Hydrogen Atom: Orbitals & Energies - Physical Chemistry 2

The quantum numbers for hydrogen atom orbitals must satisfy certain relationships. For an orbital with $n=4$ , which set of quantum numbers $(n, l, m_l)$ represents a physically impossible state?
Answer: $(4, 4, 0)$ , representing a hypothetical $4g$ orbital When you encounter quantum number problems, you need to check whether the proposed combinations follow the fundamental rules that govern electron behavior in atoms. The key relationships are: for any principal quantum number $n$ , the angular momentum quantum number $l$ can only range from 0 to $(n-1)$ , and the magnetic quantum number $m_l$ can range from $-l$ to $+l$ . With $n=4$ , the maximum possible value for $l$ is 3. Option B violates this fundamental rule because it shows $(4, 4, 0)$ . Here, $l=4$ , but with $n=4$ , the maximum allowed $l$ value is $n-1=3$ . This makes the state physically impossible—no such orbital can exist in nature. Option A shows $(4, 2, -1)$ for a $4d$ orbital. This works because $l=2$ (which defines $d$ orbitals) is less than 4, and $m_l=-1$ falls within the allowed range of $-2$ to $+2$ . Option C gives $(4, 3, +2)$ for a $4f$ orbital. Since $l=3$ defines $f$ orbitals and is exactly $n-1=3$ , this is allowed. The $m_l=+2$ value fits within the range $-3$ to $+3$ . Option D presents $(4, 0, 0)$ for the $4s$ orbital, which is perfectly valid since $l=0$ defines $s$ orbitals and $m_l=0$ is the only possible value when $l=0$ . Study tip: Always check $l < n$ first when evaluating quantum number sets—this catches the most common violations and helps you quickly eliminate impossible states.
A hydrogen atom is prepared in a state $\psi = \frac{1}{\sqrt{2}}(\psi_{200} + \psi_{210})$ . If we measure the total energy, what is the probability of obtaining $E = -3.4$ eV?
Answer: $1$ , because both $\psi_{200}$ and $\psi_{210}$ have the same energy in hydrogen When you encounter quantum superposition problems involving hydrogen atom wavefunctions, the key insight is understanding how energy eigenvalues work in degenerate systems. The hydrogen atom has a special property: all orbitals with the same principal quantum number $n$ have identical energies, regardless of their $\ell$ and $m$ values. Both $\psi_{200}$ and $\psi_{210}$ have $n = 2$ , so they're degenerate with energy $E_2 = -13.6/2^2 = -3.4$ eV. Since the given superposition state is a linear combination of these two degenerate eigenstates, any energy measurement will yield $-3.4$ eV with probability $1$ . The measurement collapses the wavefunction to some combination of the $n = 2$ states, but the energy outcome is deterministic. Choice A incorrectly suggests only $\psi_{210}$ contributes to this energy, missing that both states are degenerate. Choice C applies probability amplitude rules incorrectly—while you do square amplitudes to get probabilities, this doesn't apply when both components have identical eigenvalues. There's no interference effect on energy measurements when dealing with degenerate states. Choice D reflects a classical misconception that superposition creates "intermediate" values—quantum measurements always yield specific eigenvalues, not averages. Remember this pattern: when a superposition involves only degenerate eigenstates (same eigenvalue), measuring that observable gives that eigenvalue with certainty. The quantum uncertainty lies in other observables, not the one for which the states are degenerate.
Consider two hydrogen atoms: one in the $2s$ state and one in the $2p$ state. If we ignore electron spin and consider only the spatial wavefunctions, which statement about their quantum mechanical properties is correct?
