Quantum Numbers & Quantization - Physical Chemistry 2

Consider a hydrogen atom in a uniform electric field $\mathbf{E} = E_0 \hat{z}$ (Stark effect). For the $n = 2$ level, the degeneracy is partially lifted. If we consider only first-order perturbation theory, how many distinct energy levels result from the original four-fold degenerate $n = 2$ state?
Answer: $2$ levels corresponding to states with different parities under inversion When you encounter Stark effect problems, remember that the key is analyzing how an external electric field affects degenerate energy levels through symmetry considerations and selection rules. For the hydrogen atom's $n = 2$ level, you start with four degenerate states: $|2,0,0\rangle$ , $|2,1,0\rangle$ , $|2,1,1\rangle$ , and $|2,1,-1\rangle$ . The electric field $\mathbf{E} = E_0 \hat{z}$ creates a perturbation $H' = eE_0z$ . In first-order perturbation theory, the energy shifts depend on matrix elements $\langle \psi | z | \psi \rangle$ between the degenerate states. The crucial insight is parity: the $z$ operator has odd parity, so it only connects states with opposite parities. The $2s$ state ( $l = 0$ ) has even parity, while the $2p$ states ( $l = 1$ ) have odd parity. This mixing creates linear combinations that split into exactly two energy levels: one higher and one lower than the original $n = 2$ energy. Option A is wrong because the field doesn't shift all states equally—it breaks the degeneracy. Option C incorrectly suggests $m_l$ values determine the splitting, but the $2p_{\pm 1}$ states remain degenerate due to cylindrical symmetry about the $z$ -axis. Option D assumes complete degeneracy removal, but first-order perturbation theory only partially lifts it. The correct answer is B: two levels emerge from the parity-based splitting between $s$ and $p$ character states. Study tip: For Stark effect problems, always check the symmetry of both the perturbation operator and the unperturbed states—this determines which states can mix and how degeneracy is lifted.
Consider a particle in a three-dimensional cubic box with sides of length $L$ . If the particle is in the quantum state described by quantum numbers $(n_x, n_y, n_z) = (2, 3, 1)$ , what is the ratio of the energy of this state to the energy of the ground state?
Answer: $\frac{14}{3}$ When dealing with a particle in a three-dimensional box, you're working with one of the fundamental models in quantum mechanics where a particle is confined to move freely within a cubic container but cannot escape. The key insight is that the total energy depends on all three quantum numbers. For a 3D cubic box, the energy of any state is given by $E_{n_x,n_y,n_z} = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)$ . The ground state has quantum numbers (1,1,1), giving $E_{ground} = \frac{h^2}{8mL^2}(1^2 + 1^2 + 1^2) = \frac{3h^2}{8mL^2}$ . For the state (2,3,1), the energy is $E_{2,3,1} = \frac{h^2}{8mL^2}(2^2 + 3^2 + 1^2) = \frac{h^2}{8mL^2}(4 + 9 + 1) = \frac{14h^2}{8mL^2}$ . The ratio is therefore $\frac{E_{2,3,1}}{E_{ground}} = \frac{14h^2/8mL^2}{3h^2/8mL^2} = \frac{14}{3}$ , confirming answer A. Answer B ( $\frac{7}{2}$ ) comes from incorrectly calculating the sum of squares as 10.5 instead of 14. Answer C ( $\frac{11}{3}$ ) results from adding the quantum numbers linearly (2+3+1=6) instead of using their squares, then making an arithmetic error. Answer D ( $\frac{13}{4}$ ) appears to miscalculate both the numerator and use the wrong ground state energy. Remember: for particle-in-a-box problems, always square the quantum numbers first, then add them. The energy is proportional to the sum of squares, not the sum of the numbers themselves.
A quantum system has four energy levels with degeneracies: $E_1$ (non-degenerate), $E_2$ (doubly degenerate), $E_3$ (triply degenerate), and $E_4$ (non-degenerate). If $E_2 - E_1 = E_3 - E_2 = E_4 - E_3 = \Delta E$ , how many distinct transition frequencies are possible between all pairs of levels?
