Apply Quantum Physics - IB Physics

An X-ray photon scatters from a free electron at an angle of 180° relative to its initial direction. What is the change in the photon's wavelength?
Answer: 2h / (mₑc) The Compton scattering formula is Δλ = (h/(mₑc))(1 - cosθ). For a scattering angle of θ = 180° (backscattering), cos(180°) = -1. Therefore, the change in wavelength is Δλ = (h/(mₑc))(1 - (-1)) = (h/(mₑc))(2) = 2h/(mₑc). This represents the maximum possible wavelength shift in Compton scattering.
A particle with de Broglie wavelength $\lambda = 1.5 \times 10^{-10}$ m approaches a single slit of width $a = 3.0 \times 10^{-10}$ m. What is the angular width of the central maximum in the resulting diffraction pattern?
Answer: $60°$ representing the full central maximum width When you encounter single-slit diffraction with matter waves, you're dealing with quantum mechanics where particles exhibit wave properties. The key relationship is that the first minimum occurs when $a \sin \theta = \lambda$ , where $a$ is the slit width, $\theta$ is the angle to the first minimum, and $\lambda$ is the de Broglie wavelength. Let's find the angle to the first minimum: $\sin \theta = \frac{\lambda}{a} = \frac{1.5 \times 10^{-10}}{3.0 \times 10^{-10}} = 0.5$ This gives $\theta = 30°$ for the first minimum on one side of the central maximum. Since the diffraction pattern is symmetric, there's another first minimum at $-30°$ on the other side. The angular width of the central maximum is the total angle between these two first minima: $30° - (-30°) = 60°$ . Answer choice A ( $30°$ ) incorrectly gives only the angle to one first minimum rather than the full width. Choice B ( $120°$ ) appears to double the correct answer, possibly confusing this with a different diffraction scenario. Choice C ( $90°$ ) might come from incorrectly assuming the first minimum occurs at $45°$ , perhaps by misapplying the diffraction formula. The correct answer is D ( $60°$ ), representing the full angular width of the central maximum between the first minima on either side. Study tip: Always remember that diffraction patterns are symmetric, so when calculating the "width" of the central maximum, you need the total angle between minima on both sides, not just the angle to one minimum.
In the context of quantum physics, what does the principle of complementarity, as applied to the electron, imply?
Answer: An experiment designed to measure the wave properties of an electron will obscure its particle properties, and vice versa. The principle of complementarity, formulated by Niels Bohr, states that certain pairs of physical properties of a quantum system, like wave and particle aspects, are complementary. This means they cannot be simultaneously observed or measured with full precision. An experimental setup that demonstrates the wave nature of an electron (e.g., a double-slit experiment showing interference) will inherently prevent the observation of its particle nature (e.g., knowing its precise path through one slit). Conversely, observing the particle aspect destroys the wave aspect (the interference pattern).
In a photoelectric effect experiment, monochromatic light of frequency f illuminates a metal surface with work function Φ, and the stopping potential is measured to be Vₛ. If the frequency of the incident light is doubled to 2f, what is the new stopping potential?
Answer: Vₛ + hf/e The governing equation for the photoelectric effect is Eₖ,ₘₐₓ = hf - Φ, and the stopping potential is related by Eₖ,ₘₐₓ = eVₛ. Therefore, eVₛ = hf - Φ. In the initial case, Vₛ = (hf - Φ)/e. In the second case, with frequency 2f, the new stopping potential Vₛ' is given by eVₛ' = h(2f) - Φ. So, Vₛ' = (2hf - Φ)/e. We can rewrite this as Vₛ' = (hf - Φ + hf)/e = (hf - Φ)/e + hf/e. Since Vₛ = (hf - Φ)/e, the new stopping potential is Vₛ' = Vₛ + hf/e.
A quantum harmonic oscillator has energy levels $E_n = \hbar\omega(n + \frac{1}{2})$ . If the oscillator absorbs a photon and transitions from $n = 2$ to $n = 5$ , and then emits two photons in cascade transitions back to $n = 2$ , what is the ratio of frequencies of the two emitted photons if the intermediate state is $n = 3$ ?