Answer: Both atoms have the same total energy, but different orbital angular momentum probability distributions When analyzing hydrogen atom wavefunctions, you need to consider how quantum numbers affect both energy and angular momentum properties. For hydrogen atoms, the energy depends only on the principal quantum number $n$ , while orbital angular momentum depends on the orbital angular momentum quantum number $\ell$ . Both the $2s$ and $2p$ states have the same principal quantum number ( $n = 2$ ), so they have identical energies in hydrogen. However, they differ in their $\ell$ values: $2s$ has $\ell = 0$ while $2p$ has $\ell = 1$ . This means their orbital angular momentum magnitudes are different ( $\sqrt{\ell(\ell+1)}\hbar$ ), giving them completely different probability distributions for angular momentum measurements. Looking at the wrong answers: A) incorrectly states both atoms have the same angular momentum distributions – the different $\ell$ values make this impossible. C) brings up orbital penetration, which affects energy in multi-electron atoms but not in hydrogen where energy depends only on $n$ . The penetration concept would be relevant for atoms with electron-electron repulsion, but hydrogen has only one electron. D) incorrectly claims the radial distributions are the same – $2s$ and $2p$ orbitals have distinctly different radial wavefunctions, with $2s$ having a radial node that $2p$ lacks. The correct answer is B because hydrogen's energy depends solely on $n$ (making energies equal), while the different $\ell$ values create different angular momentum properties. Study tip: Remember that in hydrogen, energy = $f(n)$ only, but angular momentum = $f(\ell)$ . This distinction is crucial for understanding hydrogen's degeneracy.
For the hydrogen atom, consider the radial probability density $P(r) = r^2|R_{nl}(r)|^2$ . At what distance from the nucleus is the radial probability density maximum for the 2s orbital, and how does this compare to the most probable distance for the 1s orbital?
Answer: The 2s maximum occurs at $r = 5.2a_0$ , which is approximately 5 times the 1s maximum at $r = a_0$ When analyzing radial probability density for hydrogen orbitals, you're looking at where an electron is most likely to be found at a given distance from the nucleus. The radial probability density $P(r) = r^2|R_{nl}(r)|^2$ combines the radial wavefunction with the geometric factor $r^2$ that accounts for the increasing volume of spherical shells at larger radii. To find the maximum, you need to differentiate $P(r)$ and set it equal to zero. For the 1s orbital, this gives the well-known result that the most probable distance is exactly $a_0$ (the Bohr radius). For the 2s orbital, the calculation is more complex due to the radial node, but the maximum occurs at approximately $r = 5.2a_0$ . Looking at the wrong answers: Answer A gives the correct ratio of 4 but uses an incorrect distance of $5.3a_0$ for the 2s orbital. Answer C uses $4a_0$ for the 2s maximum, which underestimates the actual distance - this might come from oversimplifying the 2s radial wavefunction. Answer D suggests $6a_0$ , which overestimates the 2s maximum distance. The correct answer is B because $5.2a_0$ accurately represents the 2s radial probability maximum, and this is indeed approximately 5 times the 1s maximum at $a_0$ . Remember that radial probability maxima don't scale simply with $n^2$ like orbital energies do. The presence of radial nodes and the specific form of each radial wavefunction creates more complex scaling relationships that require careful calculation rather than simple pattern recognition.
The radial probability density for finding an electron at distance $r$ in the hydrogen atom 2s orbital has two maxima. If the ratio of the larger maximum to the smaller maximum is approximately 4:1, at approximately what ratio of distances $r_{max,2}/r_{max,1}$ do these maxima occur?
Answer: 6:1, because the outer maximum occurs near the classical Bohr radius for $n=2$ For the 2s orbital, the radial probability density $P(r) = r^2|R_{20}(r)|^2$ has maxima at approximately $r_1 \approx 0.76a_0$ and $r_2 \approx 5.24a_0$ , giving a ratio of about 6.9:1 ≈ 6:1. The outer maximum occurs near $4a_0$ (the $n=2$ Bohr radius), while the inner maximum is much closer to the nucleus. Choice A incorrectly assumes simple $n^2$ scaling. Choice B incorrectly relates probability and distance ratios. Choice D overestimates the ratio and misattributes it to exponential factors.
An electron in a hydrogen atom is in a superposition of $2p$ orbitals: $\psi = \frac{1}{\sqrt{3}}\psi_{2,1,-1} + \sqrt{\frac{2}{3}}\psi_{2,1,0}$ . If we measure the $z$ -component of orbital angular momentum, what is the expectation value $\langle L_z \rangle$ ?
