College Physics Quiz: The Bohr Model Of Atomic Structure

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The Bohr Model Of Atomic StructureQuestion 1 of 20

According to the Bohr model, the angular momentum of an electron in the nth orbit is quantized as $L = n\hbar$ . If an electron in hydrogen has an orbital radius of $2.12 \times 10^{-10}$ m, what is the electron's orbital speed? (Use $a_0 = 5.29 \times 10^{-11}$ m for the Bohr radius)

5.48 \times 10^5

m/s

1.09 \times 10^6

m/s

2.19 \times 10^6

m/s

4.38 \times 10^6

m/s

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College Physics Quiz

College Physics Quiz: The Bohr Model Of Atomic Structure

Practice The Bohr Model Of Atomic Structure in College Physics 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 The Bohr Model Of Atomic Structure, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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

$5.48 \times 10^5$ m/s (correct answer)
$1.09 \times 10^6$ m/s
$2.19 \times 10^6$ m/s
$4.38 \times 10^6$ m/s

Explanation: First find n:

r = na_0

, so

n = r/a_0 = 2.12 \times 10^{-10}/(5.29 \times 10^{-11}) = 4

. From

L = mvr = n\hbar

, we get

v = n\hbar/(mr)

. Using

\hbar = 1.055 \times 10^{-34}

J⋅s and

m_e = 9.11 \times 10^{-31}

kg:

v = 4(1.055 \times 10^{-34})/[(9.11 \times 10^{-31})(2.12 \times 10^{-10})] = 5.48 \times 10^5

m/s. Choice B uses n=2. Choice C uses n=1. Choice D incorrectly doubles the calculation.

Question 2

According to the Bohr model, the angular momentum of an electron in the $n = 3$ orbit of hydrogen is quantized. What is the magnitude of this angular momentum in terms of $\hbar$ ?

$3\hbar$ (correct answer)
$9\hbar$
$\frac{\hbar}{3}$
$6\hbar$
$\sqrt{3}\hbar$

Explanation: When you encounter questions about the Bohr model of the atom, you're dealing with the foundational concept of quantized angular momentum. The Bohr model introduced the revolutionary idea that electrons can only occupy specific orbits with discrete angular momentum values. The key formula for angular momentum quantization in the Bohr model is

L = n\hbar

, where

n

is the principal quantum number and

\hbar

is the reduced Planck constant. For an electron in the

n = 3

orbit, you simply substitute:

L = 3\hbar

. This direct relationship shows that angular momentum increases linearly with the quantum number. Looking at the incorrect options: Choice B (

9\hbar

) represents a common error where students mistakenly square the quantum number, perhaps confusing this with energy calculations where

E_n \propto 1/n^2

. Choice C (

\frac{\hbar}{3}

) suggests the misconception that angular momentum decreases with higher orbits, which contradicts the physics—electrons in higher orbits have more angular momentum. Choice D (

6\hbar

) might arise from incorrectly multiplying

n

by 2, possibly confusing orbital angular momentum formulas from quantum mechanics where

L = \sqrt{\ell(\ell+1)}\hbar

. Remember this simple pattern: in the Bohr model, angular momentum is always

n\hbar

—just multiply the orbit number by

\hbar

. Don't overthink it with squares or complex formulas. This straightforward relationship was Bohr's key insight that explained hydrogen's discrete energy levels and remains a cornerstone concept in atomic physics.

Question 3

In the Bohr model, the speed of an electron in the $n$ th orbit is inversely proportional to $n$ . If the speed of an electron in the first Bohr orbit is $v_1$ , what is the speed in the fourth orbit?

$\frac{v_1}{4}$ (correct answer)
$\frac{v_1}{16}$
$\frac{v_1}{2}$
$4v_1$
$\frac{v_1}{8}$

Explanation: When you encounter Bohr model problems, focus on the key relationships between quantum number

n

and the physical properties of electron orbits. The problem states that electron speed is inversely proportional to

n

, which means

v \propto \frac{1}{n}

v = \frac{k}{n}

where

k

is a constant. Since the speed in the first orbit is

v_1

, we can write

v_1 = \frac{k}{1} = k

. For the fourth orbit, the speed becomes

v_4 = \frac{k}{4} = \frac{v_1}{4}

. Looking at the answer choices: (A) $\frac{v_1}{4}$ is correct because it properly applies the inverse relationship—when

n

increases by a factor of 4, the speed decreases by the same factor. (B) $\frac{v_1}{16}$ incorrectly applies an inverse square relationship (