Answer: 3 distinct frequencies: $\Delta E$ , $2\Delta E$ , and $3\Delta E$ When analyzing spectroscopic transitions in quantum systems, the key principle is that transition frequencies depend only on energy differences between levels, not on the degeneracy of those levels. Degeneracy refers to multiple quantum states having the same energy, but since they're at identical energies, transitions between degenerate sublevels within different energy levels all have the same frequency. In this system, you have four energy levels separated by equal intervals of $\Delta E$ . The possible energy differences between any two levels are: adjacent levels differ by $\Delta E$ (like $E_2 - E_1$ ), levels separated by one give $2\Delta E$ (like $E_3 - E_1$ ), and levels separated by two give $3\Delta E$ (like $E_4 - E_1$ ). Each energy difference corresponds to exactly one transition frequency, regardless of how many ways you can achieve that difference due to degeneracy. Answer A incorrectly suggests 6 frequencies by somehow double-counting the three basic energy differences. Answer B falls into the trap of thinking degeneracy creates new frequencies—while degeneracy affects transition intensities (making some transitions stronger), it doesn't create new frequencies. Answer C makes a similar error, suggesting that individual sublevel transitions within degenerate levels produce distinct frequencies, which is impossible since sublevels within a degenerate level have identical energies. The correct answer is D: only 3 distinct frequencies exist, corresponding to $\Delta E$ , $2\Delta E$ , and $3\Delta E$ . Study tip: Remember that degeneracy affects transition intensities and selection rules, but never creates additional frequencies. The frequency depends solely on energy differences between levels.
A particle in a three-dimensional cubic box has quantum numbers $n_x = 2$ , $n_y = 3$ , and $n_z = 1$ . If the box dimensions are $L_x = L_y = L_z = a$ , how many other quantum states have the same total energy as this state?
Answer: 2 additional degenerate states beyond the given state Energy depends on $n_x^2 + n_y^2 + n_z^2 = 2^2 + 3^2 + 1^2 = 4 + 9 + 1 = 14$ . We need all permutations of (2,3,1): (2,3,1), (2,1,3), (3,2,1), (3,1,2), (1,2,3), (1,3,2). That's 6 total states, so 2 additional beyond the given state. Choice A miscounts total states. Choice B incorrectly counts 5 additional states. Choice D misses several permutations.
In the Bohr model, an electron transitions from $n=5$ to $n=3$ . If this transition occurs in a hydrogen-like ion with nuclear charge $Z$ , how does the wavelength of the emitted photon compare to the same transition in hydrogen ( $Z=1$ )?
Answer: The wavelength is $\frac{1}{Z^2}$ times shorter than in hydrogen In hydrogen-like ions, $E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2}$ . The transition energy is $\Delta E = Z^2 \cdot 13.6 \text{ eV} \left(\frac{1}{3^2} - \frac{1}{5^2}\right) = Z^2 \Delta E_{H}$ , where $\Delta E_{H}$ is the hydrogen transition energy. Since $E = \frac{hc}{\lambda}$ , we have $\lambda = \frac{hc}{E}$ . Therefore $\lambda_{ion} = \frac{hc}{Z^2 \Delta E_{H}} = \frac{1}{Z^2} \cdot \frac{hc}{\Delta E_{H}} = \frac{1}{Z^2} \lambda_{H}$ . Choice A reverses the relationship. Choice C uses $Z$ instead of $Z^2$ . Choice D has wrong power and direction.
An electron in a hydrogen atom transitions from the $n=4$ to $n=2$ level. If the orbital angular momentum quantum number changes from $l=2$ to $l=1$ , what is the total change in the magnitude of orbital angular momentum (in units of $\hbar$ )?
Answer: $\sqrt{6} - \sqrt{2}$ The magnitude of orbital angular momentum is given by $L = \sqrt{l(l+1)}\hbar$ . For $l=2$ : $L_i = \sqrt{2(3)}\hbar = \sqrt{6}\hbar$ . For $l=1$ : $L_f = \sqrt{1(2)}\hbar = \sqrt{2}\hbar$ . The change is $\Delta L = L_f - L_i = (\sqrt{2} - \sqrt{6})\hbar$ , so the magnitude of change is $|\Delta L| = (\sqrt{6} - \sqrt{2})\hbar$ . Choice B uses incorrect formula $\sqrt{l(l-1)}$ . Choice C incorrectly uses $L = l\hbar$ . Choice D incorrectly uses $L = \sqrt{l^2 + l}\hbar$ .
An electron with spin quantum number $s = \frac{1}{2}$ is in an orbital with orbital angular momentum quantum number $l = 1$ . When considering spin-orbit coupling, what are the possible values of the total angular momentum quantum number $j$ ?