Answer: $2:1$ from the quantum number ratios When analyzing quantum harmonic oscillator transitions, focus on the energy differences between levels to determine photon frequencies. Since $E = h\nu$ , the frequency of an emitted photon equals the energy difference divided by Planck's constant. Let's trace the emission process. The oscillator starts at $n = 5$ and returns to $n = 2$ via two transitions: first $5 \rightarrow 3$ , then $3 \rightarrow 2$ . For the first emission ( $5 \rightarrow 3$ ): $\Delta E_1 = E_5 - E_3 = \hbar\omega(5 + \frac{1}{2}) - \hbar\omega(3 + \frac{1}{2}) = 2\hbar\omega$ For the second emission ( $3 \rightarrow 2$ ): $\Delta E_2 = E_3 - E_2 = \hbar\omega(3 + \frac{1}{2}) - \hbar\omega(2 + \frac{1}{2}) = \hbar\omega$ Since $\nu = \frac{\Delta E}{h}$ , the frequency ratio is: $\frac{\nu_1}{\nu_2} = \frac{\Delta E_1}{\Delta E_2} = \frac{2\hbar\omega}{\hbar\omega} = \frac{2}{1}$ This confirms answer C is correct. Option A ( $1:2$ ) reverses the actual ratio. Option B ( $2:3$ ) incorrectly suggests transition probabilities affect frequency ratios—while probabilities determine emission likelihood, frequencies depend only on energy differences. Option D ( $3:2$ ) appears to confuse the quantum numbers involved rather than calculating actual energy differences. Study tip: For quantum oscillator problems, always calculate energy differences directly using the given formula. Photon frequency ratios equal energy difference ratios—ignore quantum number values themselves and focus on the $\Delta E$ between levels.
A hydrogen atom in the $n = 4$ state can transition to lower energy states by emitting photons. How many different wavelengths can be observed in the emission spectrum from all possible transitions that end in the ground state ( $n = 1$ ) either directly or through intermediate levels?
Answer: 6 wavelengths from all possible transition combinations From $n=4$ , electrons can transition to $n=3,2,1$ (3 transitions). From $n=3$ , they can go to $n=2,1$ (2 transitions). From $n=2$ , they can go to $n=1$ (1 transition). Total distinct transitions: $4→3, 4→2, 4→1, 3→2, 3→1, 2→1$ = 6 different wavelengths. Each transition has a unique energy difference, producing a unique wavelength. Choice A only counts direct transitions to ground state. Choice C incorrectly multiplies pathways. Choice D assumes multiple pathways create different wavelengths for the same transition.
Two identical photons, each with energy $E = 1.5$ MeV, undergo pair production near a heavy nucleus. What is the maximum kinetic energy that one of the produced particles (electron or positron) can have?
Answer: $2.0$ MeV from total photon energy sharing When you encounter pair production problems, remember that this process requires energy conservation and momentum conservation, with specific threshold energies for particle creation. In pair production, a photon creates an electron-positron pair near a heavy nucleus. The minimum energy required is $2m_ec^2 = 2 \times 0.511 \text{ MeV} = 1.022 \text{ MeV}$ to create the rest masses of both particles. Since each photon has $1.5 \text{ MeV}$ , the total available energy is $2 \times 1.5 = 3.0 \text{ MeV}$ . After subtracting the rest mass energy ( $1.022 \text{ MeV}$ ), the remaining kinetic energy is $3.0 - 1.022 = 1.978 \text{ MeV} \approx 2.0 \text{ MeV}$ . To maximize one particle's kinetic energy, the other particle should have minimum kinetic energy (essentially at rest in the lab frame). Therefore, one particle can have nearly all $2.0 \text{ MeV}$ of available kinetic energy. Option A incorrectly uses only one photon's energy ( $1.5 \text{ MeV}$ ), ignoring that both photons participate in the process. Option B suggests $2.5 \text{ MeV}$ , which exceeds the available kinetic energy after accounting for rest masses. Option D claims $3.0 \text{ MeV}$ , which would require the created particles to have no rest mass—physically impossible. The correct answer is C: $2.0 \text{ MeV}$ represents the maximum kinetic energy available when both photons contribute their energy and rest mass requirements are satisfied. Study tip: In pair production problems, always calculate total available energy first, subtract the required rest mass energy ( $2m_ec^2$ ), then determine how the remaining kinetic energy can be distributed between particles.
A particle is described by the wave function $\psi(x) = A\sin(\frac{\pi x}{L})$ for $0 \leq x \leq L$ and $\psi(x) = 0$ elsewhere. If a measurement of position is made, what is the probability of finding the particle in the region $\frac{L}{4} \leq x \leq \frac{3L}{4}$ ?