\propto \frac{1}{n^2}

), which describes other Bohr model quantities like binding energy, but not speed. (C) $\frac{v_1}{2}$ would be correct if we were looking at the second orbit, not the fourth. (D) $4v_1$ represents a direct proportional relationship, which is the opposite of what the problem describes. Remember that "inversely proportional" means when one quantity doubles, the other halves—it's a simple

\frac{1}{n}

relationship unless specified otherwise. Don't confuse this with inverse square relationships that appear elsewhere in atomic physics. Always identify whether you're dealing with

\frac{1}{n}

\frac{1}{n^2}

dependencies.

Question 4

In the Bohr model, an electron makes a transition from $n = 6$ to $n = 3$ . The wavelength of the emitted photon is $\lambda$ . What would be the wavelength if the electron instead transitioned from $n = 3$ to $n = 6$ ?

No photon would be emitted; energy would need to be absorbed (correct answer)
The wavelength would be $\lambda$ , but with opposite polarization
The wavelength would be $2\lambda$
The wavelength would be $\lambda/2$
The wavelength would be $4\lambda$

Explanation: When you encounter questions about electron transitions in the Bohr model, remember that energy changes determine whether a photon is emitted or absorbed. The direction of the transition is crucial. In the original transition from

n = 6

n = 3

, the electron moves from a higher energy level to a lower one. This releases energy as an emitted photon with wavelength

\lambda

. The energy difference is

\Delta E = E_6 - E_3

, which is positive, so energy flows out of the atom. For a transition from

n = 3

n = 6

, the electron would move from a lower energy level to a higher one. This requires energy input—the atom must absorb a photon, not emit one. The energy difference is now

\Delta E = E_3 - E_6

, which is negative, meaning energy must flow into the atom. Therefore, answer A is correct: no photon would be emitted; energy would need to be absorbed. Answer B is wrong because polarization doesn't change the fundamental energy requirement—you still can't emit a photon when energy must be absorbed. Answer C (

2\lambda

) and answer D (

\lambda/2

) both incorrectly assume a photon would be emitted, just with different wavelengths. These ignore the basic physics that upward transitions require energy absorption. Remember this key principle: downward transitions (

n_{high} \to n_{low}

) emit photons, while upward transitions (

n_{low} \to n_{high}

) require photon absorption. The direction of the energy change determines whether emission or absorption occurs, regardless of the specific energy levels involved.

Question 5

According to the Bohr model, which of the following correctly explains why electrons in hydrogen atoms do not spiral into the nucleus due to electromagnetic radiation?

Electrons can only exist in specific quantized orbits where they do not radiate energy (correct answer)
The nuclear force exactly balances the electromagnetic force at certain distances
Electrons move too fast for electromagnetic radiation to affect their motion significantly
The magnetic field produced by orbital motion cancels the electric field acceleration
Quantum tunneling allows electrons to maintain their orbital energy indefinitely

Explanation: The Bohr model was developed to solve a fundamental problem in early atomic physics: classical electromagnetic theory predicted that electrons orbiting the nucleus should continuously radiate energy and spiral inward, causing atoms to collapse instantly. This clearly contradicted the stability of matter we observe. Bohr's revolutionary solution introduced quantum mechanics into atomic structure. He proposed that electrons can only occupy specific, discrete energy levels or "quantized orbits." In these allowed orbits, electrons somehow do not radiate electromagnetic energy despite being accelerated charges. This quantum restriction explained atomic stability and successfully predicted hydrogen's emission spectrum. Answer A correctly captures this key insight. Answer B incorrectly invokes the nuclear force (strong force), which only operates at nuclear distances much smaller than atomic radii and doesn't balance electromagnetic forces in electron orbits. Answer C misunderstands the physics entirely—electron speed doesn't make them immune to electromagnetic radiation effects. Classical theory actually predicts faster-moving charges would radiate more energy. Answer D confuses electromagnetic concepts. While moving charges do create magnetic fields, this doesn't cancel the electric field acceleration that causes radiation in classical theory. Remember that Bohr's model was a crucial stepping stone between classical and modern quantum mechanics. When you encounter questions about early atomic models, focus on how quantum restrictions (quantized energy levels, angular momentum, etc.) solved classical physics paradoxes. The key concept is always "quantization prevents what classical physics predicts should happen."