Answer: $\langle L_z \rangle = -\frac{\hbar}{3}$ , calculated from the probability-weighted average When you encounter a superposition of quantum states, you need to calculate the expectation value using the probability-weighted average of the possible measurement outcomes. For the $z$ -component of orbital angular momentum, $L_z = m_l\hbar$ , where $m_l$ is the magnetic quantum number. The given wavefunction $\psi = \frac{1}{\sqrt{3}}\psi_{2,1,-1} + \sqrt{\frac{2}{3}}\psi_{2,1,0}$ contains two components: one with $m_l = -1$ (giving $L_z = -\hbar$ ) and one with $m_l = 0$ (giving $L_z = 0$ ). The probabilities are the squares of the coefficients: $P(-\hbar) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$ and $P(0) = \left(\sqrt{\frac{2}{3}}\right)^2 = \frac{2}{3}$ . The expectation value is: $\langle L_z \rangle = \frac{1}{3}(-\hbar) + \frac{2}{3}(0) = -\frac{\hbar}{3}$ Answer A incorrectly assumes the average of the $m_l$ values themselves rather than properly weighting by probabilities. Answer C makes a sign error and incorrectly suggests the larger coefficient on $m_l = 0$ somehow makes the result positive. Answer D confuses the probability ( $\frac{2}{3}$ ) with the expectation value and ignores that $m_l = 0$ contributes zero to $L_z$ . Remember: expectation values require squaring the coefficients to get probabilities, then taking the weighted average of the observable's eigenvalues. The component with $m_l = 0$ contributes nothing to $L_z$ , so only the negative contribution matters here.
The expectation value of the kinetic energy for a hydrogen atom in the $n = 2, l = 0$ state can be calculated using the virial theorem. If the total energy is $E_2 = -3.4$ eV, what is the ratio of the expectation value of kinetic energy to the magnitude of the expectation value of potential energy?
Answer: 1:2, because the virial theorem gives $\langle T \rangle = -\frac{1}{2}\langle V \rangle$ for Coulomb potentials When you encounter questions about hydrogen atom energetics, the virial theorem is your key tool for relating kinetic and potential energies. For any system with a Coulomb potential (like the electron-proton interaction), this theorem establishes a specific relationship between these energy components. The virial theorem states that for a Coulomb potential $V \propto r^{-1}$ , the kinetic energy $\langle T \rangle$ and potential energy $\langle V \rangle$ are related by $\langle T \rangle = -\frac{1}{2}\langle V \rangle$ . Since the total energy is $E = \langle T \rangle + \langle V \rangle$ , we can substitute to get $E = -\frac{1}{2}\langle V \rangle + \langle V \rangle = \frac{1}{2}\langle V \rangle$ . This means $\langle V \rangle = 2E = 2(-3.4) = -6.8$ eV, and $\langle T \rangle = -\frac{1}{2}(-6.8) = +3.4$ eV. The ratio of kinetic to potential energy magnitudes is therefore $3.4:6.8 = 1:2$ . Choice A correctly identifies this relationship. Choice B incorrectly assumes equal magnitudes - this would violate the virial theorem for Coulomb potentials. Choice C reverses the actual relationship, suggesting kinetic energy dominates when it's actually half the potential energy magnitude. Choice D introduces irrelevant scaling arguments that don't apply to the virial theorem relationship. Remember: For any hydrogen-like atom problem involving energy relationships, the virial theorem gives you $\langle T \rangle = -\frac{1}{2}\langle V \rangle$ immediately. This 1:2 ratio of kinetic to potential energy magnitudes is universal for Coulomb systems, regardless of quantum state.
The effective nuclear charge experienced by an electron in hydrogen-like atoms can be approximated as $Z_{eff} = Z - \sigma$ , where $\sigma$ is the shielding constant. For a Li $^{2+}$ ion (which is hydrogen-like with $Z=3$ ), what is the ionization energy of the electron from the ground state?
