IB Physics Quiz: Understand Atomic Structure

20 questions · exam conditions

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Understand Atomic StructureQuestion 1 of 20

A doubly ionized lithium atom (Li $^{2+}$ ) is a hydrogen-like ion with Z=3. According to the Bohr model, what is the energy required to fully ionize this atom from its ground state (n=1)?

13.6 eV

40.8 eV

122.4 eV

27.2 eV

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IB Physics Quiz

IB Physics Quiz: Understand Atomic Structure

Practice Understand Atomic Structure in IB 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 Understand Atomic Structure, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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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.

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Question 1

A doubly ionized lithium atom (Li $^{2+}$ ) is a hydrogen-like ion with Z=3. According to the Bohr model, what is the energy required to fully ionize this atom from its ground state (n=1)?

13.6 eV
40.8 eV
122.4 eV (correct answer)
27.2 eV

Explanation: The energy of an electron in a hydrogen-like ion is given by

E_n = -13.6 \text{ eV} \cdot Z^2 / n^2

. For Li

^{2+}

, Z=3. The ground state is n=1. So,

E_1 = -13.6 \cdot 3^2 / 1^2 = -13.6 \cdot 9 = -122.4 \text{ eV}

. The ionization energy is the energy required to remove the electron completely (to E=0), which is the absolute value of the ground state energy,

|-122.4| = 122.4 \text{ eV}

Question 2

An electron in a hydrogen atom is excited to the n=4 energy level. How many distinct spectral lines can possibly be observed as the atom de-excites through various pathways to the ground state (n=1)?

3
4
6 (correct answer)
10

Explanation: The electron can cascade down through various intermediate levels. The possible transitions are: from n=4 to n=3, 2, or 1 (3 lines); from n=3 to n=2 or 1 (2 lines); and from n=2 to n=1 (1 line). The total number of distinct lines is the sum of these possibilities: 3 + 2 + 1 = 6. These transitions correspond to all possible energy differences between the n=1, 2, 3, and 4 levels.

Question 3

An atomic energy level diagram shows three possible downward transitions. Transition X represents a drop of 5.0 eV. Transition Y represents a drop of 3.0 eV. Transition Z represents a drop of 2.0 eV. Which transition results in the emission of a photon with the longest wavelength?

Transition X
Transition Y
Transition Z (correct answer)
All three have the same wavelength.

Explanation: The energy of a photon is related to its wavelength by the equation

E = hc/\lambda

. This shows that energy and wavelength are inversely proportional. Therefore, the transition with the smallest energy drop will produce the photon with the longest wavelength. Transition Z has the smallest energy drop (2.0 eV), so it corresponds to the longest wavelength.

Question 4

What is a primary distinction of the Bohr model of the atom compared to the earlier Rutherford model?

The Bohr model includes a massive, central nucleus.
The Bohr model postulates that electrons can only exist in specific, non-radiating orbits. (correct answer)
The Bohr model positions electrons orbiting the nucleus due to electrostatic attraction.
The Bohr model correctly predicts the size of the atom to be mostly empty space.

Explanation: The Rutherford model successfully proposed a small, dense, positive nucleus with orbiting electrons, but it could not explain why the electrons wouldn't radiate energy and spiral into the nucleus as predicted by classical physics. The key, non-classical innovation of the Bohr model was the postulate that electrons exist in stable, quantized orbits where they do not emit radiation, only doing so when they jump between these specific orbits.

Question 5

In the Bohr model for the hydrogen atom, the radius $r$ of an electron's orbit is proportional to $n^2$ , where $n$ is the principal quantum number. How does the speed $v$ of the electron depend on $n$ ?

$v \propto n^2$
$v \propto n$
$v \propto 1/n$ (correct answer)
$v \propto 1/n^2$

Explanation: The quantization of angular momentum in the Bohr model states that

mvr = n\hbar

. We can rearrange this for speed:

v = n\hbar / (mr)

. Since we are given that

r \propto n^2

, we can substitute this into the speed equation:

v \propto n / r \propto n / n^2 \propto 1/n

. The electron moves slower in higher energy (larger radius) orbits.

Question 6

In a Rutherford scattering experiment, an alpha particle (+2e) is directed at a gold nucleus (+79e). If the experiment were repeated using a silver nucleus (+47e) as the target, what would be the effect on the electrostatic repulsive force at a given separation distance, compared to the gold target?

It would be greater.
It would be less. (correct answer)
It would be the same.
It would change from repulsive to attractive.

Explanation: The electrostatic force is given by Coulomb's Law,

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

. Here,

q_1

is the charge of the alpha particle (+2e) and

q_2

is the charge of the target nucleus. Since the silver nucleus (+47e) has a smaller charge than the gold nucleus (+79e), the product

q_1 q_2

will be smaller. Therefore, the repulsive force at the same separation distance

r

will be less.

