Physical Chemistry 2 Quiz: Hydrogen Atom Orbitals And Energies

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Hydrogen Atom Orbitals And EnergiesQuestion 1 of 20

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?

Both orbitals have the same number of angular nodes, but

3d_{z^2}

has additional conical nodes due to its axial symmetry

3d_{x^2-y^2}

has four angular nodes (nodal planes), while

3d_{z^2}

has two angular nodes (a conical surface)

Both orbitals have exactly two angular nodes, but they differ in the geometric arrangement of these nodal surfaces

3d_{z^2}

has more angular nodes than

3d_{x^2-y^2}

because it involves both

z^2

and

(x^2+y^2)

terms

The number of angular nodes depends on the measurement direction and cannot be specified without additional information

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Hydrogen Atom Orbitals And Energies

Practice Hydrogen Atom Orbitals And Energies in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Hydrogen Atom Orbitals And Energies, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

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?

Both orbitals have the same number of angular nodes, but $3d_{z^2}$ has additional conical nodes due to its axial symmetry
$3d_{x^2-y^2}$ has four angular nodes (nodal planes), while $3d_{z^2}$ has two angular nodes (a conical surface)
Both orbitals have exactly two angular nodes, but they differ in the geometric arrangement of these nodal surfaces (correct answer)
$3d_{z^2}$ has more angular nodes than $3d_{x^2-y^2}$ because it involves both $z^2$ and $(x^2+y^2)$ terms
The number of angular nodes depends on the measurement direction and cannot be specified without additional information

Explanation: 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.

Question 2

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?

$\frac{1}{9}$ , because there are 9 total possible $m_l$ values and 3 correspond to $l=1$
$\frac{1}{3}$ , because there are 3 possible $l$ values $(0, 1, 2)$ and each is equally likely (correct answer)
$\frac{3}{9} = \frac{1}{3}$ , because there are 3 $m_l$ states for $l=1$ out of 9 total $m_l$ states
$\frac{1}{4}$ , because the probability depends on the degeneracy weighting of the $l=1$ subshell
$\frac{3}{7}$ , because we must account for the radial probability distribution weighting

Explanation: 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

\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.

Question 3

The wavefunctions for hydrogen can be written as $\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi)$ . If we consider the 2p orbitals, which statement about the angular momentum properties is correct?

All three 2p orbitals have the same $L^2$ eigenvalue but different $L_z$ eigenvalues, and they form a complete basis for $l=1$ angular momentum states (correct answer)
The 2p orbitals have different $L^2$ eigenvalues because they point in different spatial directions, but the same $L_z$ eigenvalue
Only the $2p_z$ orbital is an eigenfunction of $L_z$ , while $2p_x$ and $2p_y$ are not eigenfunctions of any angular momentum operator
All 2p orbitals have zero angular momentum because the electron probability density has definite spatial orientation rather than circular motion
The $2p_x$ and $2p_y$ orbitals have undefined $L^2$ eigenvalues because they are linear combinations of spherical harmonics with different $l$ values

Explanation: When you encounter questions about hydrogen orbital angular momentum, remember that quantum mechanical angular momentum has two key operators:

L^2

(total angular momentum squared) and

L_z

(z-component of angular momentum). For any orbital with quantum number

l

, the

L^2

eigenvalue is

\hbar^2 l(l+1)

, while

L_z

eigenvalues are

\hbar m

where

m

ranges from

-l

+l

. Since all 2p orbitals have

l=1

, they share the same

L^2

eigenvalue of

2\hbar^2

. However, the three 2p orbitals correspond to different

m

values:

m = -1, 0, +1

, giving them different

L_z

eigenvalues of

-\hbar, 0, +\hbar

respectively. Together, these three orbitals span the complete set of

l=1

states, making answer A correct. Answer B incorrectly suggests that spatial orientation affects

L^2

eigenvalues – but

L^2

depends only on the

l

quantum number, not spatial direction. Answer C reflects a common misconception about the

2p_x

and

2p_y

orbitals, which are actually linear combinations of the

m = \pm 1

eigenstates, not the eigenstates themselves. While these aren't

L_z

eigenfunctions, they're still valid angular momentum states. Answer D incorrectly assumes that definite spatial orientation means zero angular momentum – but quantum angular momentum doesn't require classical circular motion. Remember: all orbitals with the same

l

value share identical

L^2

eigenvalues, regardless of their spatial appearance or

m

quantum numbers.

Question 4

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?

The radial node in 2s means the electron never penetrates to small r values, while 2p electrons can reach the nucleus
The 2s electron has a higher probability of being found near the nucleus than 2p due to penetration through the radial node (correct answer)
The radial node creates a tunneling barrier that makes 2s electrons more tightly bound than 2p electrons in multi-electron atoms
The radial node in 2s means this orbital has zero electron density at the nucleus, while 2p has maximum density there
The presence of the radial node allows 2s electrons to avoid electron-electron repulsion more effectively than 2p electrons

Explanation: 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.

