MM 765: Knowledge Assessment 2

Progress Question 1 of 20

Separation of Variables

When solving the 1D time-dependent Schrödinger's equation, we divide both sides by $\Psi(x,t) = \psi(x)\phi(t)$ to yield a purely position-dependent LHS and a purely time-dependent RHS. Both sides must equal a separation constant $\eta$. What fundamental physical quantity does $\eta$ represent?

The potential energy ($V(x)$) of the particle within the given field.

The exact momentum ($p$) of the wave packet envelope.

The total energy ($E$) of the particle, derived from $E = h\nu$.

The probability density amplitude.

Explanation

Correct Answer: C. The separation constant $\eta$ emerges because a function of strictly $x$ can only equal a function of strictly $t$ if both are equal to a constant. By solving the time-dependent portion $\eta = j\hbar \frac{1}{\phi(t)} \frac{d\phi(t)}{dt}$, it evaluates to $h\nu$, which by Planck-Einstein relation is the total energy $E$.

Time-Independence of Probability Density

According to Max Born's formulation, the total wave function is $\Psi(x,t) = \psi(x)e^{-j\omega t}$. Why is the probability density function $|\Psi(x,t)|^2$ physically stationary (completely independent of time)?

Because time dilation neutralizes the exponent at quantum scales.

Because squaring a complex exponential $|e^{-j\omega t}|^2$ always mathematically yields exactly 1.

Because particles in stationary states inherently lose their time dependence due to tunneling.

Because Max Born's postulate inherently ignored the time variable to simplify the calculus.

Explanation

Correct Answer: B. The probability density is the squared magnitude of the wave function: $|\Psi(x,t)|^2 = |\psi(x)|^2 \cdot |e^{-j\omega t}|^2$. Since the magnitude of Euler's complex exponential $e^{-j\theta}$ (which is $\cos^2\theta + \sin^2\theta$) is perfectly equal to 1, the time component vanishes, leaving $|\psi(x)|^2$, which is purely position-dependent.

Consequences of Confinement

For a particle confined in an infinite potential well of width $a$, applying the boundary condition $\psi(a) = 0$ yields the mathematical requirement that $ka = n\pi$ (where $n = 1, 2, 3...$). What is the profound physical consequence of this?

The particle's energy spectrum becomes discrete and quantized, forbidding a continuous range of energies.

The particle behaves purely as a classical object when bouncing between the walls.

The potential energy $V(x)$ within the well must dynamically fluctuate to maintain the boundary.

The probability of the particle tunnelling out of the infinite well becomes $n\pi$.

Explanation

Correct Answer: A. The restriction $k = \frac{n\pi}{a}$ fundamentally means that only specific wave numbers (and thus specific wavelengths) can perfectly fit within the box as standing waves. Because energy is intimately tied to the wave number ($E \propto k^2$), confining the particle dictates that it can only possess specific, discrete, quantized energy levels.

Particle in a Box

Use the slider to explore the wave function (solid blue) and probability density (shaded teal) of a particle confined in a 1D box. Pay close attention to the first excited state ($n=2$).

Quantum State ($n$) n = 1 (Ground)

If the particle is in the first excited state ($n=2$), what is the probability density of finding the particle at the exact center of the well ($x = a/2$)?

It is at its maximum peak probability.

It is exactly equal to the probability in the ground state ($n=1$).

It is identically zero (a node exists there).

It fluctuates wildly over time between 0 and 1.

Explanation

Correct Answer: C. The wave function is defined as $\psi_n(x) = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a})$. For $n=2$, evaluating at the center ($x=a/2$) gives $\sin(\frac{2\pi (a/2)}{a}) = \sin(\pi) = 0$. Since the wave function is zero, the probability density $|\psi|^2$ is also zero. This point is known as a node.

Quantum Tunnelling Dominance

In the potential barrier transmission approximation, $T \approx 16(\frac{E}{V_0})(1 - \frac{E}{V_0})e^{-2k_2a}$. While all variables affect $T$, which parameter's manipulation generally dictates the most dramatic (exponential) sensitivity in tunnelling probability?

The incident energy of the particle ($E$).

The width of the potential barrier ($a$).

The pre-exponential constant (16).

The particle's charge amplitude.

Explanation

Correct Answer: B. The barrier width '$a$' resides inside the argument of the exponential decay function ($e^{-2k_2a}$). While $E$ and $V_0$ also appear in the exponent (within $k_2$), their relationship is inside a square root. Changes to the physical barrier width '$a$' create direct linear changes to the exponent, resulting in massive, exponential variations in the final transmission probability $T$.