Question 7

In the Geiger-Marsden-Rutherford experiment, most alpha particles passed through the gold foil with minimal deflection, while a very small fraction were deflected through large angles. What conclusion is most directly supported by the observation that most particles were undeflected?

The atom is mostly empty space. (correct answer)
The nucleus of the atom is positively charged.
The nucleus contains nearly all of the atom's mass.
Electrons are fundamental components of the atom.

Explanation: The fact that the vast majority of alpha particles passed straight through the foil implies that they did not encounter anything substantial. This leads to the conclusion that the atom consists mainly of empty space. The deflection of a few particles supports conclusions about the nucleus being small, massive, and positively charged, but the non-deflection of most particles points to the emptiness of the atom.

Question 8

The nucleus of Lead-208 ( ${}^{208}_{82}Pb$ ) has a radius $R_{Pb}$ . The nucleus of Oxygen-16 ( ${}^{16}_{8}O$ ) has a radius $R_O$ . Assuming the empirical formula for nuclear radius $R = R_0 A^{1/3}$ , what is the approximate ratio $R_{Pb} / R_O$ ?

13.0
10.3
3.6
2.3 (correct answer)

Explanation: The nuclear radius is proportional to the cube root of the mass number A. The ratio of the radii is

R_{Pb} / R_O = (R_0 (208)^{1/3}) / (R_0 (16)^{1/3}) = (208/16)^{1/3} = (13)^{1/3}

. The cube root of 13 is between 2 (since

2^3=8

) and 3 (since

3^3=27

). Calculating gives

(13)^{1/3} \approx 2.35

Question 9

An atom has three energy levels $E_3 > E_2 > E_1$ . The emission spectrum of this atom shows lines corresponding to frequencies $f_{31}$ , $f_{32}$ , and $f_{21}$ , for transitions from state 3 to 1, 3 to 2, and 2 to 1 respectively. What is the relationship between these frequencies?

$f_{31} = f_{32} - f_{21}$
$f_{32} = f_{31} + f_{21}$
$f_{31} = f_{32} + f_{21}$ (correct answer)
$f_{21}^2 = f_{31}^2 + f_{32}^2$

Explanation: Energy must be conserved. The energy of the photon from the direct transition (3 to 1) must equal the sum of the energies of the photons from the cascade transition (3 to 2, then 2 to 1). So,

E_{31} = E_{32} + E_{21}

. Since photon energy is given by

E=hf

, we can write

h f_{31} = h f_{32} + h f_{21}

. Dividing by Planck's constant

h

gives

f_{31} = f_{32} + f_{21}

Question 10

An atom has discrete energy levels at -6.2 eV, -3.8 eV, and -1.9 eV. An electron transitions from the -1.9 eV level to the -6.2 eV level. What is the approximate frequency of the photon emitted? (Planck's constant $h \approx 6.63 \times 10^{-34}$ J s; elementary charge $e \approx 1.60 \times 10^{-19}$ C)

$5.7 \times 10^{14}$ Hz
$1.0 \times 10^{15}$ Hz (correct answer)
$6.5 \times 10^{33}$ Hz
$1.2 \times 10^{15}$ Hz

Explanation: The energy of the emitted photon is the difference between the initial and final energy levels:

\Delta E = E_{initial} - E_{final} = (-1.9 \text{ eV}) - (-6.2 \text{ eV}) = 4.3 \text{ eV}

. Convert this energy to Joules:

\Delta E = 4.3 \text{ eV} \times (1.60 \times 10^{-19} \text{ J/eV}) = 6.88 \times 10^{-19} \text{ J}

. Use the formula

E = hf

to find the frequency:

f = E/h = (6.88 \times 10^{-19} \text{ J}) / (6.63 \times 10^{-34} \text{ J s}) \approx 1.04 \times 10^{15} \text{ Hz}

Question 11

According to the Bohr model of the hydrogen atom, the angular momentum of the electron is quantized. What is the change in the electron's angular momentum when it transitions from the n=3 state to the n=1 state? (reduced Planck constant $\hbar = h/2\pi \approx 1.05 \times 10^{-34}$ J s)

$2.10 \times 10^{-34}$ J s (correct answer)
$3.15 \times 10^{-34}$ J s
$4.20 \times 10^{-34}$ J s
$1.05 \times 10^{-34}$ J s

Explanation: The angular momentum L is quantized as

L = n\hbar

, where

n

is the principal quantum number and

\hbar = h/2\pi

. In the n=3 state,

L_3 = 3\hbar

. In the n=1 state,

L_1 = 1\hbar

. The change in angular momentum is

\Delta L = L_3 - L_1 = 3\hbar - 1\hbar = 2\hbar

. Using the given value,

\Delta L = 2 \times (1.05 \times 10^{-34} \text{ J s}) = 2.10 \times 10^{-34} \text{ J s}

. Note that this angular momentum is lost from the atom, often carried away by the emitted photon.

