Quantum Physics PDF

## 2 Quantum Physics ### Objectives: * To learn certain experimental results that can be understood only by particle theory of electromagnetic waves. * To learn the particle properties of waves and the wave properties of the particles. * To understand the uncertainty principle. ### Blackbody Radi...

## 2 Quantum Physics ### Objectives: * To learn certain experimental results that can be understood only by particle theory of electromagnetic waves. * To learn the particle properties of waves and the wave properties of the particles. * To understand the uncertainty principle. ### Blackbody Radiation and Plancks's Hypothesis A black body is an object that absorbs all incident radiation. A small hole cut into a cavity is the most popular and realistic example. None of the incident radiation escapes. The radiation is absorbed in the walls of the cavity. This causes a heating of the cavity walls. The oscillators in the cavity walls vibrate and re-radiate at wavelengths corresponding to the temperature of the cavity, thereby producing standing waves. Some of the energy from these standing waves can leave through the opening. The electromagnetic radiation emitted by the black body is called black-body radiation. **Figure 2-1:** A physical model of a blackbody. The black body is an ideal absorber of incident radiation. * A black-body reaches thermal equilibrium with the surroundings when the incident radiation power is balanced by the power re-radiated. * The emitted "thermal" radiation from a black body characterizes the equilibrium temperature of the black-body. * The nature of radiation from a blackbody does not depend on the material of which the walls are made. ### Basic laws of radiation (1) All objects emit radiant energy. (2) Hotter objects emit more energy (per unit area) than colder objects. The total power of the emitted radiation is proportional to the fourth power of temperature. This is called Stefan’s Law and is given by, $P = 𝜎𝐴𝑒T^4$ where P is power radiated from the surface of the object (W), T is equilibrium surface temperature (K), σ is Stefan-Boltzmann constant (= 5.670 × 10⁻⁸ W/m²K⁴), A is surface area of the object (m²) and e is emissivity of the surface (e=1 for a perfect blackbody). (3) The peak of the wavelength distribution shifts to shorter wavelengths as the black body temperature increases. This is Wien’s Displacement Law and is given by, $λ_mT = constant = 2.898 × 10⁻³ m.K, or λ_mœ T⁻¹$ where λ_m is the wavelength corresponding to peak intensity and T is equilibrium temperature of the blackbody. **Figure 2-2:** Intensity of blackbody radiation versus wavelength at two temperatures. (4) Rayleigh-Jeans Law: This law tries to explain the distribution of energy from a black body. The intensity or power per unit area I (λ,T)dλ, emitted in the wavelength interval a to λ+da from a blackbody is given by, $Ι(λ,Τ) = \frac{2πckT}{λ⁴}$ kB is Boltzmann’s constant, c is speed of light in vacuum, T is equilibrium blackbody temperature. It agrees with experimental measurements only for long wavelengths. It predicts an energy output that diverges towards infinity as wavelengths become smaller and is known as the ultraviolet catastrophe. **Figure 2-3:** Comparison of experimental results and the curve predicted by the Rayleigh–Jeans law for the distribution of blackbody radiation (5) Planck’s Law: Max Planck developed a theory of blackbody radiation that leads to an equation for I (λ, T) that is in complete agreement with experimental results. To derive the law, Planck made two assumptions concerning the nature of the oscillators in the cavity walls: (i) The energy of an oscillator is quantized hence it can have only certain discrete values: $E_n = nhf$ where n is a positive integer called a quantum number, f is the frequency of cavity oscillators, and h is a constant called Planck’s constant. Each discrete energy value corresponds to a different quantum state, represented by the quantum number n. (ii) The oscillators emit or absorb energy only when making a transition from one quantum state to another. Difference in energy will be integral multiples of hf. **Figure 2-4:** Allowed energy levels for an oscillator with frequency f Planck's law explains the distribution of energy from a black body which is given by, $Ι(λ,Τ) = \frac{2πhc²}{λ⁵} \frac{1}{e^{\frac{hc}{λk_BT}}-1}$ where I (λ,T) da is the intensity or power per unit area emitted