= ( Prove the nonexistence of electron in the nucleus on the basis of uncertainty principle.
|
|
- Abigayle Heath
- 5 years ago
- Views:
Transcription
1 Worked out examples (Quantum mechanics). A microscope, using photons, is employed to locate an electron in an atom within a distance of. Å. What is the uncertainty in the momentum of the electron located in this way? Also find the uncertainty in energy of electron in electron volt for this situation. Solution: Here the uncertainty in position( x) =. Å =. m, Uncertainty in momentum( p) =? We have Heisenberg s uncertainty relation x p ћ. We can take x p ћ Or, p = h x = = kg ms Now uncertainty in energy of the electron= p = (5.7 4 ).5 7 m 9. 3 =.6 9 = 95 ev. Prove the nonexistence of electron in the nucleus on the basis of uncertainty principle. Solution: We have uncertainty relation: x p ћ Or, p ћ, which implies that p must at least equal to x p = ћ = If electron lies inside the nucleus, then uncertainty in position must not greater x x than r. Hence x = r = 4, where r is the radius of the nucleus. p = = 5.7 kg. Kinectic energy of the elctron = p = (5.7 ).5 m 9. 3 =.6 3 = 95 MeV. Thus kinetic energy of the electron must at least equal to 95 MeV. But experiment shows that electrons emitted from β-decay is of the order 3 MeV. Hence it is concluded that electrons do not remain in the nucleus. 3. Find the normalization constant of the energy eigen function of the particle in a box of length L. Solution: See particle in a box. Answer: L 4. Find the commutation relation between position and momentum.
2 Solution: Since x x op and p x p op = iħ x. Let ψ be a well defined wave function, then [xp x p x x]ψ = x( iħ) ψ x ( iħ) x (xψ) = iħx ψ ψ x + iħx + iħψ x x x = iħψ xp x p x x = iħ Or, [x, p x ] = iħ. Similarly [y, p y ] = [z, p z ] = iħ. But [x, p y ] = [z, p y ] = =. It is noted that since [x, p y ] =, it results that x and p y have common eigen functions and can be measured simultaneously with accuracy. Since [x, p x ], it follows that both x and p x can not be measured both simultaneously with accuracy. 5. What are the eigen values of parity operator? Solution: Let ψ be any well behaved wave function and be the parity perator. Let α be the eigen value of the operator. We have eigen value equation: ψ(x) = α ψ(x) (i) Or, ψ(x) = {α ψ(x) } = α ψ(x) = α ψ(x) (ii) We have definition of parity operator : ψ(x) = ψ( x) Or, ψ(x) = { ψ(x)} = {ψ( x)} = ψ(x). (iii) From equation (ii) and (iii) we get:
3 α = α = ±. Hence eigen value of parity operator is + and. If the eigen function ψ(x) has parity +, then it is said to have even parity while for parity -, the eigen function is said to have odd parity. 6. If the state function is given by ψ(x, t) = e i(p x xħ iωt). Find the eigen value of momentum operator. Solution: Given ψ(x, t) = e i(p x xħ iωt). Operating both side by momentum operator p x(= iħ ) we get: x p xψ = iħ x x ei(p xħ iωt) = iħ ip x ħ = p x ψ ψ Comparing with the operator eigen value equation of a operator F op ψ = fψ, we get eigen value of the momentum operator p x is equal to p x. 7. Explain whether energy and momentum can be measured simultaneously. Solution: Consider ψ be the well behaved wavefunction, E x and P x are the energy and momentum operators of the particle, along x axis. Then we have E x = P x is the momentum itself. Now, [P x, E x ]ψ = [P x, P x m ] ψ = m [P x, P x ]ψ = m [P x, P x. P x ]ψ = m [P x (P x, P x ) + P x (P x, P x ) ]ψ = m, since eigen value of momentum operator Since the momentum commutes with energy, so they can be measured simultaneously. 8. If L = r p is the angular momentum operator, find the commutator [L x, L y ]. Solution: See operator for the solution. 9. Write the momentum and time independent energy operators in quantum mechanics. Explain with proper theory, whether these physical quantities can be measured simultaneously. Solution: we have momentum operator is, 3
