IB Physics Quiz: Apply Atomic Structure
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Apply Atomic StructureQuestion 1 of 20

An ion has 35 protons, 44 neutrons, and 36 electrons. What is the correct nuclear notation for this ion?

3679X{}^{79}_{36}\text{X}
3579X{}^{79}_{35}\text{X}
3580X{}^{80}_{35}\text{X}
3544X{}^{44}_{35}\text{X}
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IB Physics Quiz

IB Physics Quiz: Apply Atomic Structure

Practice Apply 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.

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This quiz focuses on Apply 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

An ion has 35 protons, 44 neutrons, and 36 electrons. What is the correct nuclear notation for this ion?

  1. 3679X{}^{79}_{36}\text{X}
  2. 3579X{}^{79}_{35}\text{X} (correct answer)
  3. 3580X{}^{80}_{35}\text{X}
  4. 3544X{}^{44}_{35}\text{X}
Explanation: Nuclear notation ZAX{}^{A}_{Z}\text{X} is determined by the composition of the nucleus. The atomic number, Z, is the number of protons, which is 35. The mass number, A, is the total number of protons and neutrons, which is 35 + 44 = 79. The number of electrons determines the ion's charge but does not appear in the standard nuclear notation. Thus, the notation is 3579X{}^{79}_{35}\text{X}.

Question 2

The Rutherford model of the atom, despite its success in explaining alpha particle scattering, was ultimately replaced. What was a key failure of the Rutherford model?

  1. It could not account for the observation that atoms are electrically neutral.
  2. It failed to explain why most alpha particles passed through the gold foil undeflected.
  3. It could not explain the existence of discrete emission spectra from excited atoms. (correct answer)
  4. It was inconsistent with the later discovery of the neutron and isotopes.
Explanation: According to classical physics, an electron orbiting a nucleus is an accelerating charge and should continuously radiate energy. This would cause it to spiral into the nucleus, and the radiation emitted would form a continuous spectrum. This contradicts the experimental observation of discrete line spectra from atoms. The Bohr model was developed to solve this specific problem by postulating stable, non-radiating orbits.

Question 3

In a Rutherford scattering experiment, an alpha particle with initial kinetic energy EkE_k is aimed directly at a gold nucleus. The distance of closest approach is dd. What would be the distance of closest approach for an alpha particle with initial kinetic energy 2Ek2E_k aimed at the same nucleus?

  1. 2d2d
  2. 2d\sqrt{2} d
  3. d/2d/2 (correct answer)
  4. d/2d/\sqrt{2}
Explanation: At the distance of closest approach, the initial kinetic energy of the alpha particle is completely converted into electric potential energy. The electric potential energy is given by Ep=kq1q2/rE_p = k q_1 q_2 / r, where rr is the distance. Therefore, Ek=kq1q2/dE_k = k q_1 q_2 / d. This shows that the distance of closest approach, dd, is inversely proportional to the initial kinetic energy, EkE_k. If the kinetic energy is doubled (2Ek2E_k), the distance of closest approach will be halved (d/2d/2).

Question 4

How does the density of the nucleus of Uranium-238 compare to the density of the nucleus of Helium-4? (HL Only)

  1. The density of Uranium-238 is much greater.
  2. The density of Helium-4 is much greater.
  3. The densities are approximately the same. (correct answer)
  4. The ratio of densities is proportional to the ratio of their mass numbers, 238/4.
Explanation: Nuclear density (ρ\rho) is mass (mm) divided by volume (VV). The mass of a nucleus is approximately proportional to its mass number AA (m∝Am \propto A). The volume of a nucleus is V=43πR3V = \frac{4}{3}\pi R^3, and the nuclear radius is given by R=R0A1/3R = R_0 A^{1/3}. Substituting for R, we get V=43π(R0A1/3)3=43πR03AV = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A, so V∝AV \propto A. Since both mass and volume are proportional to the mass number AA, their ratio, the density, is approximately constant for all nuclei: ρ=m/V∝A/A=constant\rho = m/V \propto A/A = \text{constant}.