Tunnelling Transmission Math

Referencing the same formula, $T \approx 16(\frac{E}{V_0})(1 - \frac{E}{V_0})e^{-2k_2a}$. If an electron approaches a barrier such that its incident energy $E$ is exactly half of the barrier's height $V_0$, what does the pre-exponential factor simplify to?

Explanation

Correct Answer: B. If $E = 0.5 \cdot V_0$, then the ratio $\frac{E}{V_0} = 0.5$. Substituting this into the pre-exponential factor yields: $16 \cdot (0.5) \cdot (1 - 0.5) = 16 \cdot 0.5 \cdot 0.5 = 16 \cdot 0.25 = 4$.

The 3D Laplacian & Quantum Numbers

When solving the Schrödinger equation for a one-electron atom (using spherical coordinates and the Laplacian $\nabla^2$), three spatial quantum numbers naturally emerge. Which quantum number directly dictates the magnitude of the orbital angular momentum?

The Principal quantum number ($n$).

The Magnetic quantum number ($m_l$).

The Azimuthal / Angular quantum number ($l$).

The Spin quantum number ($s$).

Explanation

Correct Answer: C. The Azimuthal quantum number ($l$) defines the shape of the orbital subshell (s, p, d, f) and dictates the magnitude of orbital angular momentum. $n$ dictates total energy, $m_l$ dictates orientation in a magnetic field, and $s$ was added later to account for intrinsic spin.

Radial Probability Anomaly

For the ground state ($n=1, l=0$) of a Hydrogen-like atom, the wave function magnitude $|\psi|^2$ is strictly at its maximum exactly at the nucleus ($r=0$). However, the Radial Probability Density $P(r)$ evaluates to zero at $r=0$. What resolves this apparent paradox?

Strong nuclear repulsion strictly prohibits the electron from existing within $r=a_0$.

$P(r)$ multiplies $|\psi|^2$ by the differential volume of the spherical shell ($4\pi r^2 dr$), which vanishes mathematically at $r=0$.

The Pauli exclusion principle prevents the electron from occupying the nuclear coordinate.

At $r=0$, the principal quantum number $n$ drops to zero, undefined in the Schrödinger model.

Explanation

Correct Answer: B. The probability density at a specific point $|\psi|^2$ is highest at the origin. However, the radial probability $P(r)$ measures the chance of finding the electron anywhere on a spherical shell of radius $r$. The volume of that shell shrinks to zero as $r \to 0$ ($V_{shell} \propto r^2$). Thus, $0 \cdot |\psi|^2_{max} = 0$.

The Need for Spin

The spatial Schrödinger's equation naturally yields three quantum numbers ($n, l, m_l$). Why was it necessary for physicists to introduce a fourth quantum number: Spin ($s$)?

To correct for the relativistic mass gain of electrons moving near the speed of light.

To act as an integration constant required to normalize spherical harmonics.

To account for empirical observations, such as the anomalous splitting of emission spectral lines in a magnetic field.

Because time-dependent wavefunctions mandate four degrees of freedom.

Explanation

Correct Answer: C. The standard three quantum numbers could not fully explain experimental results, like the fine structure and anomalous Zeeman effect (where spectral lines split into closely spaced pairs). This required the postulate of an intrinsic angular momentum (spin) with two possible states ($+1/2, -1/2$).

Pauli's Exclusion Principle

Based strictly on Pauli's Exclusion Principle ("no two electrons may occupy the same quantum state"), what is the absolute maximum number of electrons that can share the exact same combination of spatial quantum numbers ($n, l, m_l$) in a single atom?

One

Two

Eight

Dependent on the Azimuthal number ($2l+1$)

Explanation

Correct Answer: B. The complete quantum state includes the spin quantum number ($s$), which can only be $+1/2$ (spin up) or $-1/2$ (spin down). Therefore, for any unique set of spatial numbers ($n, l, m_l$), you can fit exactly two electrons: one with spin up, and one with spin down. Any third electron would have to duplicate a full set of 4 quantum numbers, violating Pauli's principle.

Infinite Well Transitions

A particle in a 1D infinite well of width $a$ transitions from the second excited state to the ground state, emitting a photon of energy $E_{photon}$. If the well width is subsequently doubled ($2a$), what will be the energy of a photon emitted during a transition from the first excited state to the ground state in this new well, expressed in terms of $E_{photon}$?