in the wavelength interval da from a blackbody, h is Planck's constant, kB is Boltzmann's constant, c is speed of light in vacuum and T is equilibrium temperature of blackbody. The Planck's Law gives a distribution that peaks at a certain wavelength, the peak shifts to shorter wavelengths for higher temperatures, and the area under the curve grows rapidly with increasing temperature. This law is in agreement with the experimental data. The results of Planck's law: * The denominator [exp(hc/akT)] tends to infinity faster than the numerator (2-5), thus resolving the ultraviolet catastrophe and hence arriving at experimental observation: * I (λ, Τ) → O as λ → 0. * For very large λ, * $exp(\frac{hc}{λkT} -1 = \frac{hc}{λkT} ⇒ I(λ,T) → 2πcλ₄kT$ * i.e. I (λ, Τ) → 0 as λ→ ∞. From a fit between Planck's law and experimental data, Planck’s constant was derived to be h = 6.626 × 10⁻³⁴ J-s. ### Photoelectric effect Ejection of electrons from the surface of certain metals when it is irradiated by an electromagnetic radiation of suitable frequency is known as photoelectric effect. **Figure 2-5**: (a) Apparatus (b) circuit for studying Photoelectric Effect (T – Evacuated glass/quartz tube, E – Emitter Plate / Photosensitive material / Cathode, C – Collector Plate / Anode, V – Voltmeter, A - Ammeter) ### Experimental Observations: **Figure 2-6:** Photoelectric current versus applied potential difference for two light intensities 1. When plate E is illuminated by light of suitable frequency, electrons are emitted from E and a current is detected in A (Figure 2.5). 2. Photocurrent produced vs potential difference graph shows that kinetic energy of the most energetic photoelectrons is, * Kmax = e AVs where AVs is stopping potential 3. Kinetic energy of the most energetic photoelectrons is independent of light intensity. 4. Electrons are emitted from the surface of the emitter almost instantaneously 5. No electrons are emitted if the incident light frequency falls below a cutoff frequency. 6. Kinetic energy of the most energetic photoelectrons increases with increasing light frequency. ### Classical Predictions: 1. If light is really a wave, it was thought that if one shine of light of any fixed wavelength, at sufficient intensity on the emitter surface, electrons should absorb energy continuously from the em waves and electrons should be ejected. 2. As the intensity of light is increased (made it brighter and hence classically, a more energetic wave), kinetic energy of the emitted electrons should increase. 3. Measurable / larger time interval between incidence of light and ejection of photoelectrons. 4. Ejection of photoelectron should not depend on light frequency 5. In short experimental results contradict classical predictions. 6. Photoelectron kinetic energy should not depend upon the frequency of the incident light. ### Einstein’s Interpretation of electromagnetic radiation: 1. Electromagnetic waves carry discrete energy packets (light quanta called photons now). 2. The energy E, per packet depends on frequency f: E = hf. 3. More intense light corresponds to more photons, not higher energy photons. 4. Each photon of energy E moves in vacuum at the speed of light: c = 3 x 10⁸ m/s and each photon carries a momentum, p = E/c. ### Einstein’s theory of photoelectric effect: A photon of the incident light gives all its energy hf to a single electron (absorption of energy by the electrons is not a continuous process as envisioned in the wave model) and the kinetic energy of the most energetic photoelectron, $K_{max} = hf - Φ$ (Einstein's photoelectric equation) Φ is called the work function of the metal. It is the minimum energy with which an electron is bound in the metal. All the observed features of photoelectric effect could be explained by Einstein’s photoelectric equation: 1. Equation shows that Kmax depends only on frequency of the incident light. 2. Almost instantaneous emission of photoelectrons due to one -to-one interaction between photons and electrons. 3. Ejection of electrons depends on light frequency since photons should have energy greater than the work function in order to eject an electron. 