Question 6

In the Bohr model, the circumference of the $n$ th electron orbit must equal an integer number of de Broglie wavelengths. For the $n = 3$ orbit with circumference $C_3$ , how many wavelengths fit around the orbit?

$3$ wavelengths (correct answer)
$9$ wavelengths
$6$ wavelengths
$1$ wavelength
$\frac{1}{3}$ wavelength

Explanation: When you encounter questions about the Bohr model and electron orbits, you're dealing with the quantum mechanical concept that electrons can only exist in specific, quantized energy levels. A key insight of the Bohr model is that electron orbits must satisfy a standing wave condition. In the Bohr model, the fundamental quantization condition requires that the circumference of each electron orbit equals exactly

n

de Broglie wavelengths, where

n

is the principal quantum number. This can be written as:

C_n = n\lambda

, where

C_n

is the circumference of the

n

th orbit and

\lambda

is the de Broglie wavelength. For the

n = 3

orbit, this relationship directly tells us that

C_3 = 3\lambda

. Therefore, exactly 3 wavelengths fit around the third orbit, making choice (A) 3 wavelengths correct. Let's examine why the other options are incorrect: (B) 9 wavelengths likely comes from mistakenly squaring the quantum number (

n^2 = 3^2 = 9

), but this relationship doesn't apply to the standing wave condition. (C) 6 wavelengths might result from doubling

n

(

2n = 6

), which has no physical basis in the Bohr model. (D) 1 wavelength would only be correct for the ground state (

n = 1

). Study tip: Remember that in the Bohr model, the number of de Broglie wavelengths that fit around an orbit always equals the principal quantum number

n

. This is the fundamental quantization condition that makes stable electron orbits possible.

Question 7

The Bohr model successfully explained the discrete spectral lines of hydrogen. Which of the following observations was NOT explained by the original Bohr model?

The fine structure splitting of spectral lines (correct answer)
The wavelengths of the Balmer series
The ionization energy of hydrogen
The quantized angular momentum of electrons
The stability of the ground state of hydrogen

Explanation: When you encounter questions about the Bohr model's limitations, remember that this early quantum model successfully explained hydrogen's basic features but failed to account for more subtle effects discovered later. The Bohr model correctly predicted several key observations. It explained the wavelengths of the Balmer series (choice B) by proposing that electrons orbit in discrete energy levels, with transitions between levels producing specific photon energies. The model also correctly calculated hydrogen's ionization energy (choice C) as 13.6 eV, matching experimental results. Additionally, Bohr's fundamental assumption was that angular momentum is quantized in units of

\hbar

(choice D), which became a cornerstone of quantum mechanics. However, the fine structure splitting of spectral lines (choice A) was beyond the original Bohr model's capabilities. Fine structure refers to the splitting of what appear to be single spectral lines into closely-spaced doublets or multiplets when viewed with high-resolution instruments. This phenomenon arises from relativistic effects and spin-orbit coupling – concepts not incorporated in Bohr's model. The fine structure constant

\alpha \approx 1/137

characterizes these small but measurable energy corrections. Choice B is wrong because Bohr's model precisely predicted Balmer series wavelengths using

\frac{1}{\lambda} = R_H(\frac{1}{4} - \frac{1}{n^2})

. Choice C is incorrect since the model accurately calculated ionization energy. Choice D is wrong because quantized angular momentum was Bohr's central postulate. Study tip: Remember that early atomic models like Bohr's explained gross features but missed fine details requiring relativistic quantum mechanics. Look for questions distinguishing between basic quantum effects versus subtle corrections.

Question 8

In the Bohr model, the centripetal force on an electron is provided by the Coulomb attraction to the nucleus. For a hydrogen atom, if the electron's orbital radius doubles, how does the centripetal force change?