Answer: $j = \frac{1}{2}, \frac{3}{2}$ from the vector addition of orbital and spin angular momenta When you encounter spin-orbit coupling problems, you're dealing with the vector addition of two types of angular momentum: orbital ( $l$ ) and spin ( $s$ ). The total angular momentum quantum number $j$ follows specific coupling rules. For any electron, the possible $j$ values are determined by: $j = |l \pm s|$ . With $l = 1$ and $s = \frac{1}{2}$ , you calculate both possible combinations: - $j = l + s = 1 + \frac{1}{2} = \frac{3}{2}$ (parallel coupling) - $j = |l - s| = |1 - \frac{1}{2}| = \frac{1}{2}$ (antiparallel coupling) Therefore, $j = \frac{1}{2}, \frac{3}{2}$ , making answer C correct. Answer A incorrectly suggests that $j$ is limited by the spin value alone. While individual electron spin is indeed $\frac{1}{2}$ , the total angular momentum can exceed this through vector addition with orbital momentum. Answer B assumes $j$ equals the orbital quantum number, ignoring spin entirely—this would only be true if spin-orbit coupling didn't exist. Answer D provides integer values that don't follow the proper vector addition rules; the calculation $|1 \pm \frac{1}{2}|$ cannot yield integers like 0 or 1. Remember the key rule: $j = |l \pm s|$ always gives you exactly two possible values when both $l$ and $s$ are non-zero. Practice identifying these coupling scenarios—they're fundamental to understanding atomic spectroscopy and electron configurations in multi-electron atoms.
A quantum mechanical system has energy eigenvalues $E_n = \alpha n^{3/2}$ where $\alpha$ is a positive constant and $n = 1, 2, 3, ...$ If this system is in thermal equilibrium at temperature $T$ , what is the ratio of populations $\frac{N_2}{N_1}$ between the second and first excited states?
Answer: $\exp\left(-\frac{\alpha(2\sqrt{2} - 1)}{k_B T}\right)$ from the energy difference between levels When you encounter thermal equilibrium problems with quantum systems, you need to apply the Boltzmann distribution, which states that the ratio of populations between any two energy levels depends on their energy difference: $\frac{N_j}{N_i} = \exp\left(-\frac{E_j - E_i}{k_B T}\right)$ . For this system with $E_n = \alpha n^{3/2}$ , you first calculate the specific energy levels: $E_1 = \alpha(1)^{3/2} = \alpha$ and $E_2 = \alpha(2)^{3/2} = \alpha \cdot 2\sqrt{2} = 2\sqrt{2}\alpha$ . The energy difference is $E_2 - E_1 = 2\sqrt{2}\alpha - \alpha = \alpha(2\sqrt{2} - 1)$ . Applying the Boltzmann distribution: $\frac{N_2}{N_1} = \exp\left(-\frac{\alpha(2\sqrt{2} - 1)}{k_B T}\right)$ , which matches answer A. Answer B incorrectly uses $E_2$ alone rather than the energy difference $E_2 - E_1$ . The Boltzmann distribution always requires the energy difference between states, not individual energy values. Answer C assumes linear energy spacing with some average value, but this system has $n^{3/2}$ dependence, creating non-uniform spacing that must be calculated exactly. Answer D includes an unnecessary factor of $\frac{1}{2}$ , suggesting degeneracy considerations. However, the problem gives no indication that these energy levels are degenerate—each $n$ value corresponds to a single state. Study tip: Always remember that Boltzmann ratios depend on energy differences, and only include degeneracy factors when explicitly stated or when dealing with systems where you know multiple states share the same energy.
An electron in a hydrogen atom has quantum numbers $n=3$ , $l=1$ , $m_l=0$ , and $m_s=+\frac{1}{2}$ . Which of the following statements about possible transitions is correct?
Answer: The electron can transition to $n=2$ , $l=0$ states only, since $\Delta l = \pm 1$ is required Electric dipole selection rules require $\Delta l = \pm 1$ and $\Delta m_l = 0, \pm 1$ . From $l=1$ , the electron can only transition to $l=0$ or $l=2$ . For $n=2$ , only $l=0$ and $l=1$ are possible, so only $l=0$ satisfies $\Delta l = -1$ . Choice A incorrectly allows $l=2$ for $n=2$ . Choice C ignores that $n=1$ only has $l=0$ , making $\Delta l = -1$ valid, but the statement about no $\Delta n$ restrictions is misleading. Choice D incorrectly interprets $m_l=0$ as problematic.