Answer: $\frac{1}{2} + \frac{1}{\pi}$ from the normalized probability integral First normalize: $\int_0^L |A|^2\sin^2(\frac{\pi x}{L})dx = 1$ . Using $\sin^2(u) = \frac{1-\cos(2u)}{2}$ : $|A|^2 \cdot \frac{L}{2} = 1$ , so $A = \sqrt{\frac{2}{L}}$ . Probability = $\int_{L/4}^{3L/4} \frac{2}{L}\sin^2(\frac{\pi x}{L})dx = \frac{2}{L} \int_{L/4}^{3L/4} \frac{1-\cos(\frac{2\pi x}{L})}{2}dx = \frac{1}{L}[x - \frac{L}{2\pi}\sin(\frac{2\pi x}{L})]_{L/4}^{3L/4} = \frac{1}{2} + \frac{1}{\pi}$ . Choice B has wrong sign. Choices C and D use incorrect normalization constant.
Why is the wave nature of a moving macroscopic object, such as a tennis ball, not observable in everyday life?
Answer: The de Broglie wavelength is extremely small due to the object's large mass and momentum. The de Broglie wavelength is given by λ = h/p = h/(mv). For a macroscopic object like a tennis ball, the mass 'm' is very large compared to quantum particles. Even at low speeds 'v', the resulting momentum 'p' is enormous. Since Planck's constant 'h' is extremely small (≈ 6.63 × 10⁻³⁴ J s), the calculated wavelength λ is infinitesimally small, far smaller than any atomic nucleus, making it impossible to observe wave effects like diffraction or interference.
According to the de Broglie hypothesis, an electron in a stable orbit around a nucleus must have a standing wave pattern. For an electron in the n=3 energy level, how many of its de Broglie wavelengths fit into the circumference of its orbit?
Answer: 3 The de Broglie condition for a stable orbit (a standing wave) is that the circumference of the orbit (2πr) must be an integer multiple of the electron's wavelength (λ). The integer, n, corresponds to the principal quantum number. The condition is 2πr = nλ. Therefore, for the n=3 energy level, exactly 3 de Broglie wavelengths must fit into the circumference of the orbit.
In the Compton effect, a photon collides with a stationary electron. Which statement correctly describes the energy and momentum after the collision?
Answer: Both total energy and total momentum are conserved, with the scattered photon having less energy than the incident photon. The Compton effect is modeled as an elastic collision between a photon and an electron. In any such collision, both total energy and total momentum must be conserved. During the interaction, the photon transfers some of its energy and momentum to the electron, causing the electron to recoil. Consequently, the scattered photon must have less energy (and thus a longer wavelength) than the incident photon. A is the only option that correctly states both conservation laws and the energy change of the photon.
An electron in a hydrogen atom transitions from $n = 4$ to $n = 2$ and emits a photon. This photon then strikes a metal surface with work function $\phi = 2.0$ eV. What is the maximum kinetic energy of the ejected photoelectron?
Answer: $0.55$ eV from the hydrogen transition energy Hydrogen transition energy: $E_{photon} = 13.6(\frac{1}{n_f^2} - \frac{1}{n_i^2}) = 13.6(\frac{1}{4} - \frac{1}{16}) = 13.6 \times \frac{3}{16} = 2.55$ eV. For photoelectric effect: $KE_{max} = E_{photon} - \phi = 2.55 - 2.0 = 0.55$ eV. Choice B uses wrong transition calculation. Choice C adds work function instead of subtracting. Choice D ignores work function completely.
An electron is accelerated from rest to a final speed v such that its total relativistic energy is double its rest mass energy. What is the de Broglie wavelength of this electron?
Answer: h / (√3 mₑc) The de Broglie wavelength is λ = h/p, where p is the relativistic momentum. The total relativistic energy is E = γmₑc² and the rest energy is E₀ = mₑc². We are given E = 2E₀. The relativistic energy-momentum relation is E² = (pc)² + (E₀)². Substituting E = 2E₀: (2E₀)² = (pc)² + (E₀)². This gives 4(E₀)² = (pc)² + (E₀)², so (pc)² = 3(E₀)². Taking the square root, pc = √3 E₀ = √3 mₑc². Therefore, the momentum is p = √3 mₑc. The de Broglie wavelength is λ = h/p = h / (√3 mₑc).
An electron has a de Broglie wavelength of λ. If the kinetic energy of the electron is doubled, what is its new de Broglie wavelength?
