum chemistry — particle in a box revisited; H-atom solutions

True — $\psi$ itself takes negative values (see the second lobe of ${\psi }_2$ dipping below the dotted line in Figure 1); only $|\psi {|}^2$, the probability density, must be non-negative.

False — since $E_n\propto 1/L^2$ (proportionality 2 above), doubling $L$ divides every level by $4$, not by $2$. Bigger box means weaker confinement and smaller gaps — imagine the Figure 3 left ladder compressing downward.

True in pure hydrogen — energy depends only on $n$ (the accidental $1/r$ degeneracy); this breaks in multi-electron atoms where $\ell$ splits the levels.

False — the lowest state is $n=1$ with $E_1=h^2/8m{L}^2>0$ (the bottom rung of Figure 3 left, sitting above zero); this irreducible minimum is the Zero-point Energy, forced by the boundary conditions.

True — more nodes means more curvature in $\psi$ (compare the ever-sharper wiggles up the levels of Figure 1), and the kinetic operator $-\frac{{\hslash }^2}{2m}{\nabla }^2$ charges more energy for sharper curvature.

True — total nodes $=n-1=1$ for both; $2s$ puts its one node in the radial part, $2p$ puts its one node in the angular part.

False — $|\psi {|}^2$ is a probability density (probability per unit length in 1D, per unit volume in 3D); you must multiply by a length/volume to get an actual probability.

False — $\psi (0)=0$ is a boundary condition (the wavefunction is pinned to the wall in Figure 1), so the probability density there is exactly zero, like the fixed end of a guitar string.

Wrong — ${\psi }_2$ has a node at the centre $x=L/2$ (the green dot on the $n=2$ curve of Figure 1), so the density is zero there; the electron is most likely near $x=L/4$ and $3L/4$.

Wrong — you must weight by shell volume: in Figure 2 the blue density peaks at $r=0$ but the yellow radial distribution $P(r)=4\pi r^2|\psi {|}^2$ vanishes there (no volume) and peaks at $r=a_0$.

Wrong — single-valuedness $\Phi (\varphi )=\Phi (\varphi +2\pi )$ forces $e^{im2\pi }=1$, so $m_{\ell }$ must be an integer, and it is further restricted to the range $m_{\ell }=-\ell ,\dots ,+\ell$ (that is $2\ell +1$ values) — a "whole-number-fits" condition just like the box.

Wrong — quantized energy does not quantize position; the electron is spread out as a continuous probability cloud $|\psi {|}^2$ (the smooth curves of Figures 1–2) over all allowed space.

Wrong — $\ell$ and $m_{\ell }$ come from the angular equations and appear for any central potential; the $-1/r$ form (built from the elementary charge $e$ and vacuum permittivity ${\epsilon }_0$) is only special because it makes $E$ depend on $n$ alone.

Wrong — Figure 3 shows the contrast: the box (left) has $E_n\propto n^2$ with widening gaps, while hydrogen (right) has $E_n\propto -1/n^2$ with gaps that shrink and crowd toward $0$ at the ionization limit.

Wrong — only functions satisfying the boundary conditions (and normalizable) are eigenstates; those constraints are exactly what selects the discrete $E$ values in Operators and Eigenvalues.

Wrong — being an integer is necessary but not sufficient; $|m_{\ell }|$ cannot exceed $\ell$, so the allowed set is exactly $m_{\ell }=-\ell ,-\ell +1,\dots ,0,\dots ,+\ell$.

The reference zero differs, as Figure 3 makes visible: the box (left) rises above a $V=0$ floor (so $E>0$), while hydrogen (right) is measured against the free electron at $r\to \infty$ where $V\to 0$ (the red dashed line), so a bound electron sits below it, giving $E<0$.

Because $E_n\propto 1/L^2$ (proportionality 2), squeezing $L$ raises every level and widens gaps — picture the Figure 3 left ladder stretching upward — and this size-tunable gap is exactly what colours Quantum Dots.

Because momentum is promoted to the operator $\hat{p}=-i\hslash \nabla$ (the gradient that measures how fast $\psi$ changes), and $p^2/2m$ then becomes $-\frac{{\hslash }^2}{2m}{\nabla }^2$ — the Laplacian, i.e. curvature — acting on $\psi$; energy is read off as an eigenvalue, not computed from a trajectory.

Because $\hslash =h/2\pi$, so ${\pi }^2{\hslash }^2={\pi }^2{h}^2/4{\pi }^2=h^2/4$; dividing by $2m{L}^2$ turns ${\pi }^2{\hslash }^2/2$ into $h^2/8$. Same constant, just $h$ vs $\hslash$.

Only because $V(r)$ depends on $r$ alone (central) — that symmetry lets the $r$, $\theta$, $\varphi$ equations decouple into independent factors.

It gets the energies right by luck of the $1/r$ potential, but it wrongly pictures electrons on sharp circular orbits with definite position and momentum — the quantum picture in Bohr Model vs Quantum Mechanics replaces orbits with the probability clouds sketched in Figure 2.

The walls pin $\psi$ to zero at both ends, and only sines $\psi =A\sin (kx)$ with $kL=n\pi$ hit zero at both $x=0$ and $x=L$ at once. Since a half-wave spans $\lambda /2=\pi /k$, requiring $kL=n\pi$ is literally saying "$n$ half-waves fit in $L$" — count the humps of each Figure 1 curve to see $n$ half-waves. This whole-number condition is the origin of quantization.

Both add curvature to $\psi$, and the kinetic operator ${\nabla }^2$ penalises curvature; whether the wiggle is in $r$ or in angle, extra structure costs kinetic energy.

The levels $E_n\propto 1/L^2$ collapse toward each other, becoming effectively continuous — confinement is lost and the "free particle" limit (no quantization) is recovered.

Quantization survives but the rule changes: instead of "$n$ half-waves fit" you need "$n$ whole waves fit," giving $k=2\pi n/L$ and allowing $n=0$ (a flat, constant $\psi$) plus $\pm n$ pairs. This shows quantization comes from any boundary condition that forces waves to fit — not specifically from walls.

No — for the pinned box $n=0$ gives $\psi \equiv 0$ (no particle) and it is not a hydrogen solution; both ladders start at $n=1$. (A periodic box is the exception: there $n=0$ is a legitimate constant state.)

$n^2$ orbitals — summing $2\ell +1$ over $\ell =0\dots n-1$ gives $n^2$, all degenerate in pure hydrogen (each also holds two spins, but spin isn't in this simple Hamiltonian).

$\ell =2$ ($d$), because $\ell$ runs $0$ to $n-1$; a larger $\ell$ would demand more angular structure than the radial solution can support at that $n$.

Five values: $m_{\ell }=-2,-1,0,+1,+2$ (that is $2\ell +1=5$), giving the five $d$ orbitals.

No — in Figure 2 the blue $|{\psi }_{1s}{|}^2$ is largest at $r=0$ and only decreases with $r$; it is the yellow radial distribution $4\pi r^2|\psi {|}^2$ that peaks at $r=a_0$.

$\psi$ would no longer be forced to zero at the walls — it would leak (decay exponentially) outside, allowing tunnelling and giving only a finite number of bound states.