Question 12

The Rutherford model was a major step in understanding atomic structure but was inconsistent with classical physics. The observation of discrete atomic spectra was a key problem that the subsequent Bohr model addressed. Why was the Rutherford model unable to explain discrete spectra?

It did not include the existence of neutrons within the nucleus.
It could not account for the atom being mostly empty space.
It proposed that electrons could orbit at any radius, implying a continuous emission spectrum. (correct answer)
It incorrectly assumed the nucleus was positively charged, leading to repulsion.

Explanation: According to classical electromagnetic theory, an accelerating charge (like an electron in orbit) should continuously radiate energy. In the Rutherford model, electrons could orbit at any distance from the nucleus. This would mean they would radiate a continuous spectrum of light as they spiraled into the nucleus. This contradicts the observed discrete line spectra, a problem solved by Bohr's postulate of quantized, stable orbits.

Question 13

The nucleus of a neutral uranium atom is represented by ${}^{235}_{92}U$ . This atom is then ionized to form a U $^{3+}$ ion. How many protons, neutrons, and electrons are in this ion?

92 protons, 143 neutrons, 89 electrons (correct answer)
92 protons, 143 neutrons, 95 electrons
89 protons, 146 neutrons, 89 electrons
92 protons, 235 neutrons, 92 electrons

Explanation: From the notation

{}^{235}_{92}U

: The number of protons (atomic number Z) is 92. The number of neutrons is the mass number A minus the atomic number Z, so N = 235 - 92 = 143. A neutral atom would have 92 electrons to balance the 92 protons. A U

^{3+}

ion has lost 3 electrons, so it has 92 - 3 = 89 electrons.

Question 14

What is the approximate ratio of the nuclear volume of bismuth-209 to the nuclear volume of beryllium-9?

2.9
5.2
23 (correct answer)
120

Explanation: Nuclear volume

V

is proportional to the mass number A, since

V = \frac{4}{3}\pi R^3

and

R = R_0 A^{1/3}

, which means

V = (\frac{4}{3}\pi R_0^3)A

. Therefore, the ratio of the volumes is simply the ratio of their mass numbers. Ratio =

V_{Bi} / V_{Be} = A_{Bi} / A_{Be} = 209 / 9 \approx 23.2

Question 15

In the Rutherford scattering experiment, alpha particles with kinetic energy 5.0 MeV are directed at a gold nucleus (Z = 79). At what distance of closest approach does the electrostatic potential energy equal twice the initial kinetic energy of the alpha particle?

$2.3 \times 10^{-14}$ m
$4.6 \times 10^{-14}$ m
$1.15 \times 10^{-14}$ m (correct answer)
$9.2 \times 10^{-14}$ m

Explanation: The electrostatic potential energy is

U = \frac{kq_1q_2}{r}

. For twice the kinetic energy:

\frac{ke \cdot 2e \cdot Ze}{r} = 2 \times 5.0 \text{ MeV}

. Solving:

r = \frac{2kZe^2}{2 \times 5.0 \text{ MeV}} = \frac{2 \times 8.99 \times 10^9 \times 79 \times (1.6 \times 10^{-19})^2}{2 \times 5.0 \times 1.6 \times 10^{-13}} = 1.15 \times 10^{-14}

m. Choice A doubles this value, B quadruples it, and D uses an incorrect factor of 8.

Question 16

In the photoelectric effect experiment on a metal surface, photons of energy 4.5 eV produce photoelectrons with maximum kinetic energy 1.8 eV. If the intensity of the incident light is doubled while keeping the photon energy constant, what happens to the number of photoelectrons and their maximum kinetic energy?

Number doubles, maximum kinetic energy increases to 3.6 eV
Number doubles, maximum kinetic energy remains 1.8 eV (correct answer)
Number remains constant, maximum kinetic energy doubles to 3.6 eV
Number quadruples, maximum kinetic energy remains 1.8 eV

Explanation: In the photoelectric effect,

KE_{max} = h\nu - \phi

, where

\phi

is the work function. Since photon energy (frequency) remains constant, the maximum kinetic energy is unchanged. Doubling intensity means doubling the number of photons per unit time, so the number of photoelectrons doubles. Choice A incorrectly suggests kinetic energy depends on intensity. Choice C incorrectly suggests the number is unchanged. Choice D incorrectly quadruples the number.

Question 17

In X-ray production, electrons are accelerated through a potential difference of 25 kV and strike a tungsten target. What is the minimum wavelength of the continuous X-ray spectrum produced, and what physical process determines this limit?