Question 5

Which experimental observation provides the most direct evidence for the existence of discrete electron energy levels in atoms?

  1. The scattering of alpha particles by thin metal foils.
  2. The existence of isotopes for many elements.
  3. The line spectra produced by excited gases at low pressure. (correct answer)
  4. The photoelectric effect, where electrons are emitted from a metal surface by light.
Explanation: The fact that excited gases emit light only at specific, discrete wavelengths (forming a line spectrum) is direct evidence that electron energy transitions within the atom are restricted to specific, discrete values. Each line corresponds to a transition between two allowed energy levels. The other phenomena provide evidence for other concepts: alpha scattering for the nuclear model, isotopes for nuclear composition, and the photoelectric effect for the quantization of light energy (photons).

Question 6

The Balmer series in the hydrogen emission spectrum corresponds to electron transitions that end in the n=2n=2 state. The Lyman series corresponds to transitions that end in the n=1n=1 ground state. Which statement accurately compares the photons from these two series?

  1. All photons in the Lyman series have more energy than any photon in the Balmer series. (correct answer)
  2. The longest wavelength photon in the Balmer series has more energy than the shortest wavelength photon in the Lyman series.
  3. The Lyman series consists of visible light, while the Balmer series consists of ultraviolet light.
  4. Both series have the same series limit, corresponding to a transition from n=∞n=\infty.
Explanation: Lyman series transitions (to n=1n=1) involve larger energy differences than Balmer series transitions (to n=2n=2). The lowest energy transition in the Lyman series is from n=2n=2 to n=1n=1, releasing 10.210.2 eV. The highest possible energy transition in the Balmer series is from n=∞n=\infty to n=2n=2 (the series limit), releasing 3.43.4 eV. Since the minimum energy of a Lyman photon (10.2 eV) is greater than the maximum energy of a Balmer photon (3.4 eV), all Lyman photons are more energetic than any Balmer photon.

Question 7

An atom has discrete energy levels E1<E2<E3E_1 < E_2 < E_3. An electron in the state E3E_3 can transition to the ground state E1E_1 via two pathways: a single jump emitting a photon of frequency fAf_A, or a cascade E3→E2→E1E_3 \to E_2 \to E_1 emitting two photons of frequencies fBf_B and fCf_C. What is the relationship between these frequencies?

  1. fA=fB+fCf_A = f_B + f_C (correct answer)
  2. fA2=fB2+fC2f_A^2 = f_B^2 + f_C^2
  3. fA=fB−fCf_A = f_B - f_C
  4. 1/fA=1/fB+1/fC1/f_A = 1/f_B + 1/f_C
Explanation: This is an application of conservation of energy. The energy of a photon is given by E=hfE=hf. For the single jump, the energy released is E3−E1=hfAE_3 - E_1 = hf_A. For the cascade, the total energy released is the sum of the energies of the two photons: (E3−E2)+(E2−E1)=hfB+hfC(E_3 - E_2) + (E_2 - E_1) = hf_B + hf_C. Since (E3−E2)+(E2−E1)=E3−E1(E_3 - E_2) + (E_2 - E_1) = E_3 - E_1, we can equate the energies: hfA=hfB+hfChf_A = hf_B + hf_C, which simplifies to fA=fB+fCf_A = f_B + f_C.

Question 8

According to the Bohr model for the hydrogen atom, the angular momentum of an electron in the nn-th orbit is quantized. How does the kinetic energy, EkE_k, of the electron depend on the principal quantum number nn? (HL Only)

  1. Ek∝n2E_k \propto n^2
  2. Ek∝nE_k \propto n
  3. Ek∝1/nE_k \propto 1/n
  4. Ek∝1/n2E_k \propto 1/n^2 (correct answer)
Explanation: In the Bohr model, the total energy is En∝−1/n2E_n \propto -1/n^2. For an inverse-square law force, the kinetic energy EkE_k is equal to the magnitude of the total energy (Virial theorem), so Ek=−EnE_k = -E_n. Since En∝−1/n2E_n \propto -1/n^2, it follows that Ek∝1/n2E_k \propto 1/n^2. Alternatively, one can derive that the orbital radius r∝n2r \propto n^2 and since Ek∝1/rE_k \propto 1/r, it follows that Ek∝1/n2E_k \propto 1/n^2.

Question 9

The nucleus of a neutral atom of an element is represented by ZAX{}^{A}_{Z}\text{X}. A different particle has Z+1Z+1 protons, A−Z−1A-Z-1 neutrons and Z−1Z-1 electrons. Which statement correctly describes this particle?

  1. It is an isotope of element X with a charge of +2e.
  2. It is an ion of a different element with mass number A and a charge of +2e. (correct answer)
  3. It is an isotope of element X with a charge of -2e.
  4. It is an ion of a different element with mass number A-1 and a charge of +1e.
Explanation: Let's analyze the particle. The number of protons determines the element. Since it has Z+1 protons, it is a different element from X (which has Z protons). The mass number is the sum of protons and neutrons: A′=(Z+1)+(A−Z−1)=AA' = (Z+1) + (A-Z-1) = A. So the mass number is A. The charge is the number of protons minus the number of electrons: Charge=(Z+1)−(Z−1)=Z+1−Z+1=+2Charge = (Z+1) - (Z-1) = Z+1-Z+1 = +2. Therefore, the particle is an ion of a different element with mass number A and a charge of +2e.

Question 10

The results of the Geiger-Marsden-Rutherford experiment were consistent with scattering by an inverse-square electrostatic force. If the repulsive force between the alpha particles and the nucleus had been proportional to 1/r31/r^3 instead of 1/r21/r^2, what change in the observations would have been expected?

  1. More alpha particles would have been scattered through large angles.
  2. Fewer alpha particles would be scattered overall, with deflections being rarer at all angles. (correct answer)
  3. The scattering angle would have become independent of how closely the alpha particle approached the nucleus.
  4. The alpha particles would have been attracted to the nucleus instead of repelled.
Explanation: A force law of 1/r31/r^3 drops off much more rapidly with distance than a 1/r21/r^2 law. This means the force would be significantly weaker at all but the very closest distances. Consequently, far fewer alpha particles would experience a strong enough interaction to be deflected significantly. The 'effective target area' of the nucleus for causing scattering would be much smaller, leading to fewer scattering events at all angles compared to what was observed.

Question 11

Deuterium (12H{}^{2}_{1}\text{H}) is an isotope of hydrogen (11H{}^{1}_{1}\text{H}). How does the emission spectrum of deuterium compare to that of hydrogen?

  1. The deuterium spectrum has lines at completely different wavelengths because its nucleus is more massive.
  2. The deuterium spectrum is identical to the hydrogen spectrum because both have the same number of protons.
  3. Each spectral line in the deuterium spectrum is split into two separate lines due to the presence of a neutron.
  4. The wavelengths in the deuterium spectrum are slightly shifted compared to hydrogen, but the overall pattern is the same. (correct answer)
Explanation: Electron energy levels are primarily determined by the nuclear charge (number of protons), which is the same for both hydrogen and deuterium. Therefore, the pattern of spectral lines is nearly identical. However, the mass of the nucleus has a small effect on the energy levels (the reduced mass effect), causing a very slight shift in the wavelengths of the spectral lines. So, the spectra are almost the same, but not perfectly identical.

Question 12

An electron in a hypothetical atom transitions from an energy level of -1.51 eV to a level of -3.40 eV. What is the approximate wavelength of the photon associated with this transition? (Data: h≈6.63×10−34h \approx 6.63 \times 10^{-34} J s; c≈3.00×108c \approx 3.00 \times 10^8 m s⁻¹; e≈1.60×10−19e \approx 1.60 \times 10^{-19} C)

  1. 658 nm (correct answer)
  2. 365 nm
  3. 253 nm
  4. 821 nm
Explanation: First, find the energy of the emitted photon, which is the difference between the energy levels: ΔE=Einitial−Efinal=(−1.51 eV)−(−3.40 eV)=1.89 eV\Delta E = E_{initial} - E_{final} = (-1.51 \text{ eV}) - (-3.40 \text{ eV}) = 1.89 \text{ eV}. Convert this energy to Joules: E=1.89 eV×1.60×10−19 J/eV=3.024×10−19 JE = 1.89 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 3.024 \times 10^{-19} \text{ J}. Now use the formula E=hc/λE = hc/\lambda to find the wavelength λ\lambda: λ=hc/E=(6.63×10−34 J s×3.00×108 m s⁻¹)/(3.024×10−19 J)≈6.58×10−7 m\lambda = hc/E = (6.63 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m s⁻¹}) / (3.024 \times 10^{-19} \text{ J}) \approx 6.58 \times 10^{-7} \text{ m}, which is 658 nm.

Question 13

In the photoelectric effect, light with wavelength 300 nm is incident on a metal surface with work function 2.1 eV. If the intensity of the light is doubled while keeping the wavelength constant, what happens to the maximum kinetic energy and the number of photoelectrons emitted per second?

  1. Maximum kinetic energy doubles, and the number of photoelectrons doubles accordingly
  2. Maximum kinetic energy remains at 2.03 eV, while photoelectron emission rate increases by factor of √2
  3. Maximum kinetic energy increases to 4.13 eV, while photoelectron emission rate remains constant
  4. Maximum kinetic energy remains constant at 2.03 eV, while photoelectron emission rate doubles (correct answer)
Explanation: When analyzing photoelectric effect problems, focus on two key relationships: the energy equation and how intensity affects the process. The photoelectric effect demonstrates light's particle nature, where individual photons must have sufficient energy to eject electrons. First, let's calculate the maximum kinetic energy using Einstein's photoelectric equation: KEmax=hf−ϕ=hcλ−ϕKE_{max} = hf - \phi = \frac{hc}{\lambda} - \phi. With wavelength 300 nm and work function 2.1 eV: KEmax=(4.14×10−15 eV\cdotps)(3.00×108 m/s)300×10−9 m−2.1 eV=4.13−2.1=2.03 eVKE_{max} = \frac{(4.14 × 10^{-15} \text{ eV·s})(3.00 × 10^8 \text{ m/s})}{300 × 10^{-9} \text{ m}} - 2.1 \text{ eV} = 4.13 - 2.1 = 2.03 \text{ eV} When intensity doubles while wavelength stays constant, you're increasing the number of photons per second, but each photon still carries the same energy (E=hf=hc/λE = hf = hc/\lambda). Since maximum kinetic energy depends only on individual photon energy minus work function, it remains unchanged at 2.03 eV. However, more photons means more electrons can be ejected per second, doubling the emission rate. Answer A incorrectly suggests kinetic energy doubles with intensity—this confuses intensity with photon energy. Answer B correctly identifies constant kinetic energy but incorrectly calculates a 2\sqrt{2} factor for emission rate, which has no physical basis here. Answer C dramatically overestimates the kinetic energy at 4.13 eV (this would be the photon energy, not kinetic energy) and wrongly claims constant emission rate. Remember: in photoelectric effect questions, intensity affects the number of photoelectrons, while frequency (or wavelength) determines their maximum kinetic energy. These are independent effects.

Question 14

A beam of X-rays with wavelength 0.15 nm undergoes Bragg diffraction from a crystal with atomic spacing d = 0.20 nm. What is the smallest angle θ (measured from the crystal surface) at which constructive interference occurs?

  1. 22.0°, calculated using the Bragg condition with first-order diffraction and proper geometry (correct answer)
  2. 48.6°, calculated using the Bragg condition with second-order diffraction maximum intensity
  3. 38.7°, calculated using the Bragg condition accounting for refractive index of the crystal
  4. 61.0°, calculated using the Bragg condition with consideration of multiple scattering events
Explanation: Bragg's law states nλ=2dsin⁡θn\lambda = 2d\sin\theta where n is the order of diffraction. For the smallest angle, use n=1: sin⁡θ=λ2d=0.152×0.20=0.375\sin\theta = \frac{\lambda}{2d} = \frac{0.15}{2 \times 0.20} = 0.375. Therefore θ=arcsin⁡(0.375)=22.0°\theta = \arcsin(0.375) = 22.0°. Choice B uses n=2, giving a larger angle. Choice C incorrectly introduces refractive index considerations that don't apply to X-ray crystallography. Choice D uses an even higher order or incorrect formula, giving an unreasonably large angle.

Question 15

An X-ray photon with energy 100 keV undergoes Compton scattering from a stationary electron. If the scattered photon emerges at an angle of 90° relative to the incident direction, what is the energy of the scattered photon?

  1. 66.2 keV, calculated using relativistic energy-momentum conservation for the electron-photon system
  2. 83.9 keV, calculated using the Compton scattering formula with proper rest mass energy (correct answer)
  3. 75.0 keV, calculated assuming non-relativistic treatment of the recoiling electron motion
  4. 91.8 keV, calculated using conservation of energy while neglecting the electron recoil momentum
Explanation: The Compton scattering formula is 1E′−1E=1mec2(1−cos⁡θ)\frac{1}{E'} - \frac{1}{E} = \frac{1}{m_e c^2}(1 - \cos\theta). With E=100E = 100 keV, θ=90°\theta = 90°, and mec2=511m_e c^2 = 511 keV: 1E′=1100+1511(1−0)=0.01+0.00196=0.01196\frac{1}{E'} = \frac{1}{100} + \frac{1}{511}(1-0) = 0.01 + 0.00196 = 0.01196. Therefore E′=83.6E' = 83.6 keV ≈ 83.9 keV. Choice A uses an incorrect formula or calculation. Choice C incorrectly assumes non-relativistic treatment, which significantly underestimates the energy transfer. Choice D neglects electron recoil, violating momentum conservation.

Question 16

An alpha particle (charge +2e) approaches a gold nucleus (charge +79e) with initial kinetic energy 8.0 MeV. At what distance from the gold nucleus will the alpha particle have kinetic energy equal to 2.0 MeV?

  1. 38.0 fm, found by applying conservation of energy with electrostatic potential energy (correct answer)
  2. 85.5 fm, found by applying conservation of momentum and electrostatic force balance
  3. 42.8 fm, found by applying classical mechanics with electromagnetic field interactions
  4. 14.3 fm, found by applying relativistic energy conservation with rest mass corrections
Explanation: Using conservation of energy: Ki+Ui=Kf+UfK_i + U_i = K_f + U_f. Initially at infinity: Ki=8.0K_i = 8.0 MeV, Ui=0U_i = 0. At distance r: Kf=2.0K_f = 2.0 MeV, Uf=k(2e)(79e)rU_f = \frac{k(2e)(79e)}{r}. Therefore: 8.0=2.0+1.44×2×79r8.0 = 2.0 + \frac{1.44 \times 2 \times 79}{r}, giving 6.0=227.52r6.0 = \frac{227.52}{r}, so r=37.9r = 37.9 fm ≈ 38.0 fm. Choice B incorrectly applies momentum conservation. Choice C mentions classical mechanics incorrectly. Choice D unnecessarily invokes relativistic corrections for this energy range.

Question 17

A hydrogen atom initially in the ground state absorbs a photon and transitions to the n=3 excited state. Subsequently, it can decay back to the ground state through different pathways. How many distinct spectral lines can be observed from all possible decay pathways?

  1. 2 lines corresponding to direct and cascading transitions through intermediate states
  2. 6 lines corresponding to all permutations of transitions between the four energy levels
  3. 4 lines corresponding to transitions from each excited level to all lower levels
  4. 3 lines corresponding to all possible transitions between adjacent and non-adjacent levels (correct answer)
Explanation: When analyzing atomic transitions, you need to identify all possible pathways an electron can take when falling from an excited state to lower energy levels. Each unique transition between energy levels produces a distinct spectral line with a specific wavelength. Starting from n=3, the electron can decay through three distinct pathways: it can fall directly to n=1 (ground state), fall to n=2 then to n=1, or fall directly from n=3 to n=2. However, what matters for spectral lines isn't the pathways—it's the unique transitions between energy levels. The possible transitions are: n=3→n=2, n=3→n=1, and n=2→n=1. Each represents a different energy gap (ΔE=13.6 eV×(1/nf2−1/ni2)\Delta E = 13.6 \text{ eV} \times (1/n_f^2 - 1/n_i^2)), producing photons with different frequencies. This gives us exactly 3 distinct spectral lines, making D correct. Choice A incorrectly focuses on counting pathways rather than individual transitions, missing that cascading pathways contain multiple distinct lines. Choice B mistakenly applies permutation math—spectral lines depend on energy differences, not mathematical arrangements of levels. Choice C assumes transitions from "each excited level," but we only start from n=3, and it miscounts the actual transitions possible. Remember this key principle: count the unique energy level transitions, not the decay pathways. Each transition between different energy levels produces one spectral line, regardless of whether it occurs as part of a direct decay or within a cascading sequence.

Question 18

In the Geiger-Marsden-Rutherford experiment, the observation that a very small fraction of alpha particles are scattered through angles greater than 90° is primary evidence for which conclusion?

  1. The existence of isotopes, where nuclei have the same charge but different masses.
  2. The quantization of electron energy levels within the atom.
  3. The concentration of the atom's positive charge and mass into a very small volume. (correct answer)
  4. The overall electrical neutrality of the atom.
Explanation: A large-angle scattering event requires a very strong repulsive force. This occurs when the positive alpha particle gets very close to a highly concentrated positive charge. Furthermore, for the target not to be simply knocked away, most of the atom's mass must also be concentrated in this small volume (the nucleus). Therefore, this observation is direct evidence for a small, massive, positively charged nucleus.

Question 19

An electron in a hydrogen atom is in the n=3n=3 state. What is its angular momentum? (Planck's constant h=6.63×10−34h = 6.63 \times 10^{-34} J s) (HL Only)

  1. 1.05×10−341.05 \times 10^{-34} J s
  2. 2.11×10−342.11 \times 10^{-34} J s
  3. 3.17×10−343.17 \times 10^{-34} J s (correct answer)
  4. 9.95×10−349.95 \times 10^{-34} J s
Explanation: According to the Bohr model, the angular momentum LL of an electron is quantized and is given by the formula L=nh/(2π)L = nh/(2\pi), where nn is the principal quantum number and hh is Planck's constant. For an electron in the n=3n=3 state: L=3×(6.63×10−34 J s)/(2π)≈3.17×10−34L = 3 \times (6.63 \times 10^{-34} \text{ J s}) / (2\pi) \approx 3.17 \times 10^{-34} J s.

Question 20

What is the energy required to ionize a hydrogen atom that is initially in its first excited state (n=2n=2)? (HL Only)

  1. 3.4 eV (correct answer)
  2. 10.2 eV
  3. 13.6 eV
  4. -3.4 eV
Explanation: The energy of the nn-th level in a hydrogen atom is given by En=−13.6/n2E_n = -13.6/n^2 eV. For the first excited state, n=2n=2, the energy is E2=−13.6/22=−3.4E_2 = -13.6/2^2 = -3.4 eV. Ionization means removing the electron completely, which corresponds to transitioning it to the n=∞n=\infty level, where the energy is E∞=0E_\infty = 0 eV. The energy required for this transition is ΔE=E∞−E2=0−(−3.4 eV)=3.4 eV\Delta E = E_\infty - E_2 = 0 - (-3.4 \text{ eV}) = 3.4 \text{ eV}.