$\frac{3}{32} E_{photon}$

$\frac{1}{4} E_{photon}$

$\frac{3}{8} E_{photon}$

$\frac{1}{32} E_{photon}$

Explanation

Correct Answer: A. For an infinite well, $E_n = n^2 E_1$. The first transition is from the second excited state ($n=3$) to ground ($n=1$), yielding $E_{photon} = E_3 - E_1 = 9E_1 - E_1 = 8E_1$. Thus, the original ground state energy $E_1 = E_{photon}/8$.

Doubling the width ($a \to 2a$) makes the new ground state energy $E'_1 = E_1 / 4$, because $E \propto 1/a^2$. The new transition is from the first excited state ($n=2$) to ground ($n=1$), yielding $\Delta E' = E'_2 - E'_1 = 4E'_1 - E'_1 = 3E'_1$.

Substituting $E'_1$: $\Delta E' = 3(E_1 / 4) = \frac{3}{4} E_1$. Finally, substituting $E_1$: $\Delta E' = \frac{3}{4} (E_{photon}/8) = \frac{3}{32} E_{photon}$.

Tunnelling Parameter Scaling

An electron with energy $E$ tunnels through a rectangular potential barrier of height $V_0$ and width $a$, with a transmission probability $T$. If we hypothetically double the barrier width ($2a$) but replace the incident electron with a lighter particle having exactly one-fourth the electron's mass ($m_e/4$), while keeping $E$ and $V_0$ identical, what is the new transmission probability?

Exactly $T$

$\frac{T^2}{16}$

$\sqrt{T}$

$\frac{T}{4}$

Explanation

Correct Answer: A. The transmission coefficient is $T \approx 16(\frac{E}{V_0})(1 - \frac{E}{V_0})e^{-2k_2a}$.
The pre-exponential factor depends strictly on $E$ and $V_0$, which remain unchanged, meaning the pre-factor is identical.
The exponent depends on $-2ak_2$, where $k_2 = \frac{\sqrt{2m(V_0-E)}}{\hbar}$. If mass is quartered ($m \to m/4$), $k_2$ is halved ($k_2 \to k_2/2$). If width is doubled ($a \to 2a$), the product inside the exponent becomes $(2a) \cdot (k_2/2) = a k_2$.
Thus, both the exponent and prefactor remain perfectly identical, yielding the exact same transmission probability $T$.

Time-Dependence of Superposition States

The time-dependent part of a stationary state is $\phi(t) = e^{-j\omega t}$, where $\omega = E/\hbar$. If a particle is instead in a superposition of the ground state ($E_1$) and the first excited state ($E_2$), the total wave function is $\Psi(x,t) = c_1\psi_1(x)e^{-j\omega_1 t} + c_2\psi_2(x)e^{-j\omega_2 t}$. Unlike a stationary state, the probability density $|\Psi(x,t)|^2$ of this superposition state fluctuates. What is the angular frequency of this spatial oscillation ("beating")?

$\frac{E_1 + E_2}{\hbar}$

$\frac{E_2 - E_1}{\hbar}$

$\frac{\sqrt{E_1 E_2}}{\hbar}$

It does not oscillate; superposition states also have time-independent probability densities.

Explanation

Correct Answer: B. When computing the probability density $|\Psi|^2 = \Psi\Psi^*$ for a superposition state, you must multiply the function by its complex conjugate. The cross-terms will produce mathematically oscillating components: $e^{-j(\omega_1 - \omega_2)t} + e^{j(\omega_1 - \omega_2)t}$. Using Euler's identity, this simplifies to a cosine function dependent on the difference between the angular frequencies. Thus, the beat frequency is $\omega_{beat} = |\omega_2 - \omega_1| = \frac{|E_2 - E_1|}{\hbar}$.

Maximizing Radial Probability

For the ground state of a hydrogen-like atom, the radial probability density is mathematically given by $P(r) = \frac{4}{a_0^3} r^2 e^{-2r/a_0}$. By evaluating the extremum of this function ($\frac{dP}{dr} = 0$), at what exact radial distance $r$ is the electron most likely to be found?

$r = a_0 / 2$

$r = a_0$

$r = 2a_0$

$r = 0$

Explanation

Correct Answer: B. To find the extremum of $P(r)$, set its derivative with respect to $r$ to zero using the product rule: $\frac{d}{dr}[r^2 e^{-2r/a_0}] = 2r e^{-2r/a_0} + r^2 (-2/a_0) e^{-2r/a_0} = 0$.
Factoring out the common terms gives $2r(1 - r/a_0)e^{-2r/a_0} = 0$. Discarding $r=0$ (the node at the nucleus where $P=0$) and $r \to \infty$, the maximum occurs strictly at $1 - r/a_0 = 0$, meaning $r = a_0$. This precisely matches the radius derived by the classical Bohr model.

