Quantum Physics Textbook PDF

```markdown ## 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. ### 2....

```markdown ## 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. ### 2.1 BLACKBODY RADIATION AND PLANCK'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 image shows a spherical object with a small entry point on the bottom. Inside of the object, there are various wavelengths of light being emitted in different direction from the sphere's outer walls and center.* 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. #### 2.1.1 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, $P = \sigma A e T^4$ **2-1** where $P$ is power radiated from the surface of the object (W), $T$ is equilibrium surface temperature (K), $\sigma$ is Stefan-Boltzmann constant (= $5.670 \times 10^{-8} W/m^2K^4$), $A$ is surface area of the object ($m^2$) 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, $\lambda_m T = constant = 2.898 \times 10^{-3} m.K$, or $\lambda_m \propto T^{-1}$ **2-2** where $\lambda_m$ 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.** *The image is a graph with wavelength on the x-axis and intensity on the y-axis. Two curves are shown. One demonstrating high temperature and the other low temperature.* 4. **Rayleigh-Jeans Law:** This law tries to explain the distribution of energy from a black body. The intensity or power per unit area $I (\lambda,T)d\lambda$, emitted in the wavelength interval $\lambda$ to $\lambda + d\lambda$ from a blackbody is given by, $I(\lambda,T) = \frac{2\pi c k_B T}{\lambda^4}$ **2-3** $k_B$ 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.** *The image is graph with wavelength on the x-axis and intensity on the y-axis. Two curves are shown. One demonstrating the Rayleigh-Jeans Law and the other experimental data.* 5. **Planck's Law:** Max Planck developed a theory of blackbody radiation that leads to an equation for $I (\lambda, T)$ 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: $E_n = n h f$ **2-4** 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** *The image shows a diagram that includes energy levels for an oscillator with frequency f. Energy and possible transitions are displayed.* Planck's law explains the distribution of energy from a black body which is given by, $I(\lambda,T) = \frac{2\pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1}$ **2-5** where $I (\lambda,T) d\lambda$ is the intensity or power per unit area emitted in the wavelength interval $d\lambda$ from a blackbody, $h$ is Planck's constant, $k_B$ 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 $(\lambda^-5)$, thus resolving the ultraviolet catastrophe and hence arriving at experimental observation: $I (\lambda, T) \rightarrow 0$ as $\lambda \rightarrow 0$. - For very large $\lambda$, $exp(\frac{hc}{akT}) - 1 = \frac{hc}{\lambda k T} \implies I(\lambda,T) \rightarrow 2\pi c \lambda^{-4} k T$ i.e. $I (\lambda, T) \rightarrow 0$ as $\lambda \rightarrow \infty$. From a fit between Planck's law and experimental data, Planck's constant was derived to be $h = 6.626 \times 10^{-34} J-s$. ### 2.2 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)** *An illustration shows two diagrams. On the left is a schematic diagram of the apparatus and, on the right, a diagram of the circuitry.* #### 2.2.1 Experimental observations **Figure 2.6: Photoelectric current versus applied potential difference for two light intensities** *The diagram depicts a graph with applied voltage on the x-axis and current on the y-axis. Two curves are shown. One demonstrations high intensity and the other low intensity.* 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, $K_{max} = e \Delta V_s$ **2-6** where $\Delta V_s$ 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. #### 2.2.2 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. #### 2.2.3 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: $c = 3 \times 10^8 m/s$ 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, $K_{max} = hf - \phi$ **(Einstein's photoelectric equation)** **2-7** $\phi$ 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 $K_{max}$ 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 $\phi$ in order to eject an electron. 4. The cutoff frequency $f_c$ is related to $\phi$ by $f_c = \phi/h$. If the incident frequency $f$ is less than $f_c$, there is no emission of photoelectrons. The graph of kinetic energy of the most energetic photoelectron $K_{max}$ vs frequency $f$ is a straight line, according to Einstein's equation. **Figure 2.7: A representative plot of $K_{max}$ versus frequency of incident light for three different metals** *The image is a graph that displays the relationship between $K_{max}$ and frequency as linear and proportional for three metals. The curves share a similar slope but have dissimilar y-intercepts.* ### 2.3 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. #### 2.3.1 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 $\lambda_0$ and the other at $\lambda' > \lambda_0$ . The shifted peak at $\lambda'$ is caused by the scattering of X-rays from free electrons. Shift in wavelength was predicted by Compton to depend on scattering angle as $\lambda' - \lambda = \frac{h}{mc}(1 - cos \theta)$ **2-8** 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 $\frac{h}{mc} = 2.43pm$ for electron. **Figure 2.8 Schematic diagram of Compton's apparatus. The wavelength is measured with a rotating crystal spectrometer for various scattering angles $\theta$.** *The image is a schematic diagram of Compton's apparatus, including its labelled components.* **Figure 2.9: Scattered x-ray intensity versus wavelength for Compton scattering at θ=0°, 45°, 90°, and 135° showing single frequency at a given angle** *The image contains four graphs exhibiting the relationships between x-ray intensities versus wavelength for Compton scattering at various angles (0°, 45°, 90° and 135°).* #### 2.3.2 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 $E=hf_0=hc/\lambda_0$, momentum $p=hf/c = h/\lambda$ and zero rest energy. Photons collide elastically with free electrons initially at rest and moving relativistically after collision. Let $\lambda_0$, $p_0=h/\lambda$ and $E_0 = hc/\lambda_0$ be the wavelength, momentum and energy of the incident photon respectively. $\lambda'$, $p' = h/\lambda'$ and $E' = hc/\lambda'$ be the corresponding quantities for the scattered photon. We know that, for the electron, the total relativistic energy $E = \sqrt{ p^2c^2+m^2c^4}$ Kinetic energy $ K = E - mc^2 $ And momentum $p = \gamma m \nu $, where $\gamma = \frac{1}{\sqrt{1 - \frac{\nu^2}{c^2}}}$ **Figure 2.10: Quantum model for X-ray scattering from an electron** *The image presents a drawing illustrating the scattering of an X-ray by an electron, with depictions of the scattered photon, electron at rest, and arrows showing trajectory.* 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_0 = E' + K $ i.e, $E_0 = E' + (E - mc^2) $ Or $ E_0 - E' + mc^2= E = \sqrt{ p^2c^2+m^2c^4}$ Squaring both the sides, $(E_0 - E')^2 + 2(E_0 - E') mc^2 + m^2c^4 = p^2c^2+m^2c^4$ For conservation of momentum, x-component: $ p_0 = p' \cos \theta + p \cos \phi $ y-component: $ 0 = p' \sin \theta - p \sin \phi $ Rewriting these two equations $ p_0 - p' \cos \theta = p \cos \phi $ $ p' \sin \theta = p \sin \phi$ Squaring both the sides and adding, $ p_0^2 - 2p_0p' \cos \theta + p'^2 = p^2$ Substituting this $ p^2$ in the equation : $ (E_0 - E')^2 + 2(E_0 - E') mc^2 = p^2c^2$, one gets $ (E_0 - E')^2 + 2(E_0 - E') mc^2 = (p_0^2 - 2p_0p' \cos \theta + p'^2)c^2$ Substituting photon energies and photon momenta one gets $(\frac{hc}{\lambda_0} - \frac{hc}{\lambda'})^2 + 2(\frac{hc}{\lambda_0} - \frac{hc}{\lambda'}) mc^2 = ((\frac{hc}{\lambda})^2 - 2(\frac{hc}{\lambda})(\frac{hc}{\lambda'})cos \theta + (\frac{hc}{\lambda'})^2)$ Simplifying one gets i.e., $\frac{hc}{\lambda_0 \lambda'} + (\frac{1}{\lambda'_0}- \frac{1}{\lambda})^2 mc^2= -\frac{hc}{\lambda_0 \lambda'} cos \theta$ OR, $(\frac{hc}{\lambda_0^2}) mc^2 = \frac{hc}{\lambda_0 \lambda'}(1 - cos \theta)$ **Compton shift:** $\lambda' - \lambda_0 = \frac{h}{mc}(1 - cos \theta)$ ### 2.4 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. ### 2.5 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 A related to momentum $p$ in the same way as for light. de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{mv}$ **2-9** 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 $\Delta V$, has a non-relativistic kinetic energy $ \frac{1}{2} m v^2 = e \Delta V$ where e is electron charge. Hence, the momentum $(p)$ of an electron accelerated through a potential difference of $\Delta V$ is $p = mv = \sqrt{2me\Delta V}$ **2-10** Frequency of the matter wave associated with the particle is $\frac{E}{h}$, where $E$ is total relativistic energy of the particle #### 2.5.1 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).** *The image exhibits two diagrams: a schematic of the Davisson-Germer experiment and a polar plot of the intensities after scattering.* 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 $\phi$ relative to the incident beam and is shown in the polar plot of $\phi$ verses the intensity ($\propto r$, the radius vector). The most intense reflection of electron beam at an angle $\phi = 50°$, for an accelerating potential 54 V is observed. An electron of mass $m$ accelerated to a potential $V$ has kinetic energy, $\frac{1}{2} m v^2 = e V$ where, $v$ = velocity. And the momentum, $p = m v = \sqrt{2 m e V}$ $\therefore$ de Broglie wavelength, $\lambda = \frac{h}{p} = \frac{h}{mv} = \grave{\grave{}}\frac{h}{\sqrt{2mev}}$ Substituting $V = 54 V$, we get the value for the de Broglie wavelength as $\lambda = 0.167nm$. 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: $d \sin \phi = n \lambda$ Substituting $n = 1$ (for the first order diffraction maximum), $\phi = 50°$(scattering angle corresponding to the first order diffraction maximum), $d = 0.215 nm$ (inter-atomic spacing in nickel), we get $\grave{}\lambda = 0.165nm$ 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 $p = h /\lambda$. 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. ### 2.6 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 2.12. **Figure 2.12: Section of an ideal wave of single frequency** *The image shows the section of a wave that has a single frequency.* Now to build a localized entity from an infinitely long wave, waves of same amplitude, but slightly different frequencies are superposed (Figure 2.13). **Figure 2.13: Superposition of two waves Wave 1 and Wave 2** *The image contains three waves. Wave 1, Wave 2 and the superposition of 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 2.14). **Figure 2.14: Wave packet** *The image depicts one wave packet.* #### 2.6.1 Mathematical representation of a wave packet Superposition of two waves of equal amplitude, but with slightly different frequencies, $f_1$ and $f_2$, traveling in the same direction are considered. The waves are written as $y_1 = A \cos(k_1x - w_1t)$ and $y_2 = A \cos(k_2x - w_2t)$ where $k = 2\pi/\lambda$, $\omega = 2\pi f$ The resultant wave $ y = y_1 + y_2$ $ y = 2A \cos(\frac{\Delta k}{2}x - \frac{\Delta \omega}{2}t) \cos (\frac{k_1+k_2}{2}x - \frac{\omega_1 + \omega_2}{2}t)]$ where $\Delta k = k_1 - k_2$ and $\Delta \omega = \omega_1 - \omega_2 $. **Figure 2.15: Beat pattern due to superposition of wave trains y1 and y2** *The image shows diagram of beat patterns that result from the superposition of of wave trains.* 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 $V_p = f \lambda$ or $V_p = \frac{\omega}{k}$ (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 $V_g = \frac{\Delta \omega}{\Delta k} = \frac{d \omega}{d k}$ (2.12) For a superposition of large number of waves to form a wave packet, this ratio is $V_g = \frac{d \omega}{d k}$. In general these two speeds are not the same. #### 2.6.2 Relation between group speed ($v_g$) and phase speed ($v_p$) $V_p = \frac{\omega}{k} = f \lambda$ ; $\omega = k v_p$ But $V_g = \frac{d \omega}{d k} = \frac{d(kv_p)}{dk} = k \frac{dv_p}{dk} + v_p$ Substituting for $k$ in terms of $\lambda$, we get $ V_g = v_p = \lambda (\frac{dv_p}{d \lambda})$ (2.13) #### 2.6.3 Relation between group speed ($v_g$) and particle speed ($u$) $\omega = 2 \pi f = 2 \pi \frac{E}{h}$ and $k = \frac{2 \pi}{\lambda} = \frac{2 \pi p}{h} = \frac{2 \pi}{h/p}$ $V_g = \frac{d \omega}{d k} = \frac{\frac{2 \pi}{h} dE}{\frac{2 \pi}{h}dp} = \frac{dE}{dp}$ For a classical particle moving with speed u, the kinetic energy $E$ is given by $E = \frac{1}{2}m u^2 = \frac{p^2}{2 m}$ and $ dE = \frac{2 p dp}{2 m}$ or $\frac{dE}{dp} = \frac{p}{m} = u$ $V_g = \frac{d \omega}{d k} = \frac{dE}{dp} = u$ (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. ### 2.7 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, $\lambda$ 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 $\Delta \lambda = 0$, $\Delta p = 0$; but $\Delta x= \infty$ 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 $\Delta x$ is small, but $\Delta \lambda $ is large. If $\Delta x $ is made zero, $\Delta \lambda$ and thereby $\Delta p$ will become $\infty$. In short $(\Delta x) (\Delta p_x) \geq h/4\pi$ (2.15) where $\Delta x$ is uncertainty in the measurement of position x of the particle and $\Delta p_x$ is uncertainty in the measurement of momentum $ p_x$ of the particle. One more relation expressing uncertainty principle is related to energy and time which is given by $(\Delta E) (\Delta t) \geq h/4\pi$ (2.16) where $\Delta E$ is uncertainty in the measurement of energy E of the system when the measurement is done over the time interval $\Delta t$. ### 2.8 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. 16. Explain a wave packet and represent it schematically. 17. Explain (a) group speed (b) phase speed, of a wave packet. 18. Show that the group speed of a wave packet is equal to the particle speed. 19. (a) Name any two phenomena which confirm the particle nature of light. (b) Name any two phenomena which confirm the wave nature of light. 20. Explain Heisenberg uncertainty principle. 21. Write the equations for uncertainty in (a) position and momentum (b) energy and time. 22. Mention two situations which can be well explained by the uncertainty relation. ### 2.9 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 K. Ans: 9.4 $\mu$m, 1.4 $\mu$m, 0.50 $\mu$m 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. 3. Suppose the oscillator makes a transition from the $n = 5.4 \times 10^{33}$ state to the state corresponding to $n = 5.4 \times 10^{33} - 1$. By how much does the energy of the oscillator change in this one-quantum change. Ans: 2.0 J, 0.56 Hz, $5.4 \times 10^{33}$, $3.7 \times 10^{-34}$ J 4. 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 5. 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: $1.30 \times 10^{15}/s$ 6. The radius of our Sun is $6.96 \times 10^8 m$, and its total power output is $3.77 \times 10^{26} W$. A. Assuming that the Sun's surface emits as a black body, calculate its surface temperature. B. Using the result, find $\lambda_{max}$ for the Sun. Ans: 5750 K, 504 nm 7. 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, $1.28 \times 10^{-5} eV$, $1.91 \times 10^{-7} eV$, 484 nm, 9.68 cm, 6.52 m 8. 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: $2.27 \times 10^{30}$ photons/s 9. 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 10. 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, $1.01 \times 10^{15} Hz$, 2.71 V 11. Electrons are ejected from a metallic surface with speeds up to $4.60 \times 10^5 m/s$ 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, $3.34 \times 10^{14}$ Hz 12. 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 13. Two light sources are used in a photoelectric experiment to determine