Quantum Theory and Wave–Particle Duality - College Physics

A particle with mass $m = 6.6 \times 10^{-27} \text{ kg}$ (approximately the mass of a lithium nucleus) is moving with speed $v = 1.2 \times 10^6 \text{ m/s}$ . According to the Heisenberg uncertainty principle, if the momentum is known to within $\Delta p = 1.0\%$ of its value, what is the minimum uncertainty in the particle's position?
Answer: $1.3 \times 10^{-11} \text{ m}$ The Heisenberg uncertainty principle states that Δx·Δp ≥ ℏ/2, where ℏ = h/(2π). First, calculate the momentum: p = mv = (6.6 × 10⁻²⁷ kg)(1.2 × 10⁶ m/s) = 7.92 × 10⁻²¹ kg·m/s. The uncertainty in momentum is Δp = 0.01 × 7.92 × 10⁻²¹ kg·m/s = 7.92 × 10⁻²³ kg·m/s. The minimum uncertainty in position is Δx = ℏ/(2Δp) = (1.055 × 10⁻³⁴ J·s) / [2 × (7.92 × 10⁻²³ kg·m/s)] = 6.67 × 10⁻¹³ m / (1.58 × 10⁻²² kg·m/s) = 1.3 × 10⁻¹¹ m. Choice A uses h instead of ℏ and omits the factor of 2. Choice B uses only ℏ/Δp without the factor of 2. Choice D incorrectly calculates the 1% uncertainty as 10% instead.
In the photoelectric effect, if the frequency of incident light is below the threshold frequency $f_0$ , no photoelectrons are emitted regardless of the light intensity. This observation supports which aspect of quantum theory?
Answer: Light energy is quantized in discrete packets called photons When you encounter photoelectric effect questions, focus on what makes quantum theory revolutionary compared to classical physics. The key insight is that light energy comes in discrete packets, not as a continuous wave. The threshold frequency observation directly proves that light energy is quantized in discrete packets called photons. Each photon carries energy $E = hf$ , where $h$ is Planck's constant and $f$ is frequency. When $f < f_0$ , each individual photon lacks sufficient energy to eject an electron, regardless of how many low-energy photons hit the surface. Increasing intensity just means more low-energy photons, but none can do the job alone. This "all-or-nothing" behavior only makes sense if light comes in discrete energy packets, making choice A correct. Choice B describes electron wave-particle duality (de Broglie waves), which explains electron diffraction patterns but doesn't address why there's a frequency threshold for photoemission. Choice C refers to Heisenberg's uncertainty principle about measurement precision limits, which is unrelated to the photoelectric threshold phenomenon. Choice D describes matter wave interference, again dealing with wave properties of particles rather than the quantized nature of light energy. The classical wave theory predicted that any frequency should work if you just make the light bright enough, since waves carry continuous energy. The existence of a sharp frequency cutoff, regardless of intensity, was the smoking gun that forced physicists to accept that electromagnetic radiation is quantized. Remember: photoelectric effect questions test your understanding that light energy is "chunky," not smooth—it comes in discrete photon packets.
A proton and an electron are accelerated from rest through the same potential difference $V$ . Which particle has the longer de Broglie wavelength?
Answer: The electron, because it has much smaller mass When you encounter de Broglie wavelength problems involving charged particles accelerated through electric fields, you need to connect electrostatics, kinematics, and quantum mechanics. The key insight is understanding how mass affects the final wavelength when particles gain equal energy. When both particles accelerate through the same potential difference $V$ , they gain equal kinetic energy: $KE = qV$ . Since both have the same charge magnitude, $KE_{proton} = KE_{electron}$ . From kinetics, $KE = \frac{1}{2}mv^2$ , so $v = \sqrt{\frac{2KE}{m}}$ . This means velocity is inversely proportional to the square root of mass. The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$ . Substituting our expression for velocity: $\lambda = \frac{h}{m\sqrt{\frac{2KE}{m}}} = \frac{h}{\sqrt{2mKE}}$ . Since $KE$ is the same for both particles, wavelength is inversely proportional to the square root of mass. Because the electron has much smaller mass ( $m_e \approx \frac{1}{1836}m_p$ ), it has the longer wavelength. Choice A is correct. Choice B incorrectly suggests larger mass gives longer wavelength—it's the opposite. Choice C falls into the trap of thinking equal kinetic energy means equal wavelength, ignoring the mass dependence in the momentum. Choice D incorrectly focuses on charge, but since both particles have equal charge magnitude, this doesn't affect the result. Remember: for particles with equal kinetic energy, lighter particles always have longer de Broglie wavelengths due to their smaller momentum.