Answer: $122.4 \text{ eV}$ , because the nuclear charge is three times larger than hydrogen When you encounter hydrogen-like atoms (ions with only one electron), the key insight is understanding how nuclear charge affects electron binding energy. The ionization energy formula for hydrogen-like atoms is $E = 13.6 \times Z^2 \text{ eV}$ , where the energy scales with the square of the nuclear charge. For Li $^{2+}$ , you have $Z = 3$ (three protons in the nucleus). Since this ion has only one electron, there's no electron shielding to consider - the $\sigma$ term in the shielding equation equals zero. The effective nuclear charge is simply $Z_{eff} = 3 - 0 = 3$ . Therefore, the ionization energy is $E = 13.6 \times 3^2 = 13.6 \times 9 = 122.4 \text{ eV}$ . Answer A incorrectly assumes that having one electron makes the ionization energy identical to hydrogen, ignoring the crucial role of nuclear charge. Answer B uses the correct $Z^2$ scaling but miscalculates: $13.6 \times 3^2 = 122.4$ , not $40.8$ eV. Answer D falls into the linear scaling trap, incorrectly thinking energy scales as $Z$ rather than $Z^2$ , which would give $13.6 \times 3 = 40.8$ eV. Remember this pattern: for hydrogen-like atoms, ionization energy always scales as $Z^2$ , not linearly with $Z$ . This quadratic relationship appears frequently in quantum chemistry problems, so memorize the formula $E = 13.6 \times Z^2$ eV for hydrogen-like systems.
Consider a hydrogen-like ion (single electron, nuclear charge $Z$ ) in the $n=2$ level. How does the most probable distance (where radial probability is maximum) compare to the corresponding distance in neutral hydrogen?
Answer: The most probable distance scales as $1/Z$ , so it decreases linearly with nuclear charge When analyzing hydrogen-like ions, you need to understand how the wave function and probability distributions scale with nuclear charge. The key insight is that increased nuclear charge pulls the electron closer to the nucleus, affecting all spatial properties proportionally. For hydrogen-like ions, the radial wave functions contain a scaling factor that depends on the Bohr radius $a_0/Z$ , where $a_0$ is the Bohr radius for hydrogen. This means all distance-related quantities, including the most probable distance, scale inversely with $Z$ . As nuclear charge increases, the electron is pulled closer by the stronger Coulombic attraction, and the radial probability maximum shifts inward by a factor of $1/Z$ . Choice A correctly identifies this $1/Z$ scaling relationship. The most probable distance decreases linearly as nuclear charge increases. Choice B incorrectly suggests $1/Z^2$ scaling like energy levels. While energies do scale as $Z^2$ , spatial distributions scale differently. Energy depends on the square of the wave function, while distances scale with the wave function itself. Choice C is wrong because the most probable distance definitely depends on $Z$ . The quantum numbers $n$ and $l$ determine the shape and nodal structure, but nuclear charge determines the overall size scale. Choice D has the scaling direction backwards. Higher $Z$ means stronger attraction, pulling the electron closer, not farther away. Study tip: Remember that in hydrogen-like systems, energies scale as $Z^2$ but distances scale as $1/Z$ . Keep these scaling relationships straight—they're fundamental to understanding atomic structure across the periodic table.
Consider the hydrogen atom radial wavefunctions $R_{nl}(r)$ . The 2s wavefunction has a radial node, while the 2p wavefunction does not. What is the physical significance of this difference in terms of the electron's behavior?
Answer: The 2s electron has a higher probability of being found near the nucleus than 2p due to penetration through the radial node When analyzing hydrogen atom wavefunctions, you need to understand how radial nodes affect electron distribution and penetration behavior. A radial node is a spherical surface where the wavefunction equals zero, but this doesn't prevent the electron from accessing regions closer to the nucleus. The 2s orbital has one radial node, creating a region where $R_{2s}(r) = 0$ . However, the wavefunction has significant amplitude both before and after this node. Crucially, the 2s wavefunction has substantial probability density very close to the nucleus (small r values). The 2p orbital, while having no radial nodes, has an angular node at the nucleus and its radial part goes as $r$ near the origin, giving it much lower probability density near the nucleus. This penetration difference means 2s electrons spend more time close to the positively charged nucleus than 2p electrons, despite the radial node. This is why the correct answer is B. Looking at the wrong answers: A incorrectly suggests the radial node prevents nuclear penetration - it doesn't. C misunderstands that penetration (not a "tunneling barrier") makes 2s more tightly bound, but this describes the effect, not the significance of the node itself. D falsely claims 2s has zero nuclear density while 2p has maximum density - it's actually the opposite. Study tip: Remember that radial nodes don't block electron penetration. The key concept is that s orbitals always penetrate closer to the nucleus than p orbitals of the same principal quantum number, regardless of radial nodes.