4. The cutoff frequency fo is related to Φ by fc = Φ/h. If the incident frequency fis less than fo, there is no emission of photoelectrons. **Figure 2-7:** A representative plot of Kmax versus frequency of incident light for three different metals ### Compton Effect When X-rays are scattered by free/nearly free electrons, they suffer a change in their wavelength which depends on the scattering angle. This scattering phenomenon is known as Compton Effect. ### Classical Predictions: Oscillating electromagnetic waves (classically, X-rays are em waves) incident on electrons should have two effects: i) oscillating electromagnetic field causes oscillations in electrons. Each electron first absorbs radiation as a moving particle and then re-radiates in all directions as a moving particle and thereby exhibiting two Doppler shifts in the frequency of radiation. ii) radiation pressure should cause the electrons to accelerate in the direction of propagation of the waves. Because different electrons will move at different speeds after the interaction, depending on the amount of energy absorbed from electromagnetic waves, the scattered waves at a given angle will have all frequencies (Doppler- shifted values). ### Compton’s experiment and observation: Compton measured the intensity of scattered X-rays from a solid target (graphite) as a function of wavelength for different angles. The experimental setup is shown in Figure 2.8. Contrary to the classical prediction, only one frequency for scattered radiation was seen at a given angle. This is shown in the Figure 2.9. The graphs for three nonzero angles show two peaks, one at A and the other at λ’>λο . The shifted peak at l' is caused by the scattering of X-rays from free electrons. Shift in wavelength was predicted by Compton to depend on scattering angle as $λ' − λ = \frac{h}{mc}(1 − cos θ)$ where m is the mass of the electron, c is velocity of light, h is Planck’s constant. This is known as Compton shift equation, and the factor $\frac{h}{mc}$ is called the Compton wavelength and $\frac{h}{mc}$ = 2.43 pm for electron. **Figure 2-8** Schematic diagram of Compton’s apparatus. The wavelength is measured with a rotating crystal spectrometer for various scattering angles θ. **Figure 2-9:** Scattered x-ray intensity versus wavelength for Compton scattering at 0= 0°, 45°, 90°, and 135° showing single frequency at a given angle ### Derivation of the Compton shift equation: Compton could explain the experimental result by treating the X-rays not as waves but rather as point like particles (photons) having energy E=hfo=hc/A, momentum p = hf/c = h/2_and zero rest energy. Photons collide elastically with free electrons initially at rest and moving relativistically after collision. Letλο, po = h/land_E｡ = hc/o be the wavelength, momentum and energy of the incident photon respectively. λ', p' = h/λ' and E' = hc/l' be the corresponding quantities for the scattered photon. We know that, for the electron, the total relativistic energy E = √p2c² + m²c4 Kinetic energy K = E-mc² And momentum p γmν. where Y=√1- v2 2 vand mare the speed and mass of the electron respectively. **Figure 2-10:** Quantum model for X-ray scattering from an electron In the scattering process, the total energy and total linear momentum of the system must be conserved. For conservation of energy we must have, E = E' + K Or ie, E = E' + (E - mc²) Eo - E' + mc² = E = √p2c² + m²c4 Squaring both the sides, (E° − E')² + 2(E｡ - E') mc² + m²c4 = p2c² + m²c4 For conservation of momentum, x-component: po = p' cos 0 + p cos ф y-component: 0 = p' sin 0 p sin Rewriting these two equations po - p' cos 0 = p cos ф Squaring both the sides and adding, p' sine = p sin $p² - 2pop' cos θ + p'² = p²$ Substituting this p² in the equation : (E｡ − E')² + 2(E｡ – E') mc² = p2c², one gets (E｡ − E')² + 2(E｡ – E') mc² = (p² - 2pop' cos 0 + p'²)c² Substituting photon energies and photon momenta one gets hc 2 hc (-)+2(-) mc² = (c)² - 2()()cos + (1) hc λο λ Simplifying one gets hc hc λο λ' λο λο 2 hc hc hc hc 1 1 hc hc hc λο λο λο λ' λο λο +(2 hc i.e., λολ' +(-) mc² = hc cos 0 λολ OR, () mc² = hc (1 cos 0) λολ ### Compton shift: $λ' − λ_0 = \frac{h}{mc}(1 − cos θ)$ ### Photons and Electromagnetic Waves [Dual Nature of Light] * Light exhibits diffraction and interference phenomena that are only explicable in terms of wave properties. * Photoelectric effect and Compton Effect can only be explained taking light as photons / particle. This means true nature of light is not describable in terms of any single picture, instead both wave and particle nature have to be considered. In short, the particle model and the wave model of light complement each other. ### De Broglie Hypothesis – Wave Properties of Particles We have seen that light comes in discrete units (photons) with particle properties (energy E and momentum P) that are related to the wave-like properties of frequency and wavelength. Louis de Broglie postulated that because photons have both wave and particle characteristics, perhaps all forms of matter have wave-like properties, with the wavelength λ related to momentum p in the same way as for light. ### de Broglie wavelength: $λ = \frac{h}{p} = \frac{h}{mv}$ where h is Planck’s constant and p is momentum of the quantum particle, m is mass of the particle, and v is speed of the particle. The electron accelerated through a potential difference of ΔV, has a non-relativistic kinetic energy $\frac{1}{2}mv² = eΔV$ where e is electron charge. Hence, the momentum (p) of an electron accelerated through a potential difference of ΔV is $p = mv = √2 meΔV$ $Frequency of the matter wave associated with the particle is \frac{E}{h}$ where E is total relativistic energy of the particle ### Davisson and Germer Experiment This is an experiment on diffraction of accelerated electrons by crystals to establish the wave nature of electrons. The regular spacing of the atoms in a crystal act as a grating for an electron beam producing a diffraction pattern by electron scattering. The result of this experiment resembles the result of the x ray diffraction giving experimental evidence for de Broglie hypothesis. **Figure 2-11:** Schematic of Davisson-Germer experiment (left). Polar plot of intensities after scattering for incident electron energy of 54 eV. (right). A beam of electrons from a heated filament accelerated to a potential V is collimated and allowed to strike a single crystal of nickel. Electrons are scattered in all directions by the atoms in the crystal. The intensity of the scattered electron beam is measured by a detector which can be moved to any angle & relative to the incident beam and is shown in the polar plot of o verses the intensity (xr, the radius vector). The most intense reflection of electron beam at an angle $ = 50°, for an accelerating potential 54 V is observed. An electron of mass m accelerated to a potential V has kinetic energy, mv² = e V 1 2 where, v = velocity. And the momentum, p = m v = √2meV .. de Broglie wavelength, λ p h ===mev √2 me V Substituting V = 54 V, we get the value for the de Broglie wavelength as λ = 0.167nm. Wavelength can also be obtained, in this experiment, using Bragg's law of diffraction. According to which the wavelength is given by the following relationship: d sind = ηλ Substituting n = 1 (for the first order diffraction maximum), $ = 50°(scattering angle corresponding to the first order diffraction maximum), d = 0.215 nm (inter-atomic spacing in nickel), we get a = 0.165nm The values of the wavelength from both the approaches agree very well. This verifies de Broglie's hypothesis and establishes the wave nature of electrons. Davisson-Germer experiment and G P Thomson's electron diffraction experiment confirmed de Broglie relationship p = h /λ. Subsequently it was found that atomic beams, and beams of neutrons, also exhibit diffraction when reflected from regular crystals. Thus de Broglie's formula seems to apply to any kind of matter. Now the dual nature of matter and radiation is an accepted fact and it is stated in the principle of complementarity. This states that wave and particle models of either matter or radiation complement each other. ### The Quantum Particle Quantum particle is a model by which particles having dual nature are represented. We must choose one appropriate behavior for the quantum particle (particle or wave) in order to understand a particular behavior. To represent a quantum wave, we have to combine the essential features of both an ideal particle and an ideal wave. An essential feature of a particle is that it is localized in space. But an ideal wave is infinitely long (non-localized) as shown in Figure 2.11. **Figure 2.11** Section of an ideal wave of single frequency Now to build a localized entity from an infinitely long wave, waves of same amplitude, but slightly different frequencies are superposed (Figure 2.12). **Figure 2.12**: Superposition of two waves Wave 1 and Wave 2 If we add up large number of waves in a similar way, the small localized region of space where constructive interference takes place is called a wave packet, which represents a quantum particle (Figure 2.13). **Figure 2.13:** Wave packet ### Mathematical representation of a wave packet: Superposition of two waves of equal amplitude, but with slightly different frequencies, f₁, and f₂, traveling in the same direction are considered. The waves are written as * У1 = A cos(k₁x - w₁t) and y2 = A cos(k2x - w2t) where k = 2π/λ, W = 2πf The resultant wave y = y1 + y2 y = 2A [cos(x-t) cos (K₁+kzx - 1+w2t)] 2 where Ak = k1 - k2 and Δω = 01 – 02. **Figure 2.14:** Beat pattern due to superposition of wave trains yl and y2 The resulting wave oscillates with the average frequency, and its amplitude envelope (in square brackets, shown by the blue dotted curve in Figure 2.14) varies according to the difference frequency. A realistic wave (one of finite extent in space) is characterized by two different speeds. The phase speed, the speed with which wave crest of individual wave moves, is given by * Vp = fλ or Up = k (2.11) The envelope of group of waves can travel through space with a different speed than the individual waves. This speed is called the group speed or the speed of the wave packet which is given by * Vg = Δω Ak =$[\frac{dω}{dK}]$ (2.12) For a superposition of large number of waves to form a wave packet, this ratio is Vg = dw dk In general these two speeds are not the same. ### Relation between group speed (Vg) and phase speed (vp): * But * Vp = ω k = dw dk = fλ = **.. ω = k vp** * d(kvp) = k dvp dk + Vp Vg Substituting for k in terms of λ, we get * Vg = Vp - dk dvp λ αλ ### Relation between group speed (Vg) and particle speed (u): * ω = 2πf = E 2 π h and * k = 2 π λ = 2 π = h/p 2 πρ h * dw 2π de d E h Vg = = π dk 2 dp dp h For a classical particle moving with speed u, the kinetic energy E is given by * E = mu² 2 m = p2 2 m and dE = 2 p dp 2 m d E or = u dp m * Vg dw dk = d E dp = u (2.13) (2.14) i.e., we should identify the group speed with the particle speed, speed with which the energy moves. To represent a realistic wave packet, confined to a finite region in space, we need the superposition of large number of harmonic waves with a range of k-values. ### Double-Slit Experiment Revisited One way to confirm our ideas about the electron’s wave-particle duality is through an experiment in which electrons are fired at a double slit. Consider a parallel beam of mono-energetic electrons incident on a double slit as in Figure 3.15. Let’s assume the slit widths are small compared with the electron wavelength so that diffraction effects are negligible. An electron detector screen (acts like the “viewing screen” of Young’s double-slit experiment) is positioned far from the slits at a distance much greater than d, the separation distance of the slits. If the detector screen collects electrons for a long enough time, we find a typical wave interference pattern for the counts per minute, or probability of arrival of electrons. Such an interference pattern would not be expected if the electrons behaved as classical particles, giving clear evidence that electrons are interfering, a distinct wave-like behavior. **Figure 2.15:** (a) Schematic of electron beam interference experiment, (b) Photograph of a double-slit interference pattern produced by electrons If we measure the angle 0 at which the maximum intensity of the electrons arrives at the detector screen, we find they are described by exactly the same equation as that for light: dsine = mλ, where m is the order number and λ is the electron wavelength. Therefore, the dual nature of the electron is clearly shown in this experiment: the electrons are detected as particles at a localized spot on the detector screen at some instant of time, but the probability of arrival at the spot is determined by finding the intensity of two interfering waves. ### Uncertainty Principle It is fundamentally impossible to make simultaneous measurements of a particle’s position and momentum with infinite accuracy. This is known as Heisenberg uncertainty principle. The uncertainties arise from the quantum structure of matter. For a particle represented by a single wavelength wave existing throughout space, λ is precisely known, and according to de Broglie hypothesis, its p is also known accurately. But the position of the particle in this case becomes completely uncertain. This means Δλ=0, Δp =0; but 4x = ∞ In contrast, if a particle whose momentum is uncertain (combination of waves / a range of wavelengths are taken to form a wave packet), so that Ax is small, but Al is large. If Ax is made zero, Δλ and thereby Ap will become ∞. In short (4x)(4px) ≥ h/4π (2.15) where Ax is uncertainty in the measurement of position x of the particle and Apx is uncertainty in the measurement of momentum px of the particle. One more relation expressing uncertainty principle is related to energy and time which is given by (ΔΕ) (At) > h/4π (2.16) where ΔΕ is uncertainty in the measurement of energy E of the system when the measurement is done over the time interval At. ### Questions 1. Explain (a) Stefan's law (b) Wien's displacement law (c) Rayleigh-Jeans law. 2. Sketch schematically the graph of wavelength vs intensity of radiation from a blackbody. 3. Explain Planck's radiation law. 4. Write the assumptions made in Planck's hypothesis of blackbody radiation. 5. Explain photoelectric effect. 6. What are the observations in the experiment on photoelectric effect? 7. What are the classical predictions about the photoelectric effect? 8. Explain Einstein's photoelectric equation. 9. Which are the features of photoelectric effect-experiment explained by Einstein's photoelectric equation? 10. Sketch schematically the following graphs with reference to the photoelectric effect: (a) photoelectric current vs applied voltage (b) kinetic energy of most-energetic electron vs frequency of incident light. 11. Explain Compton effect. 12. Explain the experiment on Compton effect. 13. Derive the Compton shift equation. 14. Explain the wave properties of the particles. 15. How Davison and Germer experiment validated the de Broglie hypothesis? Explain. 15. Explain a wave packet and represent it schematically. 16. Explain (a) group speed (b) phase speed, of a wave packet. 17. Show that the group speed of a wave packet is equal to the particle speed. 18. (a) Name any two phenomena which confirm the particle nature of light. (b) Name any two phenomena which confirm the wave nature of light. 19. Explain Heisenberg uncertainty principle. 20. Write the equations for uncertainty in (a) position and momentum (b) energy and time. 21. Mention two situations which can be well explained by the uncertainty relation. ### Problems 1 Find the peak wavelength of the blackbody radiation emitted by each of the following. A. The human body when the skin temperature is 35°C B. The tungsten filament of a light bulb, which operates at 2000 K C. The Sun, which has a surface temperature of about 5800 Κ. Ans: 9.4 µm, 1.4 µm, 0.50 μm 2 A 2.0- kg block is attached to a spring that has a force constant of k = 25 N/m. The spring is stretched 0.40 m from its equilibrium position and released. A. Find the total energy of the system and the frequency of oscillation according to classical calculations. B. Assuming that the energy is quantized, find the quantum number n for the system oscillating with this amplitude. C. Suppose the oscillator makes a transition from the_n = 5.4 x 1033 state to the state corresponding to n = 5.4 x 1033 – 1. By how much does the energy of the oscillator change in this one-quantum change. Ans: 2.0 J, 0.56 Hz, 5.4 x 1033, 3.7 x 10-34 J 3 The human eye is most sensitive to 560 nm light. What is the temperature of a black body that would radiate most intensely at this wavelength? Ans: 5180 K 4 A blackbody at 7500 K consists of an opening of diameter 0.050 mm, looking into an oven. Find the number of photons per second escaping the hole and having wavelengths between 500 nm and 501 nm. Ans: 1.30 x 1015/s 5 The radius of our Sun is 6.96 x 108 m, and its total power output is 3.77 x 1026 W. (a) Assuming that the Sun's surface emits as a black body, calculate its surface temperature. (b) Using the result, find Amax for the Sun. Ans: 5750 K, 504 nm 6 Calculate the energy in electron volts, of a photon whose frequency is (a) 620 THz, (b) 3.10 GHz, (c) 46.0 MHz. (d) Determine the corresponding wavelengths for these photons and state the classification of each on the electromagnetic spectrum. Ans: 2.57 eV, 1.28 x 10-5 eV, 1.91 x 10-7 eV, 484 nm, 9.68 cm, 6.52 m 7 An FM radio transmitter has a power output of 150 kW and operates at a frequency of 99.7 MHz. How many photons per second does the transmitter emit? Ans: 2.27 x 1030 photons/s 8 A sodium surface is illuminated with light having a wavelength of 300 nm. The work function for sodium metal is 2.46 eV. Find A. The maximum kinetic energy of the ejected photoelectrons and B. The