Question 5

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?

Energy = $2.55 \text{ eV}$ ; transition allowed because $\Delta n = -2$ and $\Delta l = -1$ both satisfy selection rules
Energy = $2.55 \text{ eV}$ ; transition allowed because $\Delta l = -1$ satisfies the selection rule, regardless of $\Delta n$ (correct answer)
Energy = $1.89 \text{ eV}$ ; transition forbidden because $\Delta n = -2$ violates the selection rule requiring $\Delta n = \pm 1$
Energy = $2.55 \text{ eV}$ ; transition forbidden because both $\Delta n$ and $\Delta l$ must equal $\pm 1$ simultaneously
Energy = $3.4 \text{ eV}$ ; transition allowed because $\Delta l = -1$ is the only requirement for electric dipole transitions

Explanation: 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.

Question 6

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?

$13.6 \text{ eV}$ , same as hydrogen because both have one electron
$40.8 \text{ eV}$ , because $Z_{eff} = 3$ and energy scales as $Z^2$
$122.4 \text{ eV}$ , because the nuclear charge is three times larger than hydrogen (correct answer)
$27.2 \text{ eV}$ , because the energy scales linearly with nuclear charge
$81.6 \text{ eV}$ , because we must account for relativistic corrections at high $Z$

Explanation: 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.

Question 7

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?

Only $L_z = 0$ , because the $d_{xy}$ orbital has no angular momentum component along the $z$ -axis
$L_z = -2\hbar, -\hbar, 0, +\hbar, +2\hbar$ with equal probabilities of $1/5$ each
$L_z = -2\hbar$ and $+2\hbar$ with probabilities $1/2$ each, because $d_{xy}$ is a linear combination of $m = \pm 2$ (correct answer)
$L_z = -\hbar$ and $+\hbar$ with probabilities $1/2$ each, because $d_{xy}$ corresponds to $m = \pm 1$ states
The measurement is impossible because $d_{xy}$ is not an eigenfunction of $L_z$ , so angular momentum is undefined

Explanation: 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

-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.

Question 8

The radial distribution function for hydrogen $P(r) = r^2|R_{nl}(r)|^2$ gives the probability of finding an electron in a spherical shell at distance $r$ . For the $3d$ orbital, this function has a single maximum. What is the approximate location of this maximum, and how does it compare to the classical Bohr orbit for $n=3$ ?

Maximum at $r \approx 7.5a_0$ , which is less than the Bohr radius $r_{Bohr} = 9a_0$ for $n=3$
Maximum at $r \approx 9a_0$ , which exactly matches the Bohr radius for $n=3$
Maximum at $r \approx 12a_0$ , which is greater than the Bohr radius $r_{Bohr} = 9a_0$ for $n=3$ (correct answer)
Maximum at $r \approx 6a_0$ , which is significantly less than the Bohr radius for $n=3$
Maximum at $r \approx 15a_0$ , which is much greater than the classical expectation

Explanation: When you encounter questions about radial distribution functions, you're dealing with the probability of finding an electron at specific distances from the nucleus. The key insight is understanding how quantum mechanical orbitals differ from classical Bohr orbits. For the 3d orbital, the radial distribution function

P(r) = r^2|R_{31}(r)|^2

reaches its maximum at approximately

r \approx 12a_0

. This occurs because the 3d radial wavefunction has a specific nodal structure that pushes the probability density to larger radii. The

r^2

factor in the radial distribution function also favors larger distances, since it represents the increasing volume of spherical shells at greater radii. The classical Bohr radius for

n=3

r_{Bohr} = n^2a_0 = 9a_0

. The quantum mechanical maximum being larger than this classical prediction reflects the probabilistic nature of electron distributions in real atoms. Option A incorrectly suggests the maximum occurs at a smaller radius than the Bohr orbit, which contradicts the actual electron distribution. Option B incorrectly claims the quantum and classical predictions match exactly—this would be a remarkable coincidence that doesn't occur for d orbitals. Option D significantly underestimates the radial maximum, placing it even closer to the nucleus than option A. Study tip: Remember that quantum mechanical electron distributions often extend beyond classical Bohr predictions, especially for higher angular momentum states like d and f orbitals. The radial maxima typically occur at larger distances than the corresponding Bohr orbits due to the combined effects of the radial wavefunction and the

r^2

volume factor.

Question 9

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?

$\frac{1}{2}$ , because only the $\psi_{210}$ component contributes to this energy
$1$ , because both $\psi_{200}$ and $\psi_{210}$ have the same energy in hydrogen (correct answer)
$\frac{1}{4}$ , because we must square the probability amplitude and account for interference
$0$ , because the superposition creates a state with intermediate energy between the components
$\frac{3}{4}$ , because the measurement collapses the wavefunction with enhanced probability

Explanation: 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.

Question 10

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?

$(4, 2, -1)$ , representing a $4d$ orbital with specific $m_l$ value
$(4, 4, 0)$ , representing a hypothetical $4g$ orbital (correct answer)
$(4, 3, +2)$ , representing a $4f$ orbital with specific orientation
$(4, 0, 0)$ , representing the $4s$ orbital
$(4, 1, -1)$ , representing a $4p$ orbital with specific $m_l$ value

Explanation: 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

+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

+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

+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.

Question 11

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?

The most probable distance scales as $1/Z$ , so it decreases linearly with nuclear charge (correct answer)
The most probable distance scales as $1/Z^2$ , following the same scaling as the energy levels
The most probable distance is independent of $Z$ because it depends only on the quantum numbers $n$ and $l$
The most probable distance scales as $Z$ , increasing with nuclear charge due to electron-nucleus attraction
The most probable distance scales as $\sqrt{Z}$ , representing a compromise between energy and size scaling

Explanation: 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.

Question 12

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 $\langle L_z \rangle$ ?

$\langle L_z \rangle = 0$ , because the average of $-\hbar$ and $0$ weighted by probabilities is zero
$\langle L_z \rangle = -\frac{\hbar}{3}$ , calculated from the probability-weighted average (correct answer)
$\langle L_z \rangle = +\frac{\hbar}{3}$ , with positive sign due to the larger coefficient on $m_l = 0$
$\langle L_z \rangle = \frac{2\hbar}{3}$ , because the $\psi_{2,1,0}$ component dominates
$\langle L_z \rangle$ is undefined because this is not an eigenstate of $L_z$

Explanation: 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.

Question 13

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 $n=2$ ?

The quantum most probable distance is exactly $4a_0$ , matching the Bohr radius for $n=2$
The quantum most probable distance is $5a_0$ , which is $25\%$ larger than the Bohr radius (correct answer)
The quantum most probable distance is $3a_0$ , which is $25\%$ smaller than the Bohr radius
The quantum and classical predictions cannot be compared because they represent fundamentally different concepts
The quantum most probable distance is $6a_0$ , which is $50\%$ larger than the Bohr radius

Explanation: 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

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.

Question 14

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?

$\lambda = 102.6$ nm; ultraviolet region, specifically in the Lyman series (correct answer)
$\lambda = 656.3$ nm; visible region, specifically in the Balmer series
$\lambda = 121.6$ nm; ultraviolet region, but this is the Lyman $\alpha$ line
$\lambda = 486.1$ nm; visible region, corresponding to blue-green light
$\lambda = 97.2$ nm; far ultraviolet region, beyond the atmospheric cutoff

Explanation: 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

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.

Question 15

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?

The 2s maximum occurs at $r = 5.3a_0$ , which is exactly 4 times the 1s maximum at $r = a_0$
The 2s maximum occurs at $r = 5.2a_0$ , which is approximately 5 times the 1s maximum at $r = a_0$ (correct answer)
The 2s maximum occurs at $r = 4a_0$ , which is exactly 4 times the 1s maximum at $r = a_0$
The 2s maximum occurs at $r = 6a_0$ , which is exactly 6 times the 1s maximum at $r = a_0$
The 2s maximum occurs at $r = 2a_0$ , which is exactly 2 times the 1s maximum at $r = a_0$

Explanation: 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.

Question 16

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?

Both atoms have the same total energy and the same probability distribution for orbital angular momentum measurements
Both atoms have the same total energy, but different orbital angular momentum probability distributions (correct answer)
The atoms have different total energies because the $2s$ orbital penetrates closer to the nucleus than the $2p$ orbital
Both atoms have the same radial probability distribution but different angular probability distributions
The $2p$ atom has higher energy because it has non-zero orbital angular momentum, while $2s$ has zero angular momentum

Explanation: 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.

Question 17

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?

0.6, because the measurement renormalizes the $l = 1$ subspace to unity (correct answer)
0.3, because the original probability of the $m_l = 0$ state remains unchanged
0.5, because the measurement creates an equal superposition of $m_l$ states
0.4, because the measurement reduces the total probability by the s-orbital component

Explanation: 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.

Question 18

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?

1:2, because the virial theorem gives $\langle T \rangle = -\frac{1}{2}\langle V \rangle$ for Coulomb potentials (correct answer)
1:1, because kinetic and potential energies are equal in magnitude for bound states
2:1, because kinetic energy dominates in the quantum mechanical treatment of hydrogen
1:4, because the potential energy scales as $r^{-1}$ while kinetic energy scales as $n^{-2}$

Explanation: 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.

Question 19

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?

Both states have the same energy but different orbital angular momentum quantum numbers
Both states have the same energy and identical probability distributions for all observables
State A has higher energy than state B due to constructive interference effects
Both states have the same energy but different angular probability distributions (correct answer)

Explanation: 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.

Question 20

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?

2:1, because the maxima scale with the principal quantum number squared
4:1, because the probability ratio equals the distance ratio for radial functions
6:1, because the outer maximum occurs near the classical Bohr radius for $n=2$ (correct answer)
8:1, because radial nodes separate the maxima by exponential distance factors

Explanation: 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.