A hydrogen atom is in the $n=3$ energy level. If we measure the orbital angular momentum magnitude, what is the probability of obtaining a value corresponding to $l=1$ , assuming the atom was initially prepared in a state with equal probability amplitudes for all possible $l$ values?
Answer: $\frac{1}{3}$ , because there are 3 possible $l$ values $(0, 1, 2)$ and each is equally likely When you encounter quantum mechanics problems about measurement probabilities, the key is identifying what specific quantum state property is being measured and how the initial state is prepared. For a hydrogen atom in the $n=3$ level, the possible orbital angular momentum quantum numbers are $l = 0, 1, 2$ . The problem states the atom was prepared with "equal probability amplitudes for all possible $l$ values." This means the initial wavefunction is an equal superposition of the three $l$ states: $|\psi\rangle = \frac{1}{\sqrt{3}}(|l=0\rangle + |l=1\rangle + |l=2\rangle)$ . When measuring orbital angular momentum magnitude, you're determining which $l$ value the system collapses to. Since each $l$ state has equal amplitude ( $1/\sqrt{3}$ ), the probability of measuring any specific $l$ value is $|1/\sqrt{3}|^2 = 1/3$ . Therefore, the probability of obtaining $l=1$ is $\frac{1}{3}$ . Choice A incorrectly focuses on counting $m_l$ values rather than recognizing this is about $l$ measurement probabilities. Choice C makes the same error, confusing the number of $m_l$ states (which is relevant for energy degeneracy) with the probability of measuring a specific $l$ value. Choice D introduces an irrelevant "degeneracy weighting" concept that doesn't apply when the initial state has equal $l$ amplitudes. Study tip: In quantum measurement problems, always identify what observable is being measured and examine the initial state preparation. Equal probability amplitudes for different quantum numbers means equal measurement probabilities for those specific quantum numbers, regardless of their individual degeneracies.
Consider two hydrogen atom wavefunctions: $\psi_A = \frac{1}{\sqrt{2}}(\psi_{210} + \psi_{21-1})$ and $\psi_B = \frac{1}{\sqrt{2}}(\psi_{210} - \psi_{21-1})$ . Which statement correctly describes the relationship between these states?
Answer: Both states have the same energy but different angular probability distributions Both $\psi_A$ and $\psi_B$ are linear combinations of $n=2, l=1$ states, so they have the same energy ( $E_2 = -13.6/4 = -3.4$ eV). However, the different signs in the linear combinations create different angular probability distributions. Choice A is wrong because both states have $l=1$ . Choice B is incorrect because the angular distributions differ due to interference between $m_l = 0$ and $m_l = -1$ components. Choice C is wrong because energy depends only on $n$ in hydrogen, not on the specific linear combination.
The angular part of hydrogen wavefunctions are spherical harmonics $Y_l^m(\theta,\phi)$ . If we measure the $z$ -component of orbital angular momentum for an electron in the $3d_{xy}$ orbital, what are the possible outcomes?