$4.96 \times 10^{-11}$ m; determined by complete electron kinetic energy conversion (correct answer)
$2.48 \times 10^{-11}$ m; determined by the work function of tungsten
$4.96 \times 10^{-11}$ m; determined by the K-shell binding energy
$9.92 \times 10^{-11}$ m; determined by Bragg diffraction conditions

Explanation: The minimum wavelength (maximum energy) occurs when all electron kinetic energy converts to photon energy:

E = eV = h\nu_{max} = \frac{hc}{\lambda_{min}}

. Therefore:

\lambda_{min} = \frac{hc}{eV} = \frac{1240 \text{ eV·nm}}{25,000 \text{ eV}} = 0.0496 \text{ nm} = 4.96 \times 10^{-11}

m. This represents complete conversion of kinetic energy to electromagnetic energy (bremsstrahlung limit). Choice B uses half the voltage. Choice C correctly calculates wavelength but incorrectly attributes it to K-shell binding. Choice D doubles the wavelength and misidentifies the physical process.

Question 18

In the Franck-Hertz experiment with mercury vapor, electrons are accelerated through increasing voltages. The first significant drop in current occurs at 4.9 V, corresponding to the first excited state of mercury. If an electron with exactly this energy collides inelastically with a mercury atom in the ground state, what happens to the electron's kinetic energy and the mercury atom?

Electron stops completely; mercury atom emits 4.9 eV photon immediately
Electron retains small kinetic energy; mercury atom becomes ionized
Electron stops completely; mercury atom enters excited state and may emit photon (correct answer)
Electron bounces back elastically; mercury atom remains in ground state

Explanation: In an inelastic collision where the electron has exactly the excitation energy (4.9 eV), all the electron's kinetic energy transfers to the mercury atom's internal energy, exciting it to the first excited state. The electron comes to rest (or nearly so). The excited mercury atom may subsequently emit a photon to return to the ground state. Choice A incorrectly suggests immediate photon emission. Choice B confuses excitation with ionization (which requires more energy). Choice D describes elastic scattering, which doesn't occur at the excitation threshold.

Question 19

A muonic hydrogen atom consists of a proton and a muon (mass = 207 times electron mass) in orbit. Compared to regular hydrogen, what is the approximate ratio of the Bohr radius of muonic hydrogen to regular hydrogen?

207:1, because the Bohr radius is proportional to particle mass
$\sqrt{207}$ :1, because the radius scales with the square root of mass
1:206, accounting for the finite proton mass in reduced mass calculation
1:207, because the Bohr radius is inversely proportional to reduced mass (correct answer)

Explanation: When you encounter atomic physics problems involving different particles, focus on how the Bohr model depends on the reduced mass of the system. The Bohr radius formula is

r = \frac{\hbar^2}{k e^2 \mu}

, where

\mu

is the reduced mass. For hydrogen-like atoms, the reduced mass is

\mu = \frac{m_1 m_2}{m_1 + m_2}

, where

m_1

is the nuclear mass and

m_2

is the orbiting particle mass. In regular hydrogen, since the electron mass is much smaller than the proton mass,

\mu \approx m_e

. In muonic hydrogen, since the muon is 207 times heavier than an electron but still much lighter than a proton,

\mu \approx 207 m_e

. Since the Bohr radius is inversely proportional to reduced mass, muonic hydrogen has a radius that is

\frac{1}{207}

times that of regular hydrogen, giving us the ratio 1:207. Answer A incorrectly states the radius is proportional to mass when it's actually inversely proportional. Answer B uses the wrong mathematical relationship—there's no square root dependence in the Bohr radius formula. Answer C attempts to account for finite proton mass effects, but this correction is negligible since both the electron and muon are much lighter than the proton, making the "206" factor irrelevant. Remember: In atomic physics problems, always check whether quantities are directly or inversely proportional to mass. The Bohr radius decreases as the orbiting particle becomes more massive, making the atom more compact.

Question 20

A hydrogen atom in the n=4 state can transition to lower energy states. If we observe only transitions that result in photons with wavelengths between 400 nm and 700 nm (visible light), which energy level(s) can the electron transition to?

Only n=2, corresponding to the Balmer series (correct answer)
Both n=2 and n=3, with different probabilities
Only n=1, corresponding to the Lyman series
Only n=3, as n=2 produces ultraviolet radiation

Explanation: For hydrogen,

\lambda = \frac{hc}{13.6 \text{ eV} \times (\frac{1}{n_f^2} - \frac{1}{n_i^2})}

. From n=4 to n=2:

\lambda = \frac{1240 \text{ eV·nm}}{13.6 \times (\frac{1}{4} - \frac{1}{16})} = 486

nm (visible). From n=4 to n=3:

\lambda = 1875

nm (infrared). From n=4 to n=1:

\lambda = 97

nm (UV). Only the n=4→n=2 transition falls in the visible range. Choice B incorrectly includes n=3. Choice C gives UV radiation. Choice D incorrectly identifies the n=4→n=3 transition wavelength.