Apply Atomic Structure - IB Physics

Which experimental observation provides the most direct evidence for the existence of discrete electron energy levels in atoms?
Answer: The line spectra produced by excited gases at low pressure. 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).
An electron in a hydrogen atom is in the $n=3$ state. What is its angular momentum? (Planck's constant $h = 6.63 \times 10^{-34}$ J s) (HL Only)
Answer: $3.17 \times 10^{-34}$ J s According to the Bohr model, the angular momentum $L$ of an electron is quantized and is given by the formula $L = nh/(2\pi)$ , where $n$ is the principal quantum number and $h$ is Planck's constant. For an electron in the $n=3$ state: $L = 3 \times (6.63 \times 10^{-34} \text{ J s}) / (2\pi) \approx 3.17 \times 10^{-34}$ J s.
Deuterium ( ${}^{2}_{1}\text{H}$ ) is an isotope of hydrogen ( ${}^{1}_{1}\text{H}$ ). How does the emission spectrum of deuterium compare to that of hydrogen?
Answer: The wavelengths in the deuterium spectrum are slightly shifted compared to hydrogen, but the overall pattern is the same. 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.
The nucleus of a neutral atom of an element is represented by ${}^{A}_{Z}\text{X}$ . A different particle has $Z+1$ protons, $A-Z-1$ neutrons and $Z-1$ electrons. Which statement correctly describes this particle?
Answer: It is an ion of a different element with mass number A and a charge of +2e. 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: $A' = (Z+1) + (A-Z-1) = A$ . So the mass number is A. The charge is the number of protons minus the number of electrons: $Charge = (Z+1) - (Z-1) = Z+1-Z+1 = +2$ . Therefore, the particle is an ion of a different element with mass number A and a charge of +2e.
The nucleus of iron-56 ( ${}^{56}\text{Fe}$ ) has a radius $R$ . What is the approximate radius of a nucleus of germanium-70 ( ${}^{70}\text{Ge}$ ) in terms of $R$ ? (HL Only)
Answer: 1.08 $R$ The nuclear radius $R_{nucleus}$ is related to the mass number $A$ by the formula $R_{nucleus} = R_0 A^{1/3}$ , where $R_0$ is a constant. We can set up a ratio: $\frac{R_{Ge}}{R_{Fe}} = \frac{R_0 (70)^{1/3}}{R_0 (56)^{1/3}} = \left(\frac{70}{56}\right)^{1/3} = (1.25)^{1/3} \approx 1.077$ . Therefore, the radius of the germanium-70 nucleus is approximately 1.08 $R$ .
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?
Answer: It could not explain the existence of discrete emission spectra from excited atoms. 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.
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?
Answer: Maximum kinetic energy remains constant at 2.03 eV, while photoelectron emission rate doubles 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: $KE_{max} = hf - \phi = \frac{hc}{\lambda} - \phi$ . With wavelength 300 nm and work function 2.1 eV: $KE_{max} = \frac{(4.14 × 10^{-15} \text{ eV}\cdot\text{s})(3.00 × 10^8 \text{ m/s})}{300 × 10^{-9} \text{ m}} - 2.1 \text{ eV} = 4.13 - 2.1 = 2.03 \text{ eV}$ When intensity doubles while wavelength stays constant, you're increasing the number of photons per second, but each photon still carries the same energy ( $E = hf = hc/\lambda$ ). 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 $\sqrt{2}$ 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.
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?
Answer: 83.9 keV, calculated using the Compton scattering formula with proper rest mass energy The Compton scattering formula is $\frac{1}{E'} - \frac{1}{E} = \frac{1}{m_e c^2}(1 - \cos\theta)$ . With $E = 100$ keV, $\theta = 90°$ , and $m_e c^2 = 511$ keV: $\frac{1}{E'} = \frac{1}{100} + \frac{1}{511}(1-0) = 0.01 + 0.00196 = 0.01196$ . Therefore $E' = 83.6$ 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.
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 $1/r^3$ instead of $1/r^2$ , what change in the observations would have been expected?
Answer: Fewer alpha particles would be scattered overall, with deflections being rarer at all angles. A force law of $1/r^3$ drops off much more rapidly with distance than a $1/r^2$ 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.
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?
Answer: The concentration of the atom's positive charge and mass into a very small volume. 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.
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: $h \approx 6.63 \times 10^{-34}$ J s; $c \approx 3.00 \times 10^8$ m s⁻¹; $e \approx 1.60 \times 10^{-19}$ C)
Answer: 658 nm First, find the energy of the emitted photon, which is the difference between the energy levels: $\Delta E = E_{initial} - E_{final} = (-1.51 \text{ eV}) - (-3.40 \text{ eV}) = 1.89 \text{ eV}$ . Convert this energy to Joules: $E = 1.89 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 3.024 \times 10^{-19} \text{ J}$ . Now use the formula $E = hc/\lambda$ to find the wavelength $\lambda$ : $\lambda = hc/E = (6.63 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m s⁻¹}) / (3.024 \times 10^{-19} \text{ J}) \approx 6.58 \times 10^{-7} \text{ m}$ , which is 658 nm.