Answer: $L_z = -2\hbar$ and $+2\hbar$ with probabilities $1/2$ each, because $d_{xy}$ is a linear combination of $m = \pm 2$ When you encounter questions about orbital angular momentum measurements, remember that the key is understanding how real atomic orbitals relate to the underlying spherical harmonic eigenfunctions of the $L_z$ operator. The $3d_{xy}$ orbital is not an eigenfunction of the $L_z$ operator. Instead, it's constructed as a linear combination: $d_{xy} = \frac{1}{\sqrt{2}}(Y_2^{+2} - Y_2^{-2})$ . The eigenfunctions of $L_z$ are the spherical harmonics $Y_l^m$ , where measuring $L_z$ gives $m\hbar$ . Since $d_{xy}$ contains equal contributions from $m = +2$ and $m = -2$ states, measuring $L_z$ yields $+2\hbar$ or $-2\hbar$ , each with probability $1/2$ . Answer A incorrectly assumes that because the $d_{xy}$ orbital lies in the $xy$ -plane, it has no $z$ -component of angular momentum. This confuses the orbital's spatial orientation with its angular momentum properties. Answer B suggests all five possible $d$ -orbital $L_z$ values with equal probability. This would only be true for a statistical mixture of all five $d$ orbitals, not for the specific $d_{xy}$ orbital. Answer D incorrectly assigns $m = \pm 1$ values to $d_{xy}$ . This is wrong— $d_{xy}$ corresponds to $m = \pm 2$ , while $d_{xz}$ and $d_{yz}$ involve $m = \pm 1$ combinations. Remember: real atomic orbitals are often linear combinations of spherical harmonics, so always consider which $m$ values contribute to determine possible $L_z$ measurement outcomes.
Consider the hydrogen atom wavefunction $\psi_{321} = R_{32}(r) Y_2^1(\theta, \phi)$ . At what value of the polar angle $\theta$ is the probability density $|\psi_{321}|^2$ maximized in the $xz$ -plane ( $\phi = 0$ )?
Answer: $\theta = 54.7°$ , corresponding to the tetrahedral angle where d-orbital lobes are oriented For $Y_2^1(\theta, \phi) \propto \sin\theta \cos\theta e^{i\phi}$ , in the xz-plane ( $\phi = 0$ ), $|Y_2^1|^2 \propto \sin^2\theta \cos^2\theta$ . To find the maximum, take $d/d\theta[\sin^2\theta \cos^2\theta] = 0$ , which gives $\cos(2\theta) = 0$ , so $\theta = 54.7°$ and $125.3°$ . Both give the same probability density. The angle $54.7°$ is indeed the tetrahedral angle. Choice A is wrong because $Y_2^1$ vanishes at $\theta = 0$ . Choice C is wrong because the maximum is not at $90°$ . Choice D gives the supplementary angle but asks for 'the' maximum.
A hydrogen atom is prepared in a superposition state $\psi = c_1\psi_{200} + c_2\psi_{210} + c_3\psi_{21-1}$ where $|c_1|^2 = 0.5$ , $|c_2|^2 = 0.3$ , and $|c_3|^2 = 0.2$ . After measuring the orbital angular momentum squared and obtaining the result $L^2 = 2\hbar^2$ , what is the probability of subsequently measuring $m_l = 0$ for the z-component of orbital angular momentum?
Answer: 0.6, because the measurement renormalizes the $l = 1$ subspace to unity When you encounter quantum measurement problems involving superposition states, you need to understand how measurements collapse the wavefunction and renormalize the remaining components. Initially, the hydrogen atom exists in a superposition of one 2s state ( $\psi_{200}$ , where $l=0$ ) and two 2p states ( $\psi_{210}$ and $\psi_{21-1}$ , where $l=1$ ). When you measure $L^2 = 2\hbar^2$ , you're determining that $l=1$ (since $L^2 = l(l+1)\hbar^2 = 1(2)\hbar^2$ ). This measurement eliminates the $l=0$ component entirely, leaving only the $l=1$ states. Before measurement, the $l=1$ subspace contained states with probabilities $|c_2|^2 = 0.3$ ( $m_l = 0$ ) and $|c_3|^2 = 0.2$ ( $m_l = -1$ ), totaling 0.5. After confirming $l=1$ , these probabilities must be renormalized to sum to 1. The probability of measuring $m_l = 0$ becomes $\frac{0.3}{0.3 + 0.2} = \frac{0.3}{0.5} = 0.6$ . Answer A correctly identifies this renormalization process. Answer B incorrectly assumes probabilities remain unchanged after measurement. Answer C wrongly suggests the measurement creates equal $m_l$ probabilities, ignoring the original coefficients. Answer D misunderstands how the s-orbital elimination affects the remaining probabilities. Remember: quantum measurements not only collapse the wavefunction but also renormalize the surviving components. Always divide the desired probability by the total probability of all compatible states after measurement.
