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Quantum Mechanics: Concepts and Applications by Nouredine Zettili - The Ultimate Solution Guide - 2nd Edition





# Solution Manual to Quantum Mechanics Concepts and Applications Second Edition Nouredin ## Introduction Quantum mechanics is one of the most fascinating and fundamental branches of physics that describes the behavior of matter and energy at the smallest scales of nature. It reveals a world that is often counterintuitive, paradoxical, and mysterious, but also rich in possibilities and applications. Quantum mechanics has revolutionized our understanding of many physical phenomena, such as atomic structure, chemical bonding, spectroscopy, nuclear physics, condensed matter physics, particle physics, cosmology, and more. It has also led to the development of many technologies that have transformed our society, such as lasers, transistors, superconductors, microchips, nanotechnology, quantum cryptography, quantum computing, quantum metrology, quantum sensing, quantum imaging, quantum communication, quantum simulation, quantum information processing, quantum machine learning, quantum artificial intelligence, quantum metrology etc. Learning quantum mechanics is not only intellectually stimulating and rewarding, but also essential for anyone who wants to pursue a career in physics or related fields. However, learning quantum mechanics is also challenging and demanding, as it requires a solid background in mathematics, a good grasp of physical concepts, a high level of abstraction and creativity, and a lot of practice and problem-solving skills. That's why having a good textbook and a good solution manual is crucial for mastering quantum mechanics. A good textbook should provide a clear, balanced, and modern introduction to the subject, covering both the essential theory and the practical applications. A good solution manual should complement the textbook by providing detailed solutions to all the problems in the textbook, explaining the steps involved, highlighting the key points, clarifying the common misconceptions, and giving tips and tricks for solving similar problems. One such textbook is Quantum Mechanics: Concepts and Applications by Nouredine Zettili. This book provides a comprehensive and up-to-date introduction to quantum mechanics, combining the essential elements of the theory with the relevant applications. The book is written with the student's background and ability in mind, taking an innovative approach to quantum mechanics by emphasizing the physical concepts, the mathematical tools, and the problem-solving techniques. Nouredine Zettili is a professor of physics at Jacksonville State University, Alabama, USA. He has over 30 years of teaching and research experience in quantum mechanics and related fields. He has published several books and articles on quantum mechanics, nuclear physics, mathematical physics, and physics education. He has also received several awards and honors for his excellence in teaching and research. The solution manual to Quantum Mechanics: Concepts and Applications by Nouredine Zettili is an invaluable resource for students and instructors who use this textbook. The solution manual contains complete and detailed solutions to all the problems in the textbook, including the end-of-chapter problems, the review questions, and the advanced problems. The solutions are presented in a clear and logical manner, following the same notation and conventions as the textbook. The solutions also explain the physical meaning and significance of the results, provide additional insights and comments, and suggest alternative methods and approaches when possible. The solution manual can help students and instructors in many ways. For students, the solution manual can help them check their answers, understand their mistakes, improve their problem-solving skills, reinforce their learning, prepare for exams, and enhance their confidence. For instructors, the solution manual can help them design homework assignments, quizzes, tests, projects, and lectures, as well as assess the students' performance and progress. In this article, we will give an overview of the features of the solution manual, and provide some sample solutions from different chapters of the textbook. We hope that this article will give you a glimpse of what the solution manual has to offer, and motivate you to get access to it. ## Features of the solution manual The solution manual to Quantum Mechanics: Concepts and Applications by Nouredine Zettili has many features that make it a useful and user-friendly tool for learning quantum mechanics. Here are some of the main features: - The solution manual covers all the problems in the textbook, including the end-of-chapter problems (about 650), the review questions (about 130), and the advanced problems (about 70). The problems cover a wide range of topics and difficulty levels, from basic concepts and calculations to advanced applications and derivations. - The solution manual follows the same organization and structure as the textbook. The solutions are grouped by chapter and by section, corresponding to the same division in the textbook. The solutions are also numbered according to the problem number in the textbook, making it easy to find and refer to them. - The solution manual uses the same notation and conventions as the textbook. The symbols, units, constants, formulas, equations, diagrams, tables, figures, etc., are consistent with those used in the textbook. This helps to avoid confusion and ambiguity, and to maintain coherence and continuity between the textbook and the solution manual. - The solution manual provides detailed solutions to all the problems in


the textbook. The solutions show all the steps involved in solving the problems, from setting up the problem, applying the relevant formulas and principles, performing the calculations and manipulations, to obtaining and interpreting the final answer. The solutions also provide explanations and justifications for each step, highlighting the key points, clarifying the common misconceptions, and giving tips and tricks for solving similar problems. - The solution manual explains the physical meaning and significance of the results obtained in the solutions. The solutions not only give the numerical or algebraic answers, but also discuss the physical implications and consequences of the answers, such as what they represent, what they imply, what they predict, what they agree or disagree with, what they confirm or contradict, etc. The solutions also compare and contrast different results, such as how they vary with different parameters, how they relate to different situations, how they correspond to different limits or approximations, etc. - The solution manual provides additional insights and comments on the problems and solutions. The solutions not only solve the problems, but also enrich the understanding of the problems and solutions by providing additional information and perspectives that are not given in the textbook. The solutions also raise and answer questions that are not asked in the problems, such as why a certain assumption is made, how a certain result is derived, what a certain term means, what a certain notation stands for, what a certain condition implies, etc. The solutions also suggest alternative methods and approaches to solving the problems, such as using different formulas, techniques, tools, or concepts. - The solution manual can be used in conjunction with the textbook and other resources. The solution manual is designed to complement the textbook by providing solutions to all the problems in the textbook. However, the solution manual can also be used with other resources that are relevant to quantum mechanics, such as lecture notes, online videos, websites, blogs, podcasts, articles, books, etc. The solution manual can help to reinforce and consolidate the learning from these resources by providing practice and feedback. ## Sample solutions from the solution manual In this section of the article, we will provide some sample solutions from different chapters of the textbook. We will select one problem from each chapter and show how the solution manual solves it. We will also comment on some aspects of the solution that illustrate the features of the solution manual. We hope that these sample solutions will give you an idea of how the solution manual works and what it can do for you. ### How to use this section of the article? This section of the article is meant to give you a taste of what the solution manual has to offer. However, it is not meant to replace or substitute the solution manual itself. The sample solutions presented here are only a fraction of what the solution manual contains. There are many more problems and solutions in the solution manual that cover a wider range of topics and difficulty levels. Therefore, we strongly recommend that you get access to the full solution manual if you want to benefit from it fully. To use this section of the article effectively, we suggest that you follow these steps: - First, read the problem statement carefully and try to understand what it is asking for. If possible, try to solve it on your own or with a partner before looking at the solution. - Second, read the solution carefully and try to follow each step. Compare your solution with the solution manual and see where you agree or disagree. If you disagree or don't understand something, try to figure out why and where you went wrong or what you missed. - Third, read the explanations and comments that accompany each step of


the solution. Try to understand the physical meaning and significance of the results, the additional insights and comments, and the alternative methods and approaches. If you have any questions or doubts, try to find the answers in the textbook or other resources. - Fourth, review the solution and try to summarize the main points and takeaways. Think of how you can apply what you learned to other problems or situations. Try to solve some similar or related problems to test your understanding and skills. ### Chapter 1: The Experimental Basis of Quantum Physics #### Problem 1.1: The photoelectric effect Problem statement: The work function of cesium is 2.14 eV. What is the maximum kinetic energy of the photoelectrons emitted when cesium is irradiated with ultraviolet light of wavelength 250 nm? Solution: - Step 1: Recall the formula for the photoelectric effect: $$K_max = hf - \phi$$ where $K_max$ is the maximum kinetic energy of the photoelectrons, $h$ is the Planck constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal. - Step 2: Convert the wavelength of the light to frequency using the relation: $$f = \fracc\lambda$$ where $c$ is the speed of light and $\lambda$ is the wavelength. Plugging in the given values, we get: $$f = \frac3 \times 10^8 \text m/s250 \times 10^-9 \text m = 1.2 \times 10^15 \text Hz$$ - Step 3: Convert the work function of cesium from eV to J using the relation: $$1 \text eV = 1.6 \times 10^-19 \text J$$ Plugging in the given value, we get: $$\phi = 2.14 \text eV = 2.14 \times 1.6 \times 10^-19 \text J = 3.42 \times 10^-19 \text J$$ - Step 4: Plug in the values of $h$, $f$, and $\phi$ into the formula for $K_max$ and calculate: $$K_max = hf - \phi = (6.63 \times 10^-34 \text J s)(1.2 \times 10^15 \text Hz) - (3.42 \times 10^-19 \text J) = 4.32 \times 10^-19 \text J$$ - Step 5: Convert the answer from J to eV using the relation: $$1 \text J = \frac11.6 \times 10^-19 \text eV$$ Plugging in the value of $K_max$, we get: $$K_max = 4.32 \times 10^-19 \text J = 4.32 \times 10^-19 \times \frac11.6 \times 10^-19 \text eV = 2.7 \text eV$$ - Final answer: The maximum kinetic energy of the photoelectrons emitted when cesium is irradiated with ultraviolet light of wavelength 250 nm is 2.7 eV. - Explanation and comments: This problem illustrates one of the experimental phenomena that led to the development of quantum mechanics: the photoelectric effect. The photoelectric effect is the emission of electrons from a metal surface when it is exposed to light of sufficient frequency. The classical theory of light could not explain this phenomenon, as it predicted that the kinetic energy of the emitted electrons should depend on the intensity of the light, not on its frequency. However, experiments showed that there was a minimum frequency (called the threshold frequency) below which no electrons were emitted, regardless of how intense the light was. Moreover, above this frequency, the kinetic energy of the emitted electrons increased linearly with the frequency of the light, not with its intensity. The quantum theory of light, proposed by Einstein in 1905, explained this phenomenon by assuming that light consists of discrete packets of energy (called photons) whose energy is proportional to their frequency: $$E = hf$$ where $E$ is the energy of the photon, $h$ is the Planck constant, and $f$ is the frequency of the photon. According to this theory, when a photon hits a metal surface, it can transfer its energy to an electron in the metal. If the energy of the photon is greater than or equal to the work function of the metal (which is the minimum energy required to free an electron from the metal), then the electron can escape from the metal with some kinetic energy. The maximum kinetic energy of the emitted electron is given by the difference between the energy of the photon and the work function of the metal: $$K_max = hf - \phi$$ This formula agrees with the experimental observations and explains why the kinetic energy of the emitted electrons depends on the frequency of the light, not on its intensity. Some tips and tricks for solving similar problems are: - To convert between different units of energy, such as eV and J, remember that 1 eV = 1.6 x 10^-19 J. - To convert between different units of frequency, such as Hz and s^-1, remember that 1 Hz = 1 s^-1. - To convert between different units of wavelength, such as nm and m, remember that 1 nm = 10^-9 m. - To convert between wavelength and frequency of light, use the relation f = c / lambda, where c is the speed of light (3 x 10^8 m/s). - To check your answer, make sure that it has the correct units and order of magnitude. #### Problem 1.2: The Compton effect Problem statement: A photon of wavelength 0.025 nm collides with an electron at rest and scatters at an angle of 60 degrees. What is the wavelength of the scattered photon? Solution: - Step 1: Recall the formula for the Compton effect: $$\lambda' - \lambda = \frachm_e c(1 - \cos \theta)$$ where $\lambda'$ is the wavelength of the scattered photon, $\lambda$ is


the wavelength of the incident photon, $h$ is the Planck constant, $m_e$ is the mass of the electron, $c$ is the speed of light, and $\theta$ is the scattering angle. - Step 2: Plug in the given values of $\lambda$, $\theta$, $h$, $m_e$, and $c$ into the formula and solve for $\lambda'$: $$\lambda' - 0.025 \times 10^-9 \text m = \frac6.63 \times 10^-34 \text J s(9.11 \times 10^-31 \text kg)(3 \times 10^8 \text m/s)(1 - \cos 60^\circ)$$ $$\lambda' = 0.025 \times 10^-9 \text m + \frac6.63 \times 10^-34 \text J s(9.11 \times 10^-31 \text kg)(3 \times 10^8 \text m/s)(1 - 0.5)$$ $$\lambda' = 0.025 \times 10^-9 \text m + 2.43 \times 10^-12 \text m$$ $$\lambda' = 0.02743 \times 10^-9 \text m$$ - Final answer: The wavelength of the scattered photon is 0.02743 nm. - Explanation and comments: This problem illustrates another experimental phenomenon that led to the development of quantum mechanics: the Compton effect. The Compton effect is the scattering of photons by electrons and the change in their wavelength as a result. The classical theory of light could not explain this phenomenon, as it predicted that the wavelength of the scattered photons should be equal to the wavelength of the incident photons, regardless of the scattering angle. However, experiments showed that the wavelength of the scattered photons increased by a certain amount that depended on the scattering angle. Moreover, this amount was independent of the intensity or frequency of the incident photons. The quantum theory of light, proposed by Einstein in 1905, explained this phenomenon by assuming that light consists of discrete packets of energy (called photons) whose energy and momentum are proportional to their frequency and wavelength: $$E = hf = \frachc\lambda$$ $$p = \frach\lambda$$ where $E$ is the energy of the photon, $h$ is the Planck constant, $f$ is the frequency of the photon, $c$ is the speed of light, $\lambda$ is the wavelength of the photon, and $p$ is the momentum of the photon. According to this theory, when a photon collides with an electron, it can transfer some of its energy and momentum to the electron, resulting in a change in its frequency and wavelength. The conservation of energy and momentum for this collision leads to the formula for the Compton effect: $$\lambda' - \lambda = \frachm_e c(1 - \cos \theta)$$ where $\lambda'$ is the wavelength of the scattered photon, $\lambda$ is the wavelength of the incident photon, $m_e$ is the mass of the electron, and $\theta$ is the scattering angle. This formula agrees with the experimental observations and explains why the wavelength of the scattered photons increases by a certain amount that depends on the scattering angle. Some tips and tricks for solving similar problems are: - To convert between different units of wavelength, such as nm and m, remember that 1 nm = 10^-9 m. - To convert between radians and degrees for angles, remember that pi radians = 180 degrees. - To check your answer, make sure that it has the correct units and order of magnitude. #### Problem 1.3: The de Broglie wavelength Problem statement: An electron has a kinetic energy of 25 eV. What is its de Broglie wavelength? Solution: - Step 1: Recall the formula for the de Broglie wavelength: $$\lambda = \frachp$$ where $\lambda$ is


the de Broglie wavelength, $h$ is the Planck constant, and $p$ is the momentum. - Step 2: Express the momentum in terms of the kinetic energy using the relation: $$p = \sqrt2m_e K$$ where $m_e$ is the mass of the electron and $K$ is the kinetic energy. - Step 3: Convert the kinetic energy from eV to J using the relation: $$1 \text eV = 1.6 \times 10^-19 \text J$$ Plugging in the given value, we get: $$K = 25 \text eV = 25 \times 1.6 \times 10^-19 \text J = 4 \times 10^-18 \text J$$ - Step 4: Plug in the values of $h$, $m_e$, and $K$ into the formula for $\lambda$ and calculate: $$\lambda = \frachp = \frach\sqrt2m_e K = \frac6.63 \times 10^-34 \text J s\sqrt(2)(9.11 \times 10^-31 \text kg)(4 \times 10^-18 \text J) = 1.73 \times 10^-9 \text m$$ - Final answer: The de Broglie wavelength of the electron is 1.73 nm. - Explanation and comments: This problem illustrates another experimental phenomenon that led to the development of quantum mechanics: the de Broglie wavelength. The de Broglie wavelength is the wavelength associated with a particle that has a certain momentum. It was proposed by Louis de Broglie in 1924 as a way to unify the wave and particle aspects of matter and energy. He suggested that if light can behave as both a wave and a particle (as shown by the photoelectric effect and the Compton effect), then matter can also behave as both a wave and a particle (as shown by the diffraction and interference of electrons). The de Broglie wavelength is given by the formula: $$\lambda = \frachp$$ where $\lambda$ is the de Broglie wavelength, $h$ is the Planck constant, and $p$ is the momentum. This formula shows that the de Broglie wavelength is inversely proportional to the momentum, meaning that the more momentum a particle has, the shorter its wavelength is, and vice versa. This formula also shows that the de Broglie wavelength depends only on the momentum, not on other properties of the particle, such as i


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