Physical Chemistry 2 Quiz: Quantum Numbers And Quantization

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Quantum Numbers And QuantizationQuestion 1 of 14

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?

The quantum numbers remain unchanged, and all energy levels increase by

V_0

with no change in level spacing

The quantum numbers shift by

+1

, but the energy differences between adjacent levels remain

\hbar\omega

The quantum numbers remain unchanged, but the energy level spacing increases to

\hbar\omega + V_0

Both quantum numbers and energy levels remain completely unchanged since potential shifts don't affect bound systems

The quantum numbers become non-integer values, and the energy levels are no longer equally spaced by

\hbar\omega

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Quantum Numbers And Quantization

Practice Quantum Numbers And Quantization 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 Quantum Numbers And Quantization, 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

The quantum numbers remain unchanged, and all energy levels increase by $V_0$ with no change in level spacing (correct answer)
The quantum numbers shift by $+1$ , but the energy differences between adjacent levels remain $\hbar\omega$
The quantum numbers remain unchanged, but the energy level spacing increases to $\hbar\omega + V_0$
Both quantum numbers and energy levels remain completely unchanged since potential shifts don't affect bound systems
The quantum numbers become non-integer values, and the energy levels are no longer equally spaced by $\hbar\omega$

Explanation: 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.

Question 2

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?

$3$ corresponding to the three Cartesian coordinate directions
$4$ from the possible values of magnetic quantum number $m_l$
$5$ representing all allowed orientations of angular momentum vector (correct answer)
$6$ when including both orbital and spin angular momentum contributions
$9$ from the square of the principal quantum number relationship

Explanation: 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

+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

+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.

Question 3

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?

$1$ level because the electric field shifts all $n = 2$ states equally
$2$ levels corresponding to states with different parities under inversion (correct answer)
$3$ levels when grouping states by their $m_l$ quantum numbers
$4$ levels since the field completely removes all degeneracy
$0$ additional levels because the first-order correction vanishes by symmetry

Explanation: 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.

Question 4

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?

$\exp\left(-\frac{\alpha(2\sqrt{2} - 1)}{k_B T}\right)$ from the energy difference between levels (correct answer)
$\exp\left(-\frac{\alpha \sqrt{2}}{k_B T}\right)$ using the ratio of energy levels directly
$\exp\left(-\frac{3\alpha}{2k_B T}\right)$ from approximate linear energy spacing
$\frac{1}{2}\exp\left(-\frac{\alpha(2\sqrt{2} - 1)}{k_B T}\right)$ including degeneracy factors
$\frac{2}{1}\exp\left(-\frac{\alpha \sqrt{8} - \alpha}{k_B T}\right)$ with statistical weight corrections

Explanation: 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.

Question 5

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?

The measurement will yield $+2\hbar$ with certainty due to rotational symmetry
The measurement can yield values $-2\hbar, -\hbar, 0, +\hbar, +2\hbar$ with equal probabilities
The measurement will yield $0$ with certainty because $\langle L_x \rangle = 0$ in $L_z$ eigenstates
The measurement can yield multiple values determined by the total angular momentum quantum number $l$ (correct answer)
The measurement is impossible because $\hat{L}_x$ and $\hat{L}_z$ do not commute

Explanation: 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

+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.

Question 6

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?

$|3,1,1\rangle$ and $|3,3,1\rangle$ only, maintaining the principal quantum number
$|3,1,1\rangle$ only, since $l = 3$ is not allowed for $n = 3$ (correct answer)
$|2,1,1\rangle$ , $|3,1,1\rangle$ , and $|4,1,1\rangle$ with $\Delta l = -1$ transitions
$|3,1,1\rangle$ and $|3,3,1\rangle$ , but $|3,3,1\rangle$ has zero transition probability
$|2,1,1\rangle$ , $|2,3,1\rangle$ , $|4,1,1\rangle$ , and $|4,3,1\rangle$ from all allowed combinations

Explanation: 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.

Question 7

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?

$6$ states corresponding to different spin combinations and $p$ orbital orientations
$12$ states from all combinations of spin and orbital quantum numbers (correct answer)
$8$ states when accounting for indistinguishability and antisymmetrization requirements
$4$ states due to restrictions from electron-electron repulsion interactions
$24$ states including all permutations of electrons between orbitals

Explanation: 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

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.

Question 8

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 $j$ ?

$j = \frac{1}{2}$ only, since spin cannot exceed $\frac{1}{2}$ for a single electron
$j = 1$ only, determined by the orbital angular momentum value
$j = \frac{1}{2}, \frac{3}{2}$ from the vector addition of orbital and spin angular momenta (correct answer)
$j = 0, 1$ when considering both parallel and antiparallel coupling arrangements
$j = \frac{1}{2}, 1, \frac{3}{2}$ including all intermediate coupling possibilities

Explanation: 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.

Question 9

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?

$\frac{14}{3}$ (correct answer)
$\frac{7}{2}$
$\frac{11}{3}$
$\frac{13}{4}$
$\frac{5}{2}$

Explanation: 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.

Question 10

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?

6 distinct frequencies corresponding to energy differences $\Delta E$ , $2\Delta E$ , and $3\Delta E$
12 distinct frequencies accounting for degeneracy-weighted transitions
9 distinct frequencies from all possible sublevel transitions
3 distinct frequencies: $\Delta E$ , $2\Delta E$ , and $3\Delta E$ (correct answer)

Explanation: 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.

Question 11

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?

5 degenerate states total (including the given state)
4 additional degenerate states beyond the given state
2 additional degenerate states beyond the given state (correct answer)
1 additional degenerate state beyond the given state

Explanation: 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.

Question 12

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 $\hbar$ )?

$\sqrt{6} - \sqrt{2}$ (correct answer)
$\sqrt{5} - \sqrt{3}$
$2 - 1 = 1$
$\sqrt{8} - \sqrt{4} = 2\sqrt{2} - 2$

Explanation: 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

Question 13

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 ( $Z=1$ )?

The wavelength is $Z^2$ times longer than in hydrogen
The wavelength is $\frac{1}{Z^2}$ times shorter than in hydrogen (correct answer)
The wavelength is $Z$ times shorter than in hydrogen
The wavelength is $\frac{1}{Z}$ times longer than in hydrogen

Explanation: 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.

Question 14

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?

The electron can transition to any $n=2$ state with $l=0$ or $l=2$ due to selection rules
The electron can transition to $n=2$ , $l=0$ states only, since $\Delta l = \pm 1$ is required (correct answer)
The electron can transition to $n=1$ state since $\Delta n$ has no restrictions in selection rules
The electron cannot transition to any lower state since $m_l=0$ violates magnetic selection rules

Explanation: 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

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.