The force decreases by a factor of $4$ (correct answer)
The force decreases by a factor of $2$
The force increases by a factor of $2$
The force increases by a factor of $4$
The force remains constant

Explanation: When analyzing forces in the Bohr model, you need to recognize that the centripetal force on the electron comes from the electrostatic Coulomb force between the positively charged nucleus and the negatively charged electron. The Coulomb force follows an inverse square law:

F = k\frac{q_1 q_2}{r^2}

, where

r

is the distance between charges. For a hydrogen atom, this becomes

F = k\frac{e^2}{r^2}

, where

e

is the elementary charge and

r

is the orbital radius. If the orbital radius doubles (

r \rightarrow 2r

), the force becomes:

F_{new} = k\frac{e^2}{(2r)^2} = k\frac{e^2}{4r^2} = \frac{1}{4} \cdot k\frac{e^2}{r^2} = \frac{F_{original}}{4}

Therefore, the force decreases by a factor of 4, making A correct. B is wrong because it suggests the force decreases by only a factor of 2, which would be true for a linear relationship (

F \propto 1/r

) rather than the actual inverse square relationship. C and D are wrong because they claim the force increases, which contradicts the inverse relationship between force and distance. As particles move farther apart, the electrostatic attraction weakens, not strengthens. Study tip: Remember that Coulomb's law follows an inverse square relationship (

F \propto 1/r^2

). Whenever you see distance changes in electrostatic problems, square the distance change to find the force change: double the distance means

\frac{1}{4}

the force, triple means

\frac{1}{9}

, etc.

Question 9

A hydrogen atom absorbs a photon and its electron moves from $n = 2$ to $n = 5$ . Later, the electron cascades back down to $n = 2$ by emitting two photons in sequence. Which of the following represents a possible intermediate state in this cascade?

$n = 4$ or $n = 3$ (correct answer)
$n = 1$ only
$n = 6$ or $n = 7$
Any integer value from $n = 1$ to $n = 5$
$n = 2.5$ (non-integer intermediate state)

Explanation: When you encounter questions about electron transitions in hydrogen atoms, focus on the fundamental rule that electrons can only occupy specific energy levels and must follow selection rules when transitioning between them. In this scenario, the electron starts at

n = 5

after absorption and must return to

n = 2

by emitting exactly two photons. Since each photon emission corresponds to one downward transition, there must be exactly one intermediate state. The electron goes from

n = 5

to some intermediate level

n_i

, then from

n_i

n = 2

. For this to work, the intermediate state must satisfy two conditions: it must be lower than

n = 5

(since electrons cascade downward) and higher than

n = 2

(since there's still one more transition to go). Therefore,

n_i

can be

n = 3

n = 4

, making choice A correct. Choice B (

n = 1

only) is wrong because if the electron went to

n = 1

, it would have already passed through

n = 2

, making a subsequent transition to

n = 2

impossible. Choice C (

n = 6

n = 7

) is incorrect because these states are higher than the starting point of

n = 5

, which would require energy absorption, not emission. Choice D is wrong because it includes

n = 1

(impossible as explained) and

n = 5

(the starting state, not an intermediate state). Remember: in cascade processes, trace the path step by step and ensure each transition follows the physical constraints of the system.

Question 10

According to the Bohr model, which of the following statements correctly describes the relationship between the kinetic energy and potential energy of an electron in a hydrogen atom?

The kinetic energy is exactly half the magnitude of the potential energy (correct answer)
The kinetic energy equals the potential energy in magnitude
The kinetic energy is twice the magnitude of the potential energy
The kinetic energy is one-fourth the magnitude of the potential energy
The kinetic energy is four times the magnitude of the potential energy

Explanation: When you encounter Bohr model questions, you're dealing with the fundamental energy relationships that keep electrons in stable orbits around the nucleus. The key insight is understanding how kinetic and potential energies balance in these quantized orbits. In the Bohr model, an electron orbits the nucleus in a circular path where the electrostatic force provides the centripetal force. This gives us

\frac{ke^2}{r^2} = \frac{mv^2}{r}

, which simplifies to

ke^2 = mv^2r

. The kinetic energy is

KE = \frac{1}{2}mv^2 = \frac{ke^2}{2r}

, while the potential energy is

PE = -\frac{ke^2}{r}

(negative because it's attractive). Comparing these expressions, you can see that

KE = \frac{1}{2}|PE|

—the kinetic energy is exactly half the magnitude of the potential energy. This relationship comes directly from the virial theorem applied to Coulomb forces. Option B suggests equal magnitudes, which would violate the force balance requirements. Option C claims kinetic energy is twice the potential energy's magnitude, which reverses the actual relationship. Option D proposes kinetic energy is one-fourth the magnitude, which doesn't emerge from any valid orbital mechanics calculation for Coulomb forces. The correct answer is A because this 1:2 ratio between kinetic and potential energy magnitudes is a fundamental consequence of stable circular orbits under inverse-square forces. Study tip: Remember the Bohr energy ratio as "half and whole"—kinetic energy is half the magnitude of potential energy. This same relationship appears in other inverse-square force problems throughout physics.