Which of the following statements best describes the wave-particle duality of matter and radiation?
Answer: All matter and radiation exhibit both wave and particle characteristics, but not simultaneously in the same measurement Wave-particle duality is one of the most fundamental concepts in quantum mechanics, describing how all matter and electromagnetic radiation can exhibit both wave-like and particle-like properties depending on how you observe them. The key insight is Bohr's complementarity principle: wave and particle behaviors are mutually exclusive in any single measurement. When you design an experiment to detect wave properties (like interference or diffraction), you cannot simultaneously measure particle properties (like exact position or momentum), and vice versa. This isn't a limitation of our instruments—it's a fundamental feature of quantum reality. Answer A correctly captures this principle by stating that both matter and radiation exhibit both characteristics, but not simultaneously in the same measurement. Answer B is incorrect because it reflects pre-quantum thinking. We now know that matter also exhibits wave properties—electrons create interference patterns, and de Broglie showed that all matter has an associated wavelength. Answer C is wrong because no instrument, no matter how advanced, can measure complementary properties simultaneously—this violates Heisenberg's uncertainty principle. Answer D reverses the actual relationship; larger objects have shorter de Broglie wavelengths ( $\lambda = h/p$ ) making their wave properties less observable, while smaller particles more readily show wave behavior. When studying quantum mechanics, remember that duality isn't about objects "switching" between being waves or particles—they're always both, but the experimental setup determines which aspect you can observe. This complementarity principle appears frequently in physics problems involving electron diffraction, photon behavior, and quantum measurement.
A photon with wavelength $\lambda = 400$ nm strikes a metal surface with work function $\phi = 2.5$ eV. If the photoelectric effect occurs, what is the maximum kinetic energy of the ejected photoelectron?
Answer: 0.6 eV The photoelectric effect describes how photons can eject electrons from metal surfaces, and it follows Einstein's equation: $E_{photon} = \phi + KE_{max}$ , where the photon's energy equals the work function plus the maximum kinetic energy of the ejected electron. First, convert the photon's wavelength to energy using $E = \frac{hc}{\lambda}$ . With $h = 4.14 \times 10^{-15}$ eV·s and $c = 3.00 \times 10^8$ m/s: $E_{photon} = \frac{(4.14 \times 10^{-15})(3.00 \times 10^8)}{400 \times 10^{-9}} = 3.11 \text{ eV}$ Now apply Einstein's photoelectric equation: $KE_{max} = E_{photon} - \phi = 3.11 - 2.5 = 0.61 \text{ eV}$ This confirms answer A) 0.6 eV is correct. Looking at the wrong answers: B) 1.1 eV might result from calculation errors or using incorrect constants. C) 2.5 eV represents the work function itself—a common trap where students confuse the energy needed to remove an electron with the kinetic energy it gains. D) 3.1 eV is the total photon energy, ignoring that some energy must overcome the work function before any appears as kinetic energy. Study tip: Remember that in photoelectric problems, the photon energy gets "split"—part overcomes the work function (like climbing out of a well), and only the remainder becomes kinetic energy. Always check that your photon energy exceeds the work function, or no photoelectric effect occurs at all.
A photon of energy $E = 13.6 \text{ eV}$ is absorbed by a hydrogen atom initially in its ground state ( $n = 1$ ). Immediately after absorption, what is the most likely quantum state of the hydrogen atom?