Quantum State Degeneracy

Consider an electron occupying a highly excited state in a hydrogen atom with the principal quantum number $n = 4$. Accounting for all permitted values of the azimuthal ($l$), magnetic ($m_l$), and spin ($s$) quantum numbers, what is the absolute total number of unique, degenerate quantum states available at this specific energy level?

Explanation

Correct Answer: B. For a given principal quantum number $n$, the possible azimuthal numbers are $l = 0, 1, ..., n-1$. For each $l$, there are $2l+1$ magnetic states ($m_l$). The mathematical sum of all spatial states across all subshells for a given $n$ evaluates exactly to $n^2$.
For $n=4$, there are $4^2 = 16$ distinct spatial states (orbitals). Because Pauli's Exclusion Principle allows exactly 2 electrons (spin up $+1/2$, spin down $-1/2$) per orbital state, the total number of degenerate quantum states is $2n^2 = 2(16) = 32$.

Tunnelling Attenuation Factor

A beam of electrons with kinetic energy $E = 2.0$ eV approaches a rectangular potential barrier of height $V_0 = 5.0$ eV. If the physical width of the barrier is expanded by exactly $\Delta a = 0.1$ nm, by what approximate multiplicative factor does the transmitted electron probability decrease?

(Assume the pre-exponential factor is negligible in the ratio. Use $m_e = 9.11 \times 10^{-31}$ kg, $\hbar = 1.054 \times 10^{-34}$ J·s, $1$ eV = $1.602 \times 10^{-19}$ J)

~ 2.4

~ 5.9

~ 8.9

~ 34.8

Explanation

Correct Answer: B. The transmission probability is $T \propto e^{-2k_2a}$. The ratio of the old to new probability is $T_1/T_2 = e^{2k_2\Delta a}$.
First, calculate $k_2 = \frac{\sqrt{2m_e(V_0-E)}}{\hbar}$.
$V_0 - E = 3.0$ eV $= 3.0 \times 1.602 \times 10^{-19} = 4.806 \times 10^{-19}$ J.
$k_2 = \frac{\sqrt{2 \times (9.11 \times 10^{-31}) \times (4.806 \times 10^{-19})}}{1.054 \times 10^{-34}} \approx 8.878 \times 10^9$ m$^{-1}$.
The exponent is $2k_2\Delta a = 2(8.878 \times 10^9)(0.1 \times 10^{-9}) = 1.7756$.
The factor of decrease is $e^{1.7756} \approx 5.90$.

Quantum Well Expansion & Emission

An electron is initially in the ground state of a 1D infinite potential well of width $L$. The well suddenly expands symmetrically to a new width of $3L$. The electron eventually relaxes into the 3rd excited state ($n=4$) of this new widened well. From there, it transitions down to the ground state of the new well, emitting a photon. What is the energy of this emitted photon, expressed as a multiple of the electron's original ground state energy $E_1$ (before expansion)?

$\frac{5}{3} E_1$

$\frac{4}{9} E_1$

$\frac{8}{3} E_1$

$\frac{15}{8} E_1$

Explanation

Correct Answer: A. The initial ground state energy is $E_1 = \frac{\pi^2 \hbar^2}{2mL^2}$.
In the new well of width $3L$, the general energy levels are $E'_n = \frac{n^2 \pi^2 \hbar^2}{2m(3L)^2} = \frac{n^2}{9} E_1$.
The electron falls from the 3rd excited state ($n=4$) to the ground state ($n=1$) of the new well.
The photon energy is $\Delta E = E'_4 - E'_1 = \frac{16}{9}E_1 - \frac{1}{9}E_1 = \frac{15}{9}E_1 = \frac{5}{3}E_1$.

Sectional Probability

Consider an electron occupying the first excited state ($n=2$) of a 1D infinite potential well of width $a$. What is the exact mathematical probability of finding the electron strictly in the middle third of the well (i.e., between $x = a/3$ and $x = 2a/3$)?