Question 11

In the Bohr model of hydrogen, the total energy of an electron in the $n = 2$ state is $-3.4$ eV. What is the minimum energy required to completely remove this electron from the atom (ionization energy from $n = 2$ )?

$3.4$ eV (correct answer)
$10.2$ eV
$13.6$ eV
$6.8$ eV
$1.7$ eV

Explanation: When you encounter ionization energy problems in the Bohr model, remember that ionization means removing an electron completely from the atom. The key insight is understanding what the negative energy values represent and how energy conservation applies. In the Bohr model, the total energy of an electron is negative because it's bound to the nucleus. The more negative the energy, the more tightly bound the electron is. When the total energy is

-3.4

eV in the

n = 2

state, this means the electron is

3.4

eV below the reference point of zero energy. The zero energy level represents a completely free electron (infinitely far from the nucleus). To ionize the electron from

n = 2

, you must provide exactly enough energy to bring it from

-3.4

eV up to

0

eV. This requires

3.4

eV of energy input. Looking at the wrong answers: B)

10.2

eV is three times

3.4

eV, suggesting confusion about energy level relationships. C)

13.6

eV is the ionization energy from the ground state (

n = 1

), which is a common trap - students sometimes memorize this value without considering which energy level the electron starts from. D)

6.8

eV is twice

3.4

eV, possibly from incorrectly thinking you need to "overcome" the binding energy rather than simply reach zero energy. The correct answer is A)

3.4

eV. Study tip: For ionization problems, always remember that ionization energy equals the absolute value of the initial energy state. The negative sign just indicates binding - ignore it when calculating the energy needed to reach zero.

Question 12

In the Bohr model, which quantum number determines the energy of an electron in a hydrogen atom?

The principal quantum number $n$ only (correct answer)
Both the principal quantum number $n$ and orbital angular momentum quantum number $\ell$
The orbital angular momentum quantum number $\ell$ only
The magnetic quantum number $m_\ell$ only
All four quantum numbers $n$ , $\ell$ , $m_\ell$ , and $m_s$

Explanation: When you encounter questions about the Bohr model, remember that it's a simplified but foundational model of atomic structure that predates modern quantum mechanics. The key insight is understanding which quantum number controls energy in this specific model. In the Bohr model, the energy of an electron in a hydrogen atom depends solely on the principal quantum number

n

. The energy formula is

E_n = -\frac{13.6 \text{ eV}}{n^2}

, where only

n

appears. This means electrons in the same shell (same

n

value) have identical energies, regardless of their orbital shape or orientation. This makes choice A correct. Choice B incorrectly suggests that the orbital angular momentum quantum number

\ell

also affects energy. While

\ell

becomes important in more sophisticated models that account for electron-electron interactions and relativistic effects, the Bohr model doesn't include these complexities. In real atoms, different

\ell

values within the same

n

level do have slightly different energies, but this question specifically asks about the Bohr model. Choice C is wrong because

\ell

alone cannot determine energy—you need

n

for that. Choice D incorrectly identifies the magnetic quantum number

m_\ell

, which describes orbital orientation in space and has no effect on energy in any simple atomic model. Study tip: Remember that the Bohr model is intentionally simplified. When you see "Bohr model" in a question, think "energy depends only on

n

" and ignore the more complex energy relationships you might know from advanced quantum mechanics.

Question 13

In the Bohr model, an electron in the hydrogen atom transitions from the $n = 4$ energy level to the $n = 2$ energy level. If the ionization energy of hydrogen is $13.6$ eV, what is the energy of the emitted photon?

$2.55$ eV (correct answer)
$3.40$ eV
$10.2$ eV
$0.85$ eV
$12.75$ eV

Explanation: When you encounter Bohr model problems involving electron transitions, you're dealing with quantized energy levels where electrons can only exist at specific energies. The key insight is that when an electron drops from a higher to lower energy level, it emits a photon whose energy equals the difference between those levels. In hydrogen, the energy of an electron at level

n

is given by

E_n = -\frac{13.6 \text{ eV}}{n^2}

. For the

n = 4

n = 2

transition, you need to calculate both energy levels and find their difference. The initial energy is

E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85

eV. The final energy is

E_2 = -\frac{13.6}{2^2} = -\frac{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, confirming answer A. Answer B (3.40 eV) represents just the magnitude of the

n = 2

energy level, ignoring the initial state. Answer C (10.2 eV) would be the energy needed to ionize an electron from

n = 2

(since

13.6 - 3.40 = 10.2

), but that's not what's happening here. Answer D (0.85 eV) is the magnitude of the

n = 4

energy level alone. Remember: for emission problems, always calculate the energy difference between levels, not individual level energies. The emitted photon carries away the excess energy as the electron drops to a more stable state.

Question 14

The Bohr model assumes circular orbits for electrons. If an electron in the $n = 2$ orbit has an angular frequency $\omega_2$ , what is the angular frequency of an electron in the $n = 1$ orbit in terms of $\omega_2$ ?

$8\omega_2$ (correct answer)
$4\omega_2$
$2\omega_2$
$\frac{\omega_2}{2}$
$\frac{\omega_2}{4}$

Explanation: When you encounter Bohr model problems, remember that electrons orbit the nucleus like planets around the sun, following specific quantized energy levels. The key insight is understanding how orbital properties scale with the principal quantum number

n

. In the Bohr model, the angular frequency of an electron is related to its orbital energy and radius. The energy levels scale as

E_n \propto \frac{1}{n^2}

, while the orbital radius scales as

r_n \propto n^2

. For circular orbits, the angular frequency depends on both the centripetal force (provided by Coulomb attraction) and the orbital radius. Through the Bohr model equations, the angular frequency scales as

\omega_n \propto \frac{1}{n^3}

. This means

\frac{\omega_1}{\omega_2} = \frac{n_2^3}{n_1^3} = \frac{2^3}{1^3} = 8

. Therefore,

\omega_1 = 8\omega_2

. Choice A (

8\omega_2

) is correct because the

n^3

relationship gives us the factor of 8 when comparing

n=1

n=2

. Choice B (

4\omega_2

) incorrectly uses an

n^2

scaling, which applies to energy or radius but not angular frequency. Choice C (

2\omega_2

) mistakenly uses simple

n^1

proportionality, ignoring the more complex orbital mechanics. Choice D (

\frac{\omega_2}{2}

) gets the direction backwards—it suggests the inner orbit is slower than the outer orbit, which contradicts both classical physics and the Bohr model. Study tip: Remember that in the Bohr model, inner electrons orbit faster, and the scaling relationships follow specific powers of

n

: energy (

n^{-2}

), radius (

n^2

), and angular frequency (

n^{-3}

Question 15

According to the Bohr model, what is the ratio of the kinetic energy of an electron in the $n = 1$ orbit to its kinetic energy in the $n = 3$ orbit?

$9:1$ (correct answer)
$3:1$
$1:3$
$1:9$
$6:1$

Explanation: When you encounter Bohr model questions, focus on how energy scales with the principal quantum number

n

. The key insight is that kinetic energy in the Bohr model follows an inverse square relationship with

n

. In the Bohr model, the kinetic energy of an electron is given by

KE = \frac{13.6 \text{ eV}}{n^2}

. This means that as

n

increases, kinetic energy decreases dramatically. For the

n = 1

orbit:

KE_1 = \frac{13.6}{1^2} = 13.6 \text{ eV}

. For the

n = 3

orbit:

KE_3 = \frac{13.6}{3^2} = \frac{13.6}{9} = 1.51 \text{ eV}

. The ratio is

\frac{KE_1}{KE_3} = \frac{13.6}{1.51} = 9

, giving us

9:1

. Answer A (

9:1

) is correct because kinetic energy scales as

\frac{1}{n^2}

, so

\frac{KE_1}{KE_3} = \frac{3^2}{1^2} = 9

. Answer B (

3:1

) incorrectly assumes kinetic energy scales linearly with

\frac{1}{n}

rather than

\frac{1}{n^2}

. Answer C (

1:3

) reverses the relationship and uses the wrong scaling, suggesting that higher orbits have more kinetic energy. Answer D (

1:9

) uses the correct

n^2

scaling but inverts the ratio, forgetting that electrons in lower orbits move faster and have higher kinetic energy. Remember: In the Bohr model, electrons closer to the nucleus (smaller

n

) orbit faster and have higher kinetic energy. Energy relationships always involve

n^2

, not just

n

Question 16

A hydrogen atom initially in the $n = 4$ state can emit photons by transitioning to lower energy states. How many different photon energies are possible from this initial state?