Answer: The atom is ionized with the electron having zero kinetic energy, since $13.6 \text{ eV}$ exactly equals the ionization energy When you encounter hydrogen atom energy problems, think about the energy level structure and what happens when photons are absorbed. The key insight is understanding what $13.6 \text{ eV}$ represents in hydrogen's energy system. The ground state of hydrogen ( $n = 1$ ) has a binding energy of $13.6 \text{ eV}$ , which is exactly the energy needed to completely remove the electron from the atom. When a photon with this exact energy is absorbed, it provides just enough energy to ionize the atom, leaving the electron with zero kinetic energy after escaping. This makes answer D correct: the atom becomes ionized with the electron having zero kinetic energy, since the photon energy exactly matches the ionization energy. Answer A is wrong because it suggests the electron would have kinetic energy equal to $13.6 \text{ eV}$ minus the ionization energy. Since the ionization energy is $13.6 \text{ eV}$ , this would give zero kinetic energy anyway, but the reasoning is flawed. Answer B incorrectly assumes the electron must go to an excited state. The transition from $n = 1$ to $n = 2$ only requires $10.2 \text{ eV}$ , so $13.6 \text{ eV}$ is more than enough to ionize completely. Answer C misunderstands what happens when photon energy equals binding energy. This doesn't create a "resonance condition" that keeps the electron bound—it provides exactly the escape energy. Remember: when photon energy equals the ionization energy exactly, you get ionization with zero kinetic energy remaining. More energy would give the electron kinetic energy; less wouldn't ionize at all.
A particle in a one-dimensional box has quantized energy levels given by $E_n = \frac{n^2 h^2}{8mL^2}$ where $n = 1, 2, 3, ...$ If the box length is doubled, how does the ground state energy change?
Answer: It decreases by a factor of 4 This question tests your understanding of how physical parameters affect quantum mechanical systems, specifically the particle in a box model. When you encounter problems involving quantum energy levels, pay close attention to how each variable in the energy formula responds to changes in the system. Let's analyze what happens when the box length doubles. The ground state corresponds to n = 1, so initially: $E_1 = \frac{h^2}{8mL^2}$ . When the length becomes 2L, the new ground state energy is: $E_1' = \frac{h^2}{8m(2L)^2} = \frac{h^2}{8m \cdot 4L^2} = \frac{1}{4} \cdot \frac{h^2}{8mL^2} = \frac{E_1}{4}$ . The energy decreases by a factor of 4 because it's inversely proportional to L². Looking at the wrong answers: Choice B suggests the energy decreases by a factor of 2, which would be correct if energy were inversely proportional to L (not L²) - this reflects a common error of missing the squared relationship. Choice C claims no change, which ignores the L dependence entirely. Choice D suggests an increase, which contradicts the inverse relationship between energy and box size - as the box gets larger, the particle becomes less confined and its energy decreases. The correct answer is A. Remember this key pattern: in quantum mechanics, energy levels typically have inverse square relationships with spatial dimensions. When the "box" gets bigger, the particle has more room and lower energy - always check for squared terms in the denominators of energy expressions.
In a photoelectric experiment, when the frequency of incident light is doubled while keeping intensity constant, what happens to the number and maximum kinetic energy of photoelectrons?
Answer: Number decreases by half, maximum kinetic energy increases significantly When you encounter photoelectric effect problems, focus on Einstein's equation: $KE_{max} = hf - \phi$ , where $h$ is Planck's constant, $f$ is frequency, and $\phi$ is the work function. The key insight is that intensity affects the number of photons, while frequency affects each photon's energy. When frequency doubles while intensity stays constant, two things happen. First, each photon now carries twice the energy ( $E = hf$ ), so the maximum kinetic energy of ejected electrons increases significantly according to $KE_{max} = h(2f) - \phi = 2hf - \phi$ . This is much more than just doubling because you're adding $2hf$ rather than $hf$ . Second, since intensity equals the number of photons times energy per photon, keeping intensity constant while doubling photon energy means you must have half as many photons, resulting in half the number of photoelectrons. Looking at the wrong answers: Option B incorrectly assumes the number stays the same (ignoring that fewer high-energy photons maintain the same total intensity) and that kinetic energy exactly doubles (this would only be true if the work function were zero). Option C reverses the relationship, suggesting more photons when there are actually fewer. Option D correctly identifies that the number decreases by half but wrongly claims kinetic energy exactly doubles—the increase is actually greater than doubling because of the work function term. Remember: In photoelectric problems, intensity controls quantity (number of electrons), while frequency controls quality (energy per electron). These two parameters affect different aspects of the phenomenon.
In a double-slit experiment with electrons, if the slit separation is decreased while keeping all other parameters constant, what happens to the interference pattern on the screen?
Answer: The fringe spacing increases and the pattern becomes more spread out When you encounter double-slit interference problems, the key relationship to remember is that fringe spacing is inversely proportional to slit separation. This applies whether you're dealing with light, electrons, or any other wave phenomenon. The fringe spacing in a double-slit experiment is given by $y = \frac{\lambda L}{d}$ , where λ is the wavelength, L is the distance to the screen, and d is the slit separation. Since you're keeping all other parameters constant, only d changes. When d decreases, the fraction $\frac{\lambda L}{d}$ increases, meaning the fringe spacing y increases. The interference pattern spreads out more. Looking at each option: Choice A correctly describes this inverse relationship - decreasing slit separation leads to increased fringe spacing and a more spread-out pattern. Choice B gets the relationship backward, suggesting fringe spacing decreases with smaller slit separation. Choice C incorrectly assumes fringe spacing stays constant; while intensity might change due to diffraction effects, the spacing definitely changes according to the formula above. Choice D reflects a fundamental misunderstanding of quantum mechanics - electrons exhibit wave-particle duality, and the double-slit experiment famously demonstrates their wave nature through interference patterns. Study tip: Memorize the inverse relationship between slit separation and fringe spacing: smaller separation = wider spacing. This counterintuitive relationship appears frequently in wave optics problems, and many students incorrectly assume they should be directly proportional.
Which experimental observation provides the most direct evidence for the quantization of electromagnetic radiation?
Answer: The photoelectric effect shows energy threshold behavior independent of light intensity When you encounter questions about evidence for quantum mechanics, focus on which observations can only be explained by assuming energy comes in discrete packets (quanta) rather than continuous amounts. The photoelectric effect provides the most direct evidence because it demonstrates that electromagnetic radiation behaves as if it consists of discrete energy packets called photons. In this phenomenon, electrons are ejected from a metal surface only when incoming light exceeds a specific frequency threshold, regardless of the light's intensity. Crucially, increasing the brightness (intensity) of sub-threshold light never ejects electrons, but even dim light above the threshold frequency immediately does. This threshold behavior is impossible to explain with classical wave theory, which predicts that any frequency should eventually provide enough energy if made sufficiently intense. Einstein's photon explanation, where each photon carries energy $E = hf$ , perfectly accounts for this observation and directly demonstrates energy quantization. Option B is incorrect because interference patterns actually support the wave nature of light, not quantization. Option C is wrong because polarization is a classical wave property that doesn't require quantum theory to explain. Option D describes relativistic behavior of light but has nothing to do with energy quantization. Study tip: Remember that quantum mechanics emerged from phenomena that classical physics couldn't explain. Look for observations that require discrete energy levels or particle-like behavior rather than those explained by classical wave or particle theories.
In the wave-particle duality framework, consider an experiment where electrons are fired one at a time through a double-slit apparatus. After $N = 1000$ electrons have been detected, an interference pattern is clearly visible on the screen. If the experiment is repeated with the slit separation reduced by half while keeping the electron energy constant, approximately how many electrons are needed to observe the same level of pattern clarity?
Answer: Approximately $1000$ electrons, because the number needed depends only on the statistical nature of quantum measurements, not the geometry When you encounter quantum interference experiments, remember that the visibility of interference patterns depends fundamentally on statistical accumulation, not on the geometric parameters of the setup. The clarity of an interference pattern in quantum mechanics is determined by the signal-to-noise ratio, which improves as the square root of the number of particles detected. This statistical principle - that uncertainty decreases as $1/\sqrt{N}$ where $N$ is the number of measurements - is universal in quantum experiments and doesn't depend on slit separation, wavelength, or other apparatus geometry. Changing the slit separation does affect the interference pattern's characteristics: it alters fringe spacing (smaller separation creates wider fringes) and can modify the overall intensity distribution. However, these geometric changes don't affect how many particles you need to achieve a given level of statistical clarity in observing the pattern. Choice A incorrectly assumes that reduced slit separation automatically improves visibility by a geometric factor. Choice B makes a similar error, incorrectly linking pattern contrast to slit separation through a square root relationship. Choice C commits the most serious misconception, suggesting that geometric changes require quadratically more particles - this confuses spatial resolution with statistical significance. The correct answer is D because statistical clarity depends only on the number of quantum events, regardless of the experimental geometry. Whether your fringes are wide or narrow, the same number of electrons is needed to distinguish the interference pattern from random noise. Study tip: In quantum mechanics problems, always separate geometric effects (fringe spacing, wavelength) from statistical effects (measurement uncertainty, pattern clarity).
An X-ray photon with initial wavelength $\lambda_0 = 0.15$ nm undergoes Compton scattering at $\theta = 60°$ . What is the wavelength of the scattered photon?
Answer: 0.151 nm When you encounter Compton scattering problems, you're dealing with the collision between high-energy photons and electrons, where both energy and momentum are conserved. The key relationship is the Compton shift formula: $\lambda' - \lambda_0 = \frac{h}{m_ec}(1 - \cos\theta)$ , where $\lambda'$ is the scattered wavelength, $\lambda_0$ is the initial wavelength, and $\theta$ is the scattering angle. The Compton wavelength $\frac{h}{m_ec} = 2.43 \times 10^{-12}$ m = 0.00243 nm is a fundamental constant you should memorize. Substituting the given values: $\lambda' - 0.15 = 0.00243(1 - \cos 60°)$ $\lambda' - 0.15 = 0.00243(1 - 0.5) = 0.00243(0.5) = 0.001215$ nm $\lambda' = 0.15 + 0.001215 = 0.151215$ nm Rounding to three significant figures gives 0.151 nm, confirming answer A is correct. Answer B (0.149 nm) represents a common error where students subtract the Compton shift instead of adding it—the scattered photon always has longer wavelength (lower energy) than the incident photon. Answer C (0.152 nm) likely results from computational errors in the trigonometry, perhaps using an incorrect value for $\cos 60°$ . Answer D (0.147 nm) combines both the sign error and computational mistakes. Remember: in Compton scattering, the scattered photon always loses energy, so its wavelength always increases. The shift depends only on the scattering angle, not the initial wavelength.
The work function of a metal is $\phi = 3.2$ eV. What is the threshold frequency $f_0$ for the photoelectric effect in this metal?
Answer: $7.7 \times 10^{14}$ Hz When you encounter photoelectric effect problems, you're dealing with the fundamental relationship between light energy and electron emission. The threshold frequency is the minimum frequency needed to eject electrons from a metal surface. The key equation connecting work function and threshold frequency is $E = hf_0 = \phi$ , where $h$ is Planck's constant ( $6.626 \times 10^{-34}$ J·s) and $\phi$ is the work function. Solving for threshold frequency: $f_0 = \frac{\phi}{h}$ . First, convert the work function to joules: $\phi = 3.2 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 5.13 \times 10^{-19}$ J. Then calculate: $f_0 = \frac{5.13 \times 10^{-19}}{6.626 \times 10^{-34}} = 7.7 \times 10^{14}$ Hz. This confirms answer A is correct. Answer B ( $3.9 \times 10^{14}$ Hz) is exactly half the correct value, likely from using an incorrect value of Planck's constant or making an arithmetic error. Answer C ( $1.5 \times 10^{15}$ Hz) is roughly double the correct answer, possibly from unit conversion mistakes. Answer D ( $5.1 \times 10^{14}$ Hz) might result from forgetting to convert eV to joules or using rounded values incorrectly. Remember this approach: always convert energy units to joules when working with Planck's constant in SI units, and double-check your unit conversions. The photoelectric threshold frequency problems follow this same pattern every time.
A photon has energy $E = 4.0$ eV. What are the momentum and wavelength of this photon?
Answer: $p = 2.1 \times 10^{-27}$ kg⋅m/s, $\lambda = 310$ nm When you encounter photon problems, you're dealing with quantum mechanics where light behaves as both a wave and a particle. The key relationships are the photon energy equation $E = hf = \frac{hc}{\lambda}$ and the de Broglie momentum equation $p = \frac{h}{\lambda}$ . Starting with the given energy $E = 4.0$ eV, first convert to joules: $E = 4.0 \times 1.6 \times 10^{-19} = 6.4 \times 10^{-19}$ J. To find wavelength, use $E = \frac{hc}{\lambda}$ , so $\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{6.4 \times 10^{-19}} = 3.1 \times 10^{-7}$ m = 310 nm. For momentum, use $p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{3.1 \times 10^{-7}} = 2.1 \times 10^{-27}$ kg⋅m/s. This confirms answer A is correct. Answer B has the correct momentum calculation but doubles the wavelength, suggesting a factor-of-2 error in the energy-wavelength relationship. Answer C halves the wavelength (doubling the energy) and correspondingly doubles the momentum. Answer D combines the correct momentum from A with the incorrect wavelength from B, which is physically inconsistent since $p = h/\lambda$ . Remember that photon momentum and wavelength are inversely related through Planck's constant. Always check that your momentum and wavelength values are consistent with each other using $p = h/\lambda$ — this relationship catches many calculation errors in photon problems.
In the Compton scattering experiment, a photon collides with an electron at rest. If the photon is scattered at an angle of $90°$ from its original direction, how does the wavelength of the scattered photon compare to the incident photon?
Answer: The wavelength increases by exactly $\frac{h}{m_e c}$ When you encounter Compton scattering problems, you're dealing with the quantum mechanical interaction between photons and electrons, where both energy and momentum must be conserved. The key insight is that photons transfer some of their energy to the electron, which affects the photon's wavelength. The Compton scattering formula gives us the wavelength shift: $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos \theta)$ , where $\theta$ is the scattering angle. For a $90°$ scattering angle, $\cos(90°) = 0$ , so the wavelength shift becomes $\Delta \lambda = \frac{h}{m_e c}(1 - 0) = \frac{h}{m_e c}$ . Since this represents an increase in wavelength (the scattered photon has lower energy), answer A is correct. Answer B is wrong because the wavelength increases, not decreases—the photon loses energy to the electron. Answer C incorrectly suggests the shift is $\frac{2h}{m_e c}$ , which would only occur at a $180°$ scattering angle where $\cos(180°) = -1$ . Answer D is fundamentally flawed because while total energy is conserved in the collision, the photon's energy decreases as it transfers energy to the electron, necessarily changing its wavelength according to $E = \frac{hc}{\lambda}$ . Remember that in Compton scattering, the wavelength shift depends only on the scattering angle through the factor $(1 - \cos \theta)$ . At $90°$ , this factor equals 1, giving the standard Compton wavelength shift $\frac{h}{m_e c}$ .
According to de Broglie's hypothesis, what is the wavelength of an electron moving at $2.0 \times 10^6$ m/s? (Assume non-relativistic conditions)
Answer: $3.6 \times 10^{-10}$ m When you encounter a de Broglie wavelength problem, you're dealing with the wave-particle duality of matter. De Broglie proposed that all moving particles have an associated wavelength given by $\lambda = \frac{h}{p}$ , where $h$ is Planck's constant ( $6.626 \times 10^{-34}$ J·s) and $p$ is momentum. For this electron moving at $2.0 \times 10^6$ m/s, first calculate its momentum: $p = mv = (9.109 \times 10^{-31} \text{ kg})(2.0 \times 10^6 \text{ m/s}) = 1.82 \times 10^{-24}$ kg·m/s. Now apply de Broglie's equation: $\lambda = \frac{6.626 \times 10^{-34}}{1.82 \times 10^{-24}} = 3.6 \times 10^{-10}$ m. This confirms answer A is correct. Looking at the wrong answers: Answer B ( $1.8 \times 10^{-10}$ m) is exactly half the correct value, suggesting a calculation error like using half the electron's mass or velocity. Answer C ( $7.3 \times 10^{-11}$ m) is roughly one-fifth the correct answer, possibly from confusing fundamental constants or making algebraic errors. Answer D ( $1.5 \times 10^{-9}$ m) is about four times too large, which could result from using an incorrect mass value or misplacing decimal points. Remember that de Broglie wavelengths are inversely proportional to momentum—faster or more massive particles have shorter wavelengths. Always double-check that you're using the correct values for fundamental constants, especially the electron mass ( $9.109 \times 10^{-31}$ kg) and Planck's constant.
In the double-slit experiment with electrons, if the slit separation is doubled while keeping all other parameters constant, how does this affect the spacing between adjacent bright fringes on the detection screen?
Answer: The fringe spacing decreases by a factor of 2, consistent with the wave nature of electrons and the relationship λL/d for fringe spacing For a double-slit experiment, the spacing between adjacent bright fringes is given by Δy = λL/d, where λ is the de Broglie wavelength of the electrons, L is the distance to the screen, and d is the slit separation. When d is doubled, the fringe spacing becomes Δy' = λL/(2d) = (1/2)(λL/d) = Δy/2. This demonstrates the wave nature of electrons, as they interfere constructively and destructively to create the fringe pattern. Choice A incorrectly states that doubling slit separation increases fringe spacing and misinterprets the wave-particle duality. Choice C incorrectly assumes fringe spacing is independent of geometry. Choice D uses the wrong scaling factor and misapplies classical wave theory.
The Heisenberg uncertainty principle states that $\Delta x \Delta p \geq \frac{\hbar}{2}$ . If the position of an electron is known to within $\Delta x = 1.0 \times 10^{-10}$ m, what is the minimum uncertainty in its momentum?
Answer: $5.3 \times 10^{-25}$ kg⋅m/s The Heisenberg uncertainty principle is fundamental to quantum mechanics, establishing that you cannot simultaneously know both the position and momentum of a particle with perfect precision. When you see uncertainty principle problems, you're dealing with the trade-off between measuring complementary properties. To find the minimum uncertainty in momentum, you rearrange the uncertainty relation to solve for $\Delta p$ . Since we want the minimum uncertainty, we use the equality: $\Delta p = \frac{\hbar}{2\Delta x}$ . Substituting the values: $\Delta p = \frac{1.055 \times 10^{-34}}{2 \times 1.0 \times 10^{-10}} = \frac{1.055 \times 10^{-34}}{2.0 \times 10^{-10}} = 5.275 \times 10^{-25}$ kg⋅m/s, which rounds to $5.3 \times 10^{-25}$ kg⋅m/s. Looking at the wrong answers: Choice B ( $1.1 \times 10^{-24}$ ) results from using $h$ instead of $\hbar$ without the factor of 2, calculating $\frac{h}{\Delta x}$ . Choice C ( $3.3 \times 10^{-24}$ ) comes from using the full Planck constant $h$ divided by $2\Delta x$ , essentially $\frac{h}{2\Delta x}$ . Choice D ( $6.6 \times 10^{-25}$ ) appears to use an approximation where $\hbar$ is treated as $\frac{h}{2\pi} \approx \frac{h}{6}$ , leading to calculation errors. Study tip: Always remember that the uncertainty principle uses $\hbar = \frac{h}{2\pi}$ , not the full Planck constant $h$ . The minimum uncertainty occurs when you use the equality sign, and common mistakes involve confusing $h$ and $\hbar$ or incorrect algebraic manipulation.