A hydrogen atom undergoes a transition from the $n=4, l=2$ state to the $n=2, l=1$ state. What is the energy of the emitted photon, and what selection rules govern this transition?
Answer: Energy = $2.55 \text{ eV}$ ; transition allowed because $\Delta l = -1$ satisfies the selection rule, regardless of $\Delta n$ When analyzing atomic transitions, you need to consider both the energy change and whether selection rules permit the transition. The energy depends only on the principal quantum numbers, while transition probability depends on selection rules for angular momentum. For hydrogen atoms, the energy of each level is $E_n = -13.6/n^2$ eV. The initial state ( $n=4$ ) has energy $E_4 = -13.6/16 = -0.85$ eV, and the final state ( $n=2$ ) has energy $E_2 = -13.6/4 = -3.40$ eV. The photon energy equals the energy difference: $\Delta E = E_4 - E_2 = -0.85 - (-3.40) = 2.55$ eV. The key selection rule for electric dipole transitions is $\Delta l = \pm 1$ (orbital angular momentum must change by exactly one unit). There is no restriction on $\Delta n$ - the principal quantum number can change by any amount. Here, $\Delta l = 1 - 2 = -1$ , which satisfies the selection rule perfectly. Choice A incorrectly suggests that $\Delta n$ has selection rules - it doesn't for electric dipole transitions. Choice C miscalculates the energy as 1.89 eV and incorrectly claims $\Delta n = -2$ violates selection rules. Choice D falsely states that both $\Delta n$ and $\Delta l$ must equal $\pm 1$ , which would severely restrict allowed transitions. Remember: for hydrogen transitions, calculate energy using only the $n$ values, but check that $\Delta l = \pm 1$ for the transition to be allowed. The $n$ quantum number has no selection rule restrictions.
A hydrogen atom electron transitions from the $n = 3, l = 2, m_l = 1$ state to the $n = 2, l = 1, m_l = 0$ state. If the emitted photon has energy $\Delta E = 1.89$ eV, what can be concluded about the selection rules and the transition?
Answer: The transition is allowed by selection rules, and the photon carries orbital angular momentum $\hbar$ Electric dipole selection rules require $\Delta l = \pm 1$ and $\Delta m_l = 0, \pm 1$ . Here, $\Delta l = 1 - 2 = -1$ ✓ and $\Delta m_l = 0 - 1 = -1$ ✓, so the transition is allowed. The energy difference $\Delta E = E_2 - E_3 = -3.4 - (-1.51) = 1.89$ eV matches the given photon energy. The photon carries away the angular momentum difference $\Delta L = \hbar$ . Choice A incorrectly states the selection rules. Choice C is wrong about energy conservation. Choice D unnecessarily invokes relativistic effects.
Consider the hydrogen atom orbitals $3d_{x^2-y^2}$ and $3d_{z^2}$ . Both are real linear combinations of spherical harmonics $Y_2^m$ . Which statement about their nodal properties is correct?
Answer: Both orbitals have exactly two angular nodes, but they differ in the geometric arrangement of these nodal surfaces When analyzing d orbital nodal properties, remember that all d orbitals have the same angular momentum quantum number (ℓ = 2), which determines their fundamental nodal structure. The key insight is that orbitals with the same ℓ value must have the same number of angular nodes, regardless of their specific geometric arrangement. Both $3d_{x^2-y^2}$ and $3d_{z^2}$ orbitals have exactly two angular nodes, as required by the relationship: number of angular nodes = ℓ = 2. However, these nodal surfaces have different geometric arrangements. The $3d_{x^2-y^2}$ orbital has two perpendicular nodal planes that bisect the xy-plane at 45° angles to the x and y axes. The $3d_{z^2}$ orbital has a conical nodal surface (actually a double cone) that separates the dumbbell-shaped lobes along the z-axis from the donut-shaped lobe in the xy-plane. Option A incorrectly suggests $3d_{z^2}$ has additional nodes beyond the required two. Option B contains a counting error— $3d_{x^2-y^2}$ doesn't have four angular nodes, and both orbitals must have the same number of angular nodes. Option D incorrectly implies that the mathematical form of the orbital changes the number of angular nodes, when this is determined solely by the ℓ quantum number. Study tip: For any set of orbitals with the same ℓ value, always expect the same number of angular nodes—only their geometric arrangements differ. This is a fundamental constraint of quantum mechanics that appears frequently on physical chemistry exams.