$6$ different energies (correct answer)
$3$ different energies
$4$ different energies
$10$ different energies
$12$ different energies

Explanation: When you encounter questions about atomic transitions, you need to count all possible pathways an electron can take from its initial state to any lower energy level. A hydrogen atom in the

n = 4

state can transition to any lower principal quantum number:

n = 3

n = 2

, or

n = 1

. Each transition produces a photon with energy equal to the difference between energy levels:

E_{photon} = E_i - E_f

. From

n = 4

, the possible direct transitions are:

$4 \rightarrow 3$ (one transition)
$4 \rightarrow 2$ (one transition)
$4 \rightarrow 1$ (one transition)

But the electron can also make cascading transitions. After dropping to

n = 3

, it can continue to

n = 2

n = 1

. From

n = 2

, it can drop to

n = 1

. This gives us additional unique photon energies:

$3 \rightarrow 2$
$3 \rightarrow 1$
$2 \rightarrow 1$

Counting all possible transitions:

4\rightarrow3

4\rightarrow2

4\rightarrow1

3\rightarrow2

3\rightarrow1

, and

2\rightarrow1

gives us 6 different photon energies, making (A) correct. (B) only counts direct transitions from

n = 4

. (C) might represent confusion between the initial state number and possible transitions. (D) incorrectly applies a formula like

\frac{n(n-1)}{2}

for

n = 4

, which overcounts by including impossible upward transitions. Remember: always count every possible downward transition from the initial state AND from any intermediate states the electron might reach through cascading.

Question 17

In the Bohr model of hydrogen, the electron's orbital period in the $n$ th orbit scales as $T_n \propto n^3$ . If the orbital period in the ground state is $T_1 = 1.5 \times 10^{-16}$ s, what is the orbital period in the $n = 2$ state?

$1.2 \times 10^{-15}$ s (correct answer)
$6.0 \times 10^{-16}$ s
$3.0 \times 10^{-16}$ s
$7.5 \times 10^{-17}$ s
$2.4 \times 10^{-15}$ s

Explanation: When you encounter scaling relationships in quantum mechanics, you're working with proportionality laws that reveal how physical quantities change with quantum numbers. The Bohr model provides a classical framework for understanding hydrogen's energy levels and orbital mechanics. Given that the orbital period scales as

T_n \propto n^3

, you can write this relationship as

T_n = T_1 \times n^3

for any orbit relative to the ground state. For the

n = 2

state, this becomes

T_2 = T_1 \times 2^3 = T_1 \times 8

. Substituting the given ground state period:

T_2 = (1.5 \times 10^{-16} \text{ s}) \times 8 = 1.2 \times 10^{-15} \text{ s}

. This confirms answer choice A is correct. Let's examine why the other options are wrong. Choice B (

6.0 \times 10^{-16}

s) represents

T_1 \times 4 = T_1 \times 2^2

, suggesting someone incorrectly used

n^2

scaling instead of

n^3

. Choice C (

3.0 \times 10^{-16}

s) equals

T_1 \times 2

, indicating linear scaling with

n

—a significant conceptual error. Choice D (

7.5 \times 10^{-17}

s) represents

T_1/2

, which would actually correspond to a shorter period than the ground state, contradicting the physical expectation that higher orbits have longer periods. Remember that

n^3

scaling in the Bohr model reflects the combined effects of increasing orbital radius and decreasing orbital velocity. Always double-check that higher energy levels give longer periods, not shorter ones.

Question 18

A hydrogen atom in the Bohr model emits a photon when transitioning from $n = 5$ to $n = 2$ . In which region of the electromagnetic spectrum does this emission occur?