In the Bohr model of hydrogen, an electron in the n=5 orbit has angular momentum L. If this electron transitions to the n=3 orbit, by what factor does its angular momentum change?
Answer: Decreases by factor of 3/5 due to the direct proportionality between n and angular momentum In the Bohr model, angular momentum is quantized as $L = n\hbar$ . For n=5: $L_5 = 5\hbar$ . For n=3: $L_3 = 3\hbar$ . The ratio is $\frac{L_3}{L_5} = \frac{3\hbar}{5\hbar} = \frac{3}{5}$ , so angular momentum decreases by factor of 3/5. Choice A gives the inverse ratio. Choice B incorrectly uses the square of the quantum numbers. Choice D incorrectly applies a square root relationship that doesn't exist for angular momentum in the Bohr model.
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?
Answer: 38.0 fm, found by applying conservation of energy with electrostatic potential energy Using conservation of energy: $K_i + U_i = K_f + U_f$ . Initially at infinity: $K_i = 8.0$ MeV, $U_i = 0$ . At distance r: $K_f = 2.0$ MeV, $U_f = \frac{k(2e)(79e)}{r}$ . Therefore: $8.0 = 2.0 + \frac{1.44 \times 2 \times 79}{r}$ , giving $6.0 = \frac{227.52}{r}$ , so $r = 37.9$ 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.
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?
Answer: 22.0°, calculated using the Bragg condition with first-order diffraction and proper geometry Bragg's law states $n\lambda = 2d\sin\theta$ where n is the order of diffraction. For the smallest angle, use n=1: $\sin\theta = \frac{\lambda}{2d} = \frac{0.15}{2 \times 0.20} = 0.375$ . Therefore $\theta = \arcsin(0.375) = 22.0°$ . 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.
Which statement provides a correct description of a feature of the atomic models?
Answer: The Bohr model resolved a key failure of the Rutherford model by postulating that electrons in certain stable orbits do not radiate energy. A primary failure of the classical Rutherford model was that an orbiting electron should, according to Maxwell's equations, continuously radiate energy and spiral into the nucleus. Bohr's first postulate directly addressed this by stating that electrons can exist in specific 'stationary states' or 'stable orbits' without radiating energy. This postulate was a radical departure from classical physics and was necessary to explain atomic stability and line spectra.
An emission spectrum of an element consists of several discrete lines. Which feature of the spectrum corresponds to the electron transition involving the largest energy change?
Answer: The line with the shortest wavelength. The energy of an emitted photon is related to its frequency $f$ and wavelength $\lambda$ by the equation $\Delta E = hf = hc/\lambda$ . This relationship shows that the energy change ( $\Delta E$ ) is inversely proportional to the wavelength ( $\lambda$ ). Therefore, the largest energy change corresponds to the photon with the shortest wavelength.
According to the Bohr model for the hydrogen atom, the angular momentum of an electron in the $n$ -th orbit is quantized. How does the kinetic energy, $E_k$ , of the electron depend on the principal quantum number $n$ ? (HL Only)
Answer: $E_k \propto 1/n^2$ In the Bohr model, the total energy is $E_n \propto -1/n^2$ . For an inverse-square law force, the kinetic energy $E_k$ is equal to the magnitude of the total energy (Virial theorem), so $E_k = -E_n$ . Since $E_n \propto -1/n^2$ , it follows that $E_k \propto 1/n^2$ . Alternatively, one can derive that the orbital radius $r \propto n^2$ and since $E_k \propto 1/r$ , it follows that $E_k \propto 1/n^2$ .
The Balmer series in the hydrogen emission spectrum corresponds to electron transitions that end in the $n=2$ state. The Lyman series corresponds to transitions that end in the $n=1$ ground state. Which statement accurately compares the photons from these two series?
Answer: All photons in the Lyman series have more energy than any photon in the Balmer series. Lyman series transitions (to $n=1$ ) involve larger energy differences than Balmer series transitions (to $n=2$ ). The lowest energy transition in the Lyman series is from $n=2$ to $n=1$ , releasing $10.2$ eV. The highest possible energy transition in the Balmer series is from $n=\infty$ to $n=2$ (the series limit), releasing $3.4$ 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.
An ion has 35 protons, 44 neutrons, and 36 electrons. What is the correct nuclear notation for this ion?
Answer: ${}^{79}_{35}\text{X}$ Nuclear notation ${}^{A}_{Z}\text{X}$ 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 ${}^{79}_{35}\text{X}$ .
How does the density of the nucleus of Uranium-238 compare to the density of the nucleus of Helium-4? (HL Only)
Answer: The densities are approximately the same. Nuclear density ( $\rho$ ) is mass ( $m$ ) divided by volume ( $V$ ). The mass of a nucleus is approximately proportional to its mass number $A$ ( $m \propto A$ ). The volume of a nucleus is $V = \frac{4}{3}\pi R^3$ , and the nuclear radius is given by $R = R_0 A^{1/3}$ . Substituting for R, we get $V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$ , so $V \propto A$ . Since both mass and volume are proportional to the mass number $A$ , their ratio, the density, is approximately constant for all nuclei: $\rho = m/V \propto A/A = \text{constant}$ .

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