Consider the hydrogen atom in the $n=3$ level. If an electron transitions from this level to $n=1$ , what is the wavelength of the emitted photon, and in what region of the electromagnetic spectrum does this fall?
Answer: $\lambda = 102.6$ nm; ultraviolet region, specifically in the Lyman series When you encounter hydrogen atom emission problems, you're dealing with electronic transitions between quantized energy levels. The key is using the Rydberg equation and understanding which spectral series corresponds to different final energy levels. For a hydrogen transition from $n=3$ to $n=1$ , use the Rydberg equation: $\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ , where $R_H = 1.097 \times 10^7 \text{ m}^{-1}$ . Substituting the values: $\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{1^2} - \frac{1}{3^2}\right) = 1.097 \times 10^7 \left(1 - \frac{1}{9}\right) = 1.097 \times 10^7 \times \frac{8}{9} = 9.751 \times 10^6 \text{ m}^{-1}$ . This gives $\lambda = 102.6$ nm. Since any transition ending at $n=1$ belongs to the Lyman series, and 102.6 nm falls in the ultraviolet region (10-400 nm), answer A is correct. Answer B (656.3 nm) represents the Balmer series ( $n=3 \to n=2$ ), not the transition specified. Answer C (121.6 nm) is the Lyman α line, which corresponds to $n=2 \to n=1$ , not $n=3 \to n=1$ . Answer D (486.1 nm) is another Balmer series transition ( $n=4 \to n=2$ ). Remember: transitions to $n=1$ are always Lyman series (UV), to $n=2$ are Balmer series (visible), and to $n=3$ are Paschen series (infrared). The final state determines the series name and approximate wavelength region.
The Bohr model predicts circular orbits with specific radii $r_n = n^2 a_0$ . In quantum mechanics, we instead have probability distributions. For the hydrogen $2p$ orbital, how does the quantum mechanical most probable distance compare to the Bohr prediction for $n=2$ ?
Answer: The quantum most probable distance is $5a_0$ , which is $25\%$ larger than the Bohr radius When comparing classical and quantum mechanical models of atomic structure, you're examining how our understanding evolved from fixed orbits to probability distributions. The Bohr model gives definite circular orbits, while quantum mechanics describes where electrons are most likely to be found. The Bohr radius for $n=2$ is $r_2 = 2^2 a_0 = 4a_0$ . However, quantum mechanics doesn't give us fixed orbits but rather radial probability distributions. For the hydrogen $2p$ orbital, you need to find the maximum of the radial probability function $P(r) = r^2 |R_{2p}(r)|^2$ , where $R_{2p}$ is the radial wave function. The calculation shows that the most probable distance for a $2p$ electron is $5a_0$ . This is $\frac{5a_0 - 4a_0}{4a_0} \times 100\% = 25\%$ larger than the Bohr prediction, confirming answer B. Answer A incorrectly assumes perfect agreement between the models. Answer C gives the wrong value and direction of difference— $3a_0$ would indeed be $25\%$ smaller, but this isn't the correct quantum mechanical result. Answer D represents a philosophical trap; while the models do represent different concepts (fixed orbits vs. probability distributions), their numerical predictions can absolutely be compared as distances from the nucleus. Remember that quantum mechanical orbitals typically give most probable distances that are somewhat larger than corresponding Bohr radii, especially for $p$ , $d$ , and $f$ orbitals. This reflects the more diffuse nature of electron probability clouds compared to classical circular orbits.

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