Answer: λ/√2 The de Broglie wavelength is given by λ = h/p, where p is the momentum. The non-relativistic kinetic energy is Eₖ = p²/(2m). Therefore, momentum can be expressed as p = √(2mEₖ). Substituting this into the de Broglie equation gives λ = h/√(2mEₖ). This shows that λ is inversely proportional to the square root of the kinetic energy (λ ∝ 1/√Eₖ). If the kinetic energy Eₖ is doubled to 2Eₖ, the new wavelength λ' will be λ' = λ/√2.
A proton and an alpha particle are both accelerated from rest through the same potential difference. What is the ratio of the de Broglie wavelength of the proton (λₚ) to that of the alpha particle (λₐ)? (An alpha particle has mass ≈ 4mₚ and charge +2e).
Answer: 2√2 The kinetic energy gained by a particle of charge q accelerated through a potential difference V is Eₖ = qV. The de Broglie wavelength is λ = h/p = h/√(2mEₖ). So, λ = h/√(2mqV). For the proton: λₚ = h/√(2mₚeV). For the alpha particle: λₐ = h/√(2(4mₚ)(2e)V) = h/√(16mₚeV). The ratio is λₚ/λₐ = [h/√(2mₚeV)] / [h/√(16mₚeV)] = √(16mₚeV) / √(2mₚeV) = √8 = 2√2.
An X-ray photon with wavelength λ₀ scatters from an electron at an angle of 60°. A second, identical X-ray photon scatters from another electron at an angle of 120°. What is the ratio of the change in wavelength in the second case (Δλ₁₂₀) to the change in wavelength in the first case (Δλ₆₀)?
Answer: 3 The Compton shift is given by Δλ = (h/mₑc)(1 - cosθ). Let C = h/mₑc. For θ = 60°, Δλ₆₀ = C(1 - cos60°) = C(1 - 0.5) = 0.5C. For θ = 120°, Δλ₁₂₀ = C(1 - cos120°) = C(1 - (-0.5)) = 1.5C. The ratio is Δλ₁₂₀ / Δλ₆₀ = (1.5C) / (0.5C) = 3.
A photon undergoes Compton scattering from a stationary electron. The initial photon has energy E₀ and the scattered photon has energy E. The electron recoils with kinetic energy Kₑ. Which statement correctly relates these quantities?
Answer: E₀ = E + Kₑ The principle of conservation of energy must apply to the collision. The total energy before the collision is the energy of the incident photon, E₀ (the electron is stationary, so its initial kinetic energy is zero). The total energy after the collision is the sum of the energy of the scattered photon, E, and the kinetic energy of the recoiling electron, Kₑ. Therefore, E₀ = E + Kₑ.
What is the momentum of a photon of yellow light with a wavelength of 6.0 × 10⁻⁷ m? (h ≈ 6.6 × 10⁻³⁴ J s)
Answer: 1.1 × 10⁻²⁷ kg m s⁻¹ The momentum of a photon is related to its wavelength by the de Broglie relation, which applies to all particles including photons: p = h/λ. Substituting the given values: p = (6.6 × 10⁻³⁴ J s) / (6.0 × 10⁻⁷ m) = 1.1 × 10⁻²⁷ J s m⁻¹ = 1.1 × 10⁻²⁷ kg m s⁻¹.
In a double-slit experiment with electrons, an interference pattern is formed. If the accelerating voltage of the electron gun is increased, what happens to the spacing between the interference fringes?
Answer: It decreases, because the electrons' de Broglie wavelength decreases. The spacing of interference fringes (s) is given by s ≈ λD/d, where λ is the wavelength. Increasing the accelerating voltage (V) increases the kinetic energy of the electrons (Eₖ = eV). The de Broglie wavelength is λ = h/p = h/√(2mEₖ). Therefore, as V increases, Eₖ increases, and the wavelength λ decreases. Since the fringe spacing s is proportional to λ, a decrease in λ causes the fringe spacing to decrease.
Which of the following experimental observations from the photoelectric effect provides the strongest evidence against the classical wave theory of light?
Answer: There exists a threshold frequency below which no photoelectrons are emitted, regardless of the light intensity. The classical wave theory of light predicts that energy is delivered continuously. Therefore, any frequency of light, if intense enough and applied for a sufficient duration, should be able to impart enough energy to an electron to overcome the work function. The existence of a sharp threshold frequency (B) below which no emission occurs, no matter how intense the light, directly contradicts this prediction. While A, C, and D are also explained by the photon model and not the wave model, the existence of a threshold frequency is the most fundamental failure of the classical theory in this context.

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