cutoff wavelength for sodium. Ans: 1.67 eV, 504 nm 9 Molybdenum has a work function of 4.2eV. (a) Find the cut off wavelength and cut off frequency for the photoelectric effect. (b) What is the stopping potential if the incident light has wavelength of 180 nm? Ans: 296 nm, 1.01 x 1015 Hz, 2.71 V 10 Electrons are ejected from a metallic surface with speeds up to 4.60 x 105 m/s when light with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cut-off frequency for this surface? Ans: 1.38 eV, 3.34 x 1014 Hz 11 The stopping potential for photoelectrons released from a metal is 1.48 V larger compared to that in another metal. If the threshold frequency for the first metal is 40.0% smaller than for the second metal, determine the work function for each metal. Ans: 3.70 eV, 2.22 eV 12 Two light sources are used in a photoelectric experiment to determine the work function for a metal surface. When green light from a mercury lamp (λ = 546.1 nm) is used, a stopping potential of 0.376 V reduces the photocurrent to zero. (a) Based on this what is the work function of this metal? (b) What stopping potential would be observed when using the yellow light from a helium discharge tube (λ = 587.5 nm)? Ans: 1.90 eV, 0.215 V 13 X-rays of wavelength = 0.20 nm are scattered from a block of material. The scattered X-rays are observed at an angle of 45° to the incident beam. Calculate their wavelength. What if we move the detector so that scattered X-rays are detected at an angle larger than 45°? Does the wavelength of the scattered X-rays increase or decrease as the angle 0 increase? Ans: 0.200710 nm, INCREASES 14 Calculate the energy and momentum of a photon of wavelength 700 nm. Ans: 1.78 eV, 9.47 x 10-28kg.m/s 15 A 0.00160 nm photon scatters from a free electron. For what photon scattering angle does the recoiling electron have kinetic energy equal to the energy of the scattered photon? Ans: 70° 16 A 0.880 MeV photon is scattered by a free electron initially at rest such that the scattering angle of the scattered electron is equal to that of the scattered photon (0 = φ). (a) Determine the angles 0 & φ. (b) Determine the energy and momentum of the scattered electron and photon. Ans: 43°, 43°, 0.602 MeV, 3.21 x 10-22 kg.m/s, 0.278 MeV, 3.21 x 10-22 kg.m/s 17 Calculate the de- Broglie wavelength for an electron moving at 1.0 x 107 m/s. Ans: 7.28 x 10-11 m 18 A rock of mass 50 g is thrown with a speed of 40 m/s. What is its de Broglie wavelength? Ans: 3.3 x 10-34 m 19 A particle of charge q and mass m has been accelerated from rest to a nonrelativistic speed through a potential difference of AV. Find an expression for its de Broglie wavelength. Ans: λ = h √2 m q Av 20 (a) An electron has a kinetic energy of 3.0 eV. Find its wavelength. (b) Also find the wavelength of a photon having the same energy. Ans: 7.09 x 10–10 m, 4.14 x 10-7 m 21 In the Davisson-Germer experiment, 54.0 eV electrons were diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at $= 50.0°, what was the lattice spacing a between the vertical rows of atoms in the figure? Ans: 2.18 x 10-10 m 22 Consider a freely moving quantum particle with mass m and speed u. Its energy is E= K= mu²/2. Determine the phase speed of the quantum wave representing the particle and show that it is different from the speed at which the particle transports mass and energy. Ans: VGROUP = u ≠ VPHASE 23 Electrons are incident on a pair of narrow slits 0.060 µm apart. The 'bright bands' in the interference pattern are separated by 0.40 mm on a 'screen' 20.0 cm from the slits. Determine the potential difference through which the electrons were accelerated to give this pattern. Ans: 105 V 24 The speed of an electron is measured to be 5.00 x 103 m/s to an accuracy of 0.0030%. Find the minimum uncertainty in determining the position of this electron. Ans: 0.383 mm 25 The lifetime of an excited atom is given as 1.0 x 10-8 s. Using the uncertainty principle, compute the line width Af produced by this finite lifetime? Ans: 8.0 x 106 Hz 26 Use the uncertainty principle to show that if an electron were confined inside an atomic nucleus of diameter 2 x 10–15 m, it would have to be moving relativistically, while a proton confined to the same nucleus can be moving nonrelativistically. Ans: VELECTRON ≈ 0.99996 с, VPROTON = 1.8 x 107 m/s 27 Find the minimum kinetic energy of a proton confined within a nucleus having a diameter of 1.0 x 10-15 m. Ans: 5.2 MeV