Understand Atomic Structure - IB Physics

An absorption spectrum is produced by passing white light through a cool gas. Which statement best explains the resulting dark lines?
Answer: Photons of specific energies are absorbed to excite electrons, and then re-emitted isotropically. Atoms in the cool gas absorb photons whose energies exactly match the energy difference between electron shells. This excites the electrons. The excited electrons quickly de-excite, re-emitting photons of the same energy but in random directions (isotropically). Since the re-emitted photons are scattered and not directed towards the observer, the corresponding wavelengths appear as dark lines in the spectrum.
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?
Answer: Transition Z 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.
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?
Answer: It proposed that electrons could orbit at any radius, implying a continuous emission spectrum. 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.
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?
Answer: The atom is mostly empty space. 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.
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?
Answer: Number doubles, maximum kinetic energy remains 1.8 eV 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.
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?
Answer: 1:207, because the Bohr radius is inversely proportional to reduced mass 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.
A doubly-charged positive ion of an isotope of an element has 10 neutrons and 7 electrons. The element is fluorine. What is the correct nuclear notation for the nucleus of this ion?
Answer: ${}^{19}_{9}F$ A doubly-charged positive ion with 7 electrons must have 7 + 2 = 9 protons in its nucleus to have been neutral with 9 electrons. The atomic number Z is the number of protons, so Z = 9. The mass number A is the sum of protons and neutrons, so A = 9 (protons) + 10 (neutrons) = 19. The nuclear notation is ${}^{A}_{Z}X$ , which is ${}^{19}_{9}F$ .
What is the approximate ratio of the nuclear volume of bismuth-209 to the nuclear volume of beryllium-9?
Answer: 23 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 = $V_{Bi} / V_{Be} = A_{Bi} / A_{Be} = 209 / 9 \approx 23.2$ .
What is a primary distinction of the Bohr model of the atom compared to the earlier Rutherford model?
Answer: The Bohr model postulates that electrons can only exist in specific, non-radiating orbits. 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.
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)?
Answer: 6 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.
Consider a hydrogen-like ion with nuclear charge Z=3 (lithium ion, Li²⁺). Compared to hydrogen, what is the ratio of the binding energy of the ground state electron in Li²⁺ to the binding energy of the ground state electron in hydrogen?
Answer: 9:1, because binding energy scales as the square of nuclear charge For hydrogen-like ions, the binding energy is $E_n = -13.6 \times \frac{Z^2}{n^2}$ eV. For Li²⁺ (Z=3) in the ground state (n=1): $E_1 = -13.6 \times 3^2 = -122.4$ eV. For hydrogen (Z=1): $E_1 = -13.6$ eV. The ratio is $3^2 = 9$ . Choice A incorrectly uses linear scaling. Choice C incorrectly includes reduced mass effects (negligible for Li²⁺). Choice D ignores the nuclear charge dependence entirely.
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?
Answer: It would be less. 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.
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)
Answer: $2.10 \times 10^{-34}$ J s 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.
Which aspect of the Geiger-Marsden experiment provided the strongest evidence that the atomic nucleus is both very small and very massive?
Answer: The observation that a few alpha particles were deflected through very large angles, some greater than 90 degrees. A large-angle deflection implies a very strong repulsive force, which occurs only if the alpha particle gets very close to the positive charge of the nucleus (implying the nucleus is small). It also implies a large change in momentum for the alpha particle, which could only happen if it collides with a much more massive object (implying the nucleus is massive). The undeflected particles show the atom is mostly empty space, but the rare, large-angle scattering events are the key evidence for a small, massive nucleus.
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?
Answer: $1.15 \times 10^{-14}$ m 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.
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?
Answer: $f_{31} = f_{32} + f_{21}$ 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 $f_{31} = f_{32} + f_{21}$ .
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)
Answer: $1.0 \times 10^{15}$ Hz 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: $f = E/h = (6.88 \times 10^{-19} \text{ J}) / (6.63 \times 10^{-34} \text{ J s}) \approx 1.04 \times 10^{15} \text{ Hz}$ .
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)?
Answer: 122.4 eV 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, $|-122.4| = 122.4 \text{ eV}$ .
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?
Answer: 92 protons, 143 neutrons, 89 electrons 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.
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 $n$ ?
Answer: $v \propto 1/n$ 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.

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