The stopping potential $V_0$ in a photoelectric effect experiment is the potential difference needed to stop the most energetic photoelectrons. How is $V_0$ related to the maximum kinetic energy $K_{max}$ of the photoelectrons?
Answer: $eV_0 = K_{max}$ When you encounter photoelectric effect questions involving stopping potential, remember that this is fundamentally about energy conservation and the relationship between electric potential and kinetic energy. The stopping potential $V_0$ represents the minimum voltage needed to completely halt the most energetic photoelectrons before they reach the collecting electrode. To understand this relationship, consider what happens when an electron with maximum kinetic energy $K_{max}$ encounters this opposing electric field. As the photoelectron moves against the electric field created by the stopping potential, it does work against the electric force. The work done equals the change in electric potential energy: $W = eV_0$ , where $e$ is the elementary charge. For the electron to just stop, all of its initial kinetic energy must be converted to electric potential energy. By energy conservation: $K_{max} = eV_0$ , which gives us $eV_0 = K_{max}$ . Looking at the wrong answers: Option B ( $eV_0 = 2K_{max}$ ) incorrectly doubles the kinetic energy, perhaps confusing this with kinetic energy formulas involving $\frac{1}{2}mv^2$ . Option C ( $eV_0 = \frac{K_{max}}{2}$ ) makes the opposite error, incorrectly halving the relationship. Option D ( $V_0 = K_{max}$ ) omits the crucial elementary charge $e$ , mixing up energy units (joules) with potential units (volts). Study tip: Remember that stopping potential problems are always about complete energy conversion. When you see $V_0$ , immediately think "energy balance" and include the charge $e$ to convert between electrical potential and energy units.
A beam of X-rays with wavelength $\lambda_0 = 0.15 \text{ nm}$ undergoes Compton scattering from a stationary electron. If the scattered X-ray is detected at an angle of $\theta = 90°$ relative to the incident direction, what is the change in wavelength of the X-ray?
Answer: $2.4 \times 10^{-12} \text{ m}$ When you encounter Compton scattering problems, you're dealing with the quantum mechanical interaction between high-energy photons and electrons. The key insight is that photons behave like particles with momentum, and when they collide with electrons, both energy and momentum must be conserved. The Compton wavelength shift formula gives you the change in wavelength: $\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta)$ , where $h$ is Planck's constant, $m_e$ is the electron rest mass, $c$ is the speed of light, and $\theta$ is the scattering angle. For $\theta = 90°$ , $\cos(90°) = 0$ , so the formula simplifies to $\Delta\lambda = \frac{h}{m_e c}$ . The quantity $\frac{h}{m_e c}$ is called the Compton wavelength of the electron, which equals $2.43 \times 10^{-12} \text{ m}$ . This gives us answer A: $2.4 \times 10^{-12} \text{ m}$ . Notice that the initial wavelength $\lambda_0 = 0.15 \text{ nm}$ doesn't appear in the calculation—it's not needed for finding the wavelength shift. Answer B ( $1.2 \times 10^{-12} \text{ m}$ ) represents half the Compton wavelength, which might result from incorrectly using $\theta = 60°$ . Answer C ( $4.8 \times 10^{-12} \text{ m}$ ) is twice the Compton wavelength, possibly from confusing the shift with the final wavelength. Answer D has no clear physical basis in this context. Remember: For Compton scattering at 90°, the wavelength shift always equals the electron's Compton wavelength, regardless of the incident photon's initial wavelength.

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