$\frac{1}{3}$

$\frac{1}{3} - \frac{\sqrt{3}}{4\pi}$

$\frac{1}{3} + \frac{\sqrt{3}}{4\pi}$

$\frac{1}{4}$

Explanation

Correct Answer: B. The probability is $P = \int_{a/3}^{2a/3} |\psi_2(x)|^2 dx = \frac{2}{a} \int_{a/3}^{2a/3} \sin^2(\frac{2\pi x}{a}) dx$.
Substituting $u = \frac{2\pi x}{a}$, the limits become $2\pi/3$ to $4\pi/3$, and $dx = \frac{a}{2\pi}du$. Thus, $P = \frac{1}{\pi} \int_{2\pi/3}^{4\pi/3} \sin^2(u) du = \frac{1}{2\pi} \int_{2\pi/3}^{4\pi/3} (1 - \cos(2u)) du$.
Evaluating this yields $\frac{1}{2\pi} [u - \frac{1}{2}\sin(2u)]_{2\pi/3}^{4\pi/3}$.
The $u$ terms give $\frac{4\pi}{3} - \frac{2\pi}{3} = \frac{2\pi}{3}$. The sine terms give $\frac{1}{2}(\sin(8\pi/3) - \sin(4\pi/3)) = \frac{1}{2}(\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})) = \frac{\sqrt{3}}{2}$.
Combining these, $P = \frac{1}{2\pi} (\frac{2\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{1}{3} - \frac{\sqrt{3}}{4\pi} \approx 19.55\%$.

Most Probable Radius

The normalized radial wave function for the $2p$ orbital ($n=2, l=1$) of a Hydrogen atom is given by $R_{21}(r) = \frac{1}{\sqrt{24 a_0^3}} \frac{r}{a_0} e^{-r/(2a_0)}$. By maximizing the radial probability density function $P(r)$, determine the exact distance $r$ from the nucleus where the electron is most likely to be found.

$r = a_0$

$r = 2a_0$

$r = 4a_0$

$r = 5a_0$

Explanation

Correct Answer: C. The radial probability density is $P(r) = r^2 |R(r)|^2$. Substituting $R_{21}$, we get $P(r) = C r^4 e^{-r/a_0}$ (where $C$ is a constant incorporating the normalization factor).
To find the maximum, set the derivative $\frac{dP}{dr} = 0$. Using the product rule: $\frac{d}{dr}[r^4 e^{-r/a_0}] = 4r^3 e^{-r/a_0} - \frac{r^4}{a_0} e^{-r/a_0} = 0$.
Factoring out $r^3 e^{-r/a_0}$, we get $4 - \frac{r}{a_0} = 0$, which means the extremum (maximum probability) strictly occurs at $r = 4a_0$.

3D Well Degeneracy & Energy

An electron is confined to a 3D cubical potential well of side length $L$. The permitted energy levels are mathematically defined by $E = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)$. What is the exact energy of the second excited state, and what is the total number of permitted independent quantum states (accounting for intrinsic spin) available precisely at this energy level?

$E = \frac{3h^2}{4mL^2}$, Total States = 3

$E = \frac{9h^2}{8mL^2}$, Total States = 6

$E = \frac{3h^2}{8mL^2}$, Total States = 6

$E = \frac{11h^2}{8mL^2}$, Total States = 2

Explanation

Correct Answer: B. The energy is proportional to $n^2 = n_x^2 + n_y^2 + n_z^2$.
Ground state: $(1,1,1) \to n^2 = 3$.
First excited state: $(2,1,1), (1,2,1), (1,1,2) \to n^2 = 6$.
Second excited state: $(2,2,1), (2,1,2), (1,2,2) \to n^2 = 9$. Thus, $E = 9 \cdot (\frac{h^2}{8mL^2})$.
There are 3 spatially degenerate combinations. Since electrons are fermions with spin $\pm 1/2$, Pauli's Exclusion Principle allows exactly 2 electrons per spatial state. The total number of available independent quantum states is $3 \times 2 = 6$.

Assessment Complete!

Separation of Variables

Explanation

Time-Independence of Probability Density

Explanation

Consequences of Confinement

Explanation

Particle in a Box

Explanation

Quantum Tunnelling Dominance

Explanation

Tunnelling Transmission Math

Explanation

The 3D Laplacian & Quantum Numbers

Explanation

Radial Probability Anomaly

Explanation

The Need for Spin

Explanation

Pauli's Exclusion Principle

Explanation

Infinite Well Transitions

Explanation

Tunnelling Parameter Scaling

Explanation

Time-Dependence of Superposition States

Explanation

Maximizing Radial Probability

Explanation

Quantum State Degeneracy

Explanation

Tunnelling Attenuation Factor

Explanation

Quantum Well Expansion & Emission

Explanation

Sectional Probability

Explanation

Most Probable Radius

Explanation

3D Well Degeneracy & Energy

Explanation