Visible light (Balmer series) (correct answer)
Ultraviolet (Lyman series)
Infrared (Paschen series)
X-ray region
Radio wave region

Explanation: When you encounter hydrogen emission problems, you need to identify which spectral series corresponds to the transition by looking at the final energy level (lower n value). The key insight is that spectral series are named based on their final state: Lyman series ends at n=1, Balmer series ends at n=2, and Paschen series ends at n=3. Since this transition goes from n=5 to n=2, it belongs to the Balmer series. To find the photon energy, use the Rydberg formula:

\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 n=2 to n=5:

\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{25}\right) = 2.304 \times 10^6 \text{ m}^{-1}

. This gives

\lambda = 434 \text{ nm}

, which falls in the visible spectrum (blue-violet light). Choice A is correct because n=5→n=2 transitions belong to the Balmer series, which produces visible light. Choice B (Lyman series) would require transitions ending at n=1, producing UV radiation. Choice C (Paschen series) involves transitions ending at n=3, producing infrared radiation. Choice D (X-ray) would require much higher energy transitions, typically involving inner electron shells in heavier atoms. Study tip: Remember the spectral series by their final states: Lyman (n=1, UV), Balmer (n=2, visible), Paschen (n=3, IR). The final quantum number tells you immediately which series and approximate wavelength range to expect.

Question 19

In the Bohr model of hydrogen, the radius of the electron orbit is given by $r_n = n^2 a_0$ , where $a_0$ is the Bohr radius. If an electron moves from the $n = 1$ orbit to the $n = 3$ orbit, by what factor does the orbital radius change?

The radius increases by a factor of $9$ (correct answer)
The radius increases by a factor of $3$
The radius decreases by a factor of $3$
The radius increases by a factor of $6$
The radius decreases by a factor of $9$

Explanation: When you encounter Bohr model problems, you're dealing with quantized electron orbits where specific formulas relate the quantum number

n

to physical properties like radius, energy, and angular momentum. The orbital radius formula

r_n = n^2 a_0

tells you that radius scales with the square of the principal quantum number. To find how the radius changes when an electron transitions from

n = 1

n = 3

, calculate the ratio of final to initial radius:

\frac{r_3}{r_1} = \frac{3^2 a_0}{1^2 a_0} = \frac{9a_0}{a_0} = 9

The radius increases by a factor of 9, confirming answer A. Let's examine why the other options are wrong. Answer B (factor of 3) represents the common mistake of using

n

directly instead of

n^2

- this would be correct if radius were proportional to

n

rather than

n^2

. Answer C (decreases by factor of 3) not only uses the wrong proportionality but also gets the direction backward - higher quantum numbers always mean larger orbits in the Bohr model. Answer D (factor of 6) has no clear mathematical basis in the given formula and likely represents a calculation error. Study tip: In Bohr model problems, always pay careful attention to the exponents in the given formulas. Energy goes as

1/n^2

, radius as

n^2

, and velocity as

1/n

. The

n^2

dependencies are especially important since they amplify changes between quantum levels.

Question 20

The Bohr model predicts that the frequency of orbital motion of an electron in the $n$ th orbit is proportional to $n^{-3}$ . If the orbital frequency in the $n = 1$ state is $f_1$ , what is the orbital frequency in the $n = 2$ state?

$\frac{f_1}{8}$ (correct answer)
$\frac{f_1}{4}$
$\frac{f_1}{2}$
$2f_1$
$\frac{f_1}{16}$

Explanation: When you encounter Bohr model problems involving orbital frequencies, you're dealing with the fundamental relationship between quantum energy levels and electron motion. The key insight is understanding how scaling laws work in quantum mechanics. The problem states that orbital frequency is proportional to

n^{-3}

, which means

f \propto \frac{1}{n^3}

. This allows you to write the relationship as

f_n = \frac{C}{n^3}

, where

C

is a constant. For the first orbit,

f_1 = \frac{C}{1^3} = C

, so

C = f_1

. For the second orbit, you can substitute:

f_2 = \frac{f_1}{2^3} = \frac{f_1}{8}

. This confirms that answer choice A is correct. Let's examine why the other answers are wrong. Choice B (

\frac{f_1}{4}

) would correspond to a

n^{-2}

dependence, which might arise if you confused this with area scaling. Choice C (

\frac{f_1}{2}

) represents simple

n^{-1}

scaling, suggesting you might have overlooked the cubic relationship entirely. Choice D (

2f_1

) actually increases with

n

, which would violate the basic physics that higher energy levels correspond to slower orbital motion. Remember this pattern: when you see inverse power law relationships like

n^{-3}

, always substitute the specific values directly into the proportionality. Don't try to reason about "doubling" or simple ratios—the power law dominates the behavior, often leading to more dramatic changes than intuition suggests.