A quantum mechanical system has angular momentum operators $\hat{L}_x$ , $\hat{L}_y$ , and $\hat{L}_z$ satisfying the usual commutation relations. If a system is prepared in an eigenstate of $\hat{L}_z$ with eigenvalue $+2\hbar$ , and then a measurement of $\hat{L}_x$ is performed, which statement about the possible outcomes is correct?
Answer: The measurement can yield multiple values determined by the total angular momentum quantum number $l$ When you encounter angular momentum measurement problems, remember that the key is understanding how commutation relations limit simultaneous measurements and how the total angular momentum quantum number $l$ determines all possible measurement outcomes. Since the system is in a $\hat{L}_z$ eigenstate with eigenvalue $+2\hbar$ , we know $m_l = +2$ . The angular momentum operators $\hat{L}_x$ , $\hat{L}_y$ , and $\hat{L}_z$ don't commute with each other, so you cannot simultaneously know exact values of different components. However, the possible measurement outcomes for any component are constrained by the total angular momentum quantum number $l$ . For $m_l = +2$ to be possible, we need $l \geq 2$ (since $m_l$ ranges from $-l$ to $+l$ ). The minimum case is $l = 2$ , giving possible $m_l$ values of $-2, -1, 0, +1, +2$ . When measuring $\hat{L}_x$ , the possible outcomes correspond to these same quantized values: $-2\hbar, -\hbar, 0, +\hbar, +2\hbar$ . Answer A is wrong because rotational symmetry doesn't preserve measurement outcomes when switching between non-commuting operators. Answer B incorrectly assumes equal probabilities—the actual probabilities depend on the overlap between $L_z$ and $L_x$ eigenstates, which aren't equal. Answer C confuses expectation value with measurement outcomes; while $\langle L_x \rangle = 0$ , individual measurements can yield non-zero values. Answer D correctly recognizes that the possible measurement values are determined by the quantum number $l$ , which constrains all angular momentum measurements for the system. Remember: for angular momentum problems, always identify $l$ first—it determines the complete set of possible measurement outcomes for any component.
A quantum mechanical harmonic oscillator with frequency $\omega$ has its potential energy function shifted upward by a constant $V_0$ . If the original ground state energy was $E_0 = \frac{1}{2}\hbar\omega$ , what is the effect of this shift on the quantum numbers and energy levels of the system?
Answer: The quantum numbers remain unchanged, and all energy levels increase by $V_0$ with no change in level spacing When dealing with quantum mechanical systems, it's crucial to understand how changes to the potential energy function affect the overall system. The key insight here is distinguishing between absolute energy values and the underlying structure of the system. When you shift a potential energy function upward by a constant $V_0$ , you're adding the same amount of energy to every point in space. This is like raising the entire "energy landscape" uniformly. Since the Schrödinger equation depends on the shape of the potential (its derivatives and relative differences), not its absolute position, the quantum mechanical solutions remain structurally identical. The correct answer is A because adding a constant $V_0$ to the potential simply adds $V_0$ to every energy eigenvalue. The quantum numbers $n = 0, 1, 2, ...$ remain unchanged, and each energy level becomes $E_n = \hbar\omega(n + \frac{1}{2}) + V_0$ . The spacing between adjacent levels stays $\hbar\omega$ . Option B incorrectly suggests quantum numbers shift - but quantum numbers are determined by the potential's shape, not its vertical position. Option C mistakenly claims the level spacing changes to $\hbar\omega + V_0$ - this confuses adding energy to each level with changing the differences between levels. Option D is wrong because while the relative energy differences stay the same, the absolute energy values do increase. Study tip: Remember that only changes in potential shape affect quantum numbers and level spacing. Constant shifts only translate all energies uniformly upward.
An electron in a hydrogen atom is initially in the state $|n,l,m_l\rangle = |3,2,1\rangle$ . If the atom is subjected to a time-dependent perturbation that can cause transitions with $\Delta l = \pm 1$ and $\Delta m_l = 0$ , which of the following represents all possible final states after one transition?
Answer: $|3,1,1\rangle$ only, since $l = 3$ is not allowed for $n = 3$ When you encounter quantum mechanical selection rules, you need to consider both the transition rules AND the physical constraints of the quantum system. This question tests your understanding of allowed quantum states in hydrogen atoms. The initial state $|3,2,1\rangle$ can transition according to $\Delta l = \pm 1$ and $\Delta m_l = 0$ . This means $l$ can change from 2 to either 1 or 3, while $m_l$ stays at 1. However, you must also check whether the resulting states are physically allowed. For hydrogen atoms, the angular momentum quantum number $l$ must satisfy $0 \leq l \leq n-1$ . Since we're staying in the $n=3$ level, $l$ can only be 0, 1, or 2. This immediately eliminates $l=3$ as a possibility, leaving only the transition to $|3,1,1\rangle$ . Answer B is correct because $l=3$ violates the fundamental constraint $l \leq n-1$ for $n=3$ . Answer A incorrectly includes $|3,3,1\rangle$ , which is impossible since $l=3$ exceeds the maximum allowed value for $n=3$ . Answer C introduces transitions that change the principal quantum number $n$ , but the perturbation described typically doesn't cause such transitions in simple cases. Answer D acknowledges that $|3,3,1\rangle$ exists but claims zero transition probability, when actually this state doesn't exist at all. Study tip: Always verify that quantum states satisfy the fundamental constraints ( $0 \leq l \leq n-1$ and $|m_l| \leq l$ ) before applying selection rules. The physics of the atom itself can eliminate transitions that selection rules might otherwise allow.
A particle is confined to move on the surface of a sphere of radius $R$ . The quantized energy levels for this system depend on the quantum number $l$ according to $E_l = \frac{\hbar^2 l(l+1)}{2IR^2}$ , where $I$ is the moment of inertia. For a given energy level with quantum number $l = 2$ , what is the degeneracy of this level?
Answer: $5$ representing all allowed orientations of angular momentum vector When you encounter problems involving particles confined to curved surfaces like spheres, you're dealing with angular momentum quantization, where the key insight is recognizing how many distinct quantum states share the same energy. For a particle on a sphere, the energy depends only on the orbital angular momentum quantum number $l$ , but each energy level $E_l$ corresponds to multiple quantum states. This is because angular momentum is a vector quantity that can point in different directions in space while maintaining the same magnitude. The magnetic quantum number $m_l$ specifies these different orientations, and it can take values from $-l$ to $+l$ in integer steps. For $l = 2$ , the possible values of $m_l$ are: $-2, -1, 0, +1, +2$ . That's exactly 5 different quantum states, all with the same energy $E_2$ , making the degeneracy equal to 5. This represents all the allowed orientations of the angular momentum vector in space. Looking at the wrong answers: (A) confuses this with translational motion in Cartesian coordinates, which isn't relevant here. (B) miscounts the magnetic quantum numbers—if $m_l$ ranges from $-l$ to $+l$ , that's $2l+1$ values, not $l+2$ . (D) incorrectly brings in spin angular momentum, but the problem only deals with orbital motion on the sphere's surface. Remember: for any orbital angular momentum quantum number $l$ , the degeneracy is always $2l+1$ , representing all possible spatial orientations of the angular momentum vector.
Consider two electrons in a helium atom. If one electron is in the $2s$ orbital and the other is in the $2p$ orbital, how many distinct quantum states are possible for this two-electron configuration when considering the Pauli exclusion principle?
Answer: $12$ states from all combinations of spin and orbital quantum numbers When analyzing multi-electron quantum states, you need to systematically count all possible combinations of quantum numbers for each electron while ensuring each electron occupies a distinct quantum state. For this helium configuration, start by identifying the available quantum states. The $2s$ orbital corresponds to quantum numbers $n=2$ , $\ell=0$ , $m_\ell=0$ , giving two possible states when considering spin: $|2s, m_s = +\frac{1}{2}\rangle$ and $|2s, m_s = -\frac{1}{2}\rangle$ . The $2p$ orbitals have $n=2$ , $\ell=1$ , with three possible $m_\ell$ values $(-1, 0, +1)$ , each accommodating two spin states, yielding six total $2p$ states. Since the electrons occupy different orbitals ( $2s$ vs $2p$ ), the Pauli exclusion principle is automatically satisfied—no two electrons share identical quantum numbers. You can place the first electron in any of the 2 available $2s$ states and the second electron in any of the 6 available $2p$ states, giving $2 \times 6 = 12$ distinct configurations. Option A incorrectly limits the count to 6, perhaps only considering $2p$ variations. Option C suggests that antisymmetrization reduces the count to 8, but antisymmetrization doesn't eliminate states—it determines wavefunction form. Option D claims electron-electron repulsion restricts possibilities to 4, but repulsion affects energy, not the number of possible quantum states. Study tip: For multi-electron state counting, multiply the number of available single-electron states for each orbital involved, provided the Pauli principle is satisfied. Don't confuse state counting with energy considerations.

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