4 p x = iħ x (in one dimension) and time independent energy operator is H = ħ m d dx + V(x). In the case of free particle, i.e. V(x) =. Hence H = ħ Now, we can prove that [H, p x] = [See operator] Hence these observables can be measured simultaneously with high accuracy.. Calculate the probability density corresponding to ψ(x) = eikx x. Solution: The probability current density J x = ħ im (ψ ψ x ψ ψ ) (i) x d. m dx where, ψ = e ikx x, ψ = x x eikx + x eikx (ik) and ψ = x x e ikx + x e ikx ( ik). Putting these values in equation (i) we get: J x = ħ {e ikx im x ( x eikx + x eikx (ik)) eikx x ( x e ikx + x e ikx ( ik))} = ħ im [( x 3 + ik x ) + ( x 3 + ik x )] = ħ ik im x Or, J x = ħk mx This is the expression of required probability current density for the given wave function.. The wave function ψ(r, θ, φ) of the hydrogen atom for s state is ψ = ( is normalized. Solution: We have given the wave function : ψ = ( The normalized wave function satisfy the condition ψ dv =. (i) In spherical polar coordinates, the volume element: a 3) e r a a 3) e r a. Show that the wave function dv = r sin θ dr dθ dϕ 4
5 Hence equation (i) becomes a3 e r a r dr dφ sin θ dθ = () To solve this equation, put r a = x. Then dx = dr a. e r ( a Then a r dr = e x )3 x dx = ( a )3!, where we have used e x x n dx = n!. Hence left hand side of equation () becomes: a 3 ( a )3! [φ] [ cos θ] = 8 =!. Hence normalizing condition i.e. equation (), is satisfied by the given wave function ψ. Hence ψ is the normalized wave function.. Calculate the average value of /r for an electron in the s state of the hydrogen atom. Solution: The expectation value of /r is given by = r () r ψ dv () The wave function for a s state electron is ψ,, = (a 3 ) Or, ψ = e r a a3 e r a Hence equation () becomes : r = a3 = r and also volume element in spherical coordinates is given by : dv = r sin θ dr dθ dϕ ( ) r e a r dr sin θ dθ dϕ r a3 e a r dr sin θ dθ dϕ = a 3 ( a )! ()() Or, r = a 5
6 3. A particle on a straight line is described by ψ(x) = +ix +ix. Normalize it and explain where this particle is most likely found. Solution: We have given wave function ψ(x) = +ix and +ix ψ (x) = ix ix. Let M be the normalization constant. Then using normalization condition M ψ (x)ψ(x)dx = Or, M Or, M Or, M +x + x Take x x = t +x 4 dx = x + x (x x ) + +x dx = dx = ( + ) dx = dt. Hence x dt M dx =. The integration gives : (t) +( ) M [tan (t/ )] = Or, M ( + ) This gives M = ( ) Then the normalized wave function will be ) +ix. ψ(x) = ( +ix The maximum probability of finding particle at distance x for which dp =, where P is the probability density equal to ψ ψ. Hence d dx ( ψ ψ) = dx Or, d dx (+x +x 4) =. Or, (+x4 )x (+x )4x 3 (+x 4 ) = Using d dx (u v ) = u v uv Or, x =. Hence the particle will have maximum probability of finding at x =. v 4. Find the probability current density of the wave function ψ = e ikx + Ae ikx in the region x < where the potential V=. Hint : Use J x = ħ ψ ψ (ψ ψ ) where ψ = im x x eikx + Ae ikx and ψ = e ikx + Ae +ikx. The final result is obtained as J x = ħk m [ A ] 5. Compare the reflection coefficient with that of transmission coefficient in the case of a free electron moving towards a potential step of height Kev if the energy of the particle is.5 kev. Solution: Energy of the electron E=.5 Kev and that of potential height V = Kev. Hence E > V. In this case, 6
7 Reflection coefficient R = (n n ) m(e V ). Now R T = (n n ) (n +n ) and transmission coefficient T = 4n n (n +n ) where n = me and n = = ( E E V ) 4n n 4 E E V = (.5.5 ) = Normalize the harmonic oscillator wave function given by ψ n = N n H n e μ Solution: ψ n (μ) = N n H n (μ)e μ..() where N n is called a normalization constant. To obtain the value of the constant, we start from the generating function e S +Sμ = H n (μ)s n n= n! Similarly for the index t e t +tμ = H m (μ)t n m= m! Multiplying both the equations (e S +Sμ ). (e t +tμ ) = H n (μ)s n n= H m (μ)t n n! m= m! Multiply both side by e μ and then integrate with respect to μ from to + we get: () Considering only left side (e S +Sμ ). (e t +tμ ) e μ dμ = exp( s t + sμ + tμ μ )dμ + = exp { (s + t sμ tμ + μ )} dμ + = exp [ {(μ s) + t tμ + st st}] dμ = e st exp{ (μ s t) } d(μ s t) = e st. 7
8 Or, (e S +Sμ ). (e t +tμ ) e μ dμ = { + st + (st) + + (st)n + }!! n! = (st) n n= n! Now equation () becomes (3) If equal powers of st are equated in this series: n n! = n! H n! n (μ)e μ dμ, If n = m. H n (μ)e μ dμ = n! n...(#) But, H n (μ)h m (μ)e μ dμ =, if n m. By normalizing condition, ψ n (μ) dμ = N n H n (μ)e μ dμ = N n n! n = N n = ( ) n n! 7. Find the energy expectation value of particle described by the wave function Aexp (iax bt) moving between L and L. Solution: We have wave function ψ = Aexp (iax bt) and ψ = Aexp ( iax bt). The normalization condition is L ψ ψ dx = L L Or, Aexp (iax bt)aexp ( iax bt) dx = L L Or, A e bt dx = L Or, A e bt (L L ) = Or, A = ebt (L L ) Hence wave function ψ = e bt (L L ) exp (iax bt) The energy expectation value of the particle is given by L E x = ψ (iħ ) ψ dx t L 8
9 L = Aexp ( iax bt) (iħ ) t L L L L L Aexp (iax bt) dx = Aexp ( iax bt) Aexp (iax bt)( b) dx = A exp ( bt) ( iħb)dx exp ( bt)( iħb) L dx [ Putting the value of A ] (L L ) L = ebt = iħb (L L ) = iħb (L L ) 8. What is the significance of commutation relation in quantum mechanics? Hint: see operator. 9. Explain the concept of probability density, normalization and expectation value in quantum mechanics.. Show that parity operator commutes with Hamiltonian of a system if the potential V(x) = V( x) Solution: Let H be the Hamiltonian operator and be the parity operator. Then we can write H = ћ m d dx + V(x) Then H ψ(x) = ћ [H ψ(x)] = ћ Similarly, d m dx d m dx d = ћ = ћ H [ ψ(x)] = [ ћ = ћ m m dx d m dx d ψ(x) + V(x)ψ(x) ψ(x) + {V(x)ψ(x)} m dx d ψ( x) ψ( x) + V( x)ψ( x) ψ( x) + V(x)ψ( x).. (i) [ Given V(x) = V( x)] + V(x)] ψ( x) dx From equation (i) and (ii), we get: [H ψ(x)] H [ ψ(x)] = + V(x)ψ( x) (ii) Or, [ H H ] = [H, ] =. Hence the parity operator commutes with Hamiltonian of a system in the given condition V(x) = V( x). An electron of mass 9. 3 kg is constrained to move in a -D potential well of width.5 Angstrom. Compare the energy values of electron in the ground state to second excited states if the walls of the well are impenetrable. Solution: Given mass of an electron m e =9. 3 kg, width of the well L=.5 Å =.5 m. Here energy value of the electron inside the well is E n = h n For the ground state, n =, for the second excited, n = 3. Then the ratio of energy is now E E =. Estimate the de-broglie wavelength of an electron of kinetic energy of kilo electron volt. 8mL h n 8mL h n 3 8mL = n n = 3 9 Hint: Use λ = h p, where h is Planck s constant = joule; momentum p = me with energy E = KeV = 3 ev = joule. The answer is.3867 Å. 3. Under what circumstances is an atomic electron s probability density distribution spherically symmetric? Why? Solution: According to quantum mechanics, the wave function in spherical coordinate system is given by ψ(r, θ, φ) = ψ(r, Θ, φ) = R(r)Θ(Θ)Φ(φ). Then the probability density P(r, θ, φ) = ψψ = R(r)Θ(Θ)Φ(φ)R (r)θ (θ)φ (φ) 9
10 = R(r) Θ(θ) Φ(φ). (i) We have solution of φ equation in hydrogen atom is Φ = e±imφ ( see hydrogen atom) Hence Φ Φ = = constant. Hence equation (i) becomes P(r, θ, φ) = R(r) Θ(θ) Hence the probability density P depends only on the radial probability ( R(r) ) and polar probability ( Θ(θ) ) but not on the azimuthal coordinate ϕ. This results electron distribution is symmetric about z-axis. These numericals are only for practice you should also practice other numricals problems [ ] from Murugeshan and other books of quantum mechanis
Problems and Multiple Choice Questions
Problems and Multiple Choice Questions 1. A momentum operator in one dimension is 2. A position operator in 3 dimensions is 3. A kinetic energy operator in 1 dimension is 4. If two operator commute, a)
More informationQuantum Physics 130A. April 1, 2006
Quantum Physics 130A April 1, 2006 2 1 HOMEWORK 1: Due Friday, Apr. 14 1. A polished silver plate is hit by beams of photons of known energy. It is measured that the maximum electron energy is 3.1 ± 0.11
More information1 Schroenger s Equation for the Hydrogen Atom
Schroenger s Equation for the Hydrogen Atom Here is the Schroedinger equation in D in spherical polar coordinates. Note that the definitions of θ and φ are the exact reverse of what they are in mathematics.
More informationOne-electron Atom. (in spherical coordinates), where Y lm. are spherical harmonics, we arrive at the following Schrödinger equation:
One-electron Atom The atomic orbitals of hydrogen-like atoms are solutions to the Schrödinger equation in a spherically symmetric potential. In this case, the potential term is the potential given by Coulomb's
More informationA few principles of classical and quantum mechanics
A few principles of classical and quantum mechanics The classical approach: In classical mechanics, we usually (but not exclusively) solve Newton s nd law of motion relating the acceleration a of the system
More informationPhysics 228 Today: Ch 41: 1-3: 3D quantum mechanics, hydrogen atom
Physics 228 Today: Ch 41: 1-3: 3D quantum mechanics, hydrogen atom Website: Sakai 01:750:228 or www.physics.rutgers.edu/ugrad/228 Happy April Fools Day Example / Worked Problems What is the ratio of the
More informationSchrödinger equation for the nuclear potential
Schrödinger equation for the nuclear potential Introduction to Nuclear Science Simon Fraser University Spring 2011 NUCS 342 January 24, 2011 NUCS 342 (Lecture 4) January 24, 2011 1 / 32 Outline 1 One-dimensional
More informationPhysics 43 Exam 2 Spring 2018
Physics 43 Exam 2 Spring 2018 Print Name: Conceptual Circle the best answer. (2 points each) 1. Quantum physics agrees with the classical physics limit when a. the total angular momentum is a small multiple
More informationPHYS 3313 Section 001 Lecture # 22
PHYS 3313 Section 001 Lecture # 22 Dr. Barry Spurlock Simple Harmonic Oscillator Barriers and Tunneling Alpha Particle Decay Schrodinger Equation on Hydrogen Atom Solutions for Schrodinger Equation for
More informationPhysics 217 Problem Set 1 Due: Friday, Aug 29th, 2008
Problem Set 1 Due: Friday, Aug 29th, 2008 Course page: http://www.physics.wustl.edu/~alford/p217/ Review of complex numbers. See appendix K of the textbook. 1. Consider complex numbers z = 1.5 + 0.5i and
More informationMathematical Tripos Part IB Michaelmas Term Example Sheet 1. Values of some physical constants are given on the supplementary sheet
Mathematical Tripos Part IB Michaelmas Term 2015 Quantum Mechanics Dr. J.M. Evans Example Sheet 1 Values of some physical constants are given on the supplementary sheet 1. Whenasampleofpotassiumisilluminatedwithlightofwavelength3
More informationUGC ACADEMY LEADING INSTITUE FOR CSIR-JRF/NET, GATE & JAM PH 05 PHYSICAL SCIENCE TEST SERIES # 1. Quantum, Statistical & Thermal Physics
UGC ACADEMY LEADING INSTITUE FOR CSIR-JRF/NET, GATE & JAM BOOKLET CODE SUBJECT CODE PH 05 PHYSICAL SCIENCE TEST SERIES # Quantum, Statistical & Thermal Physics Timing: 3: H M.M: 00 Instructions. This test
More informationChem 452 Mega Practice Exam 1
Last Name: First Name: PSU ID #: Chem 45 Mega Practice Exam 1 Cover Sheet Closed Book, Notes, and NO Calculator The exam will consist of approximately 5 similar questions worth 4 points each. This mega-exam
More informationQuantum Mechanics & Atomic Structure (Chapter 11)
Quantum Mechanics & Atomic Structure (Chapter 11) Quantum mechanics: Microscopic theory of light & matter at molecular scale and smaller. Atoms and radiation (light) have both wave-like and particlelike
More informationWelcome back to PHY 3305
Welcome back to PHY 3305 Today s Lecture: Hydrogen Atom Part I John von Neumann 1903-1957 One-Dimensional Atom To analyze the hydrogen atom, we must solve the Schrodinger equation for the Coulomb potential
More informationQuantum Theory. Thornton and Rex, Ch. 6
Quantum Theory Thornton and Rex, Ch. 6 Matter can behave like waves. 1) What is the wave equation? 2) How do we interpret the wave function y(x,t)? Light Waves Plane wave: y(x,t) = A cos(kx-wt) wave (w,k)
More informationPHYSICS 250 May 4, Final Exam - Solutions
Name: PHYSICS 250 May 4, 999 Final Exam - Solutions Instructions: Work all problems. You may use a calculator and two pages of notes you may have prepared. There are problems of varying length and difficulty.
More information( ) = 9φ 1, ( ) = 4φ 2.
Chemistry 46 Dr Jean M Standard Homework Problem Set 6 Solutions The Hermitian operator A ˆ is associated with the physical observable A Two of the eigenfunctions of A ˆ are and These eigenfunctions are
More informationCHM320 EXAM #2 USEFUL INFORMATION
CHM30 EXAM # USEFUL INFORMATION Constants mass of electron: m e = 9.11 10 31 kg. Rydberg constant: R H = 109737.35 cm 1 =.1798 10 18 J. speed of light: c = 3.00 10 8 m/s Planck constant: 6.66 10 34 Js
More informationH atom solution. 1 Introduction 2. 2 Coordinate system 2. 3 Variable separation 4
H atom solution Contents 1 Introduction 2 2 Coordinate system 2 3 Variable separation 4 4 Wavefunction solutions 6 4.1 Solution for Φ........................... 6 4.2 Solution for Θ...........................
More informationIntroduction to Quantum Mechanics (Prelude to Nuclear Shell Model) Heisenberg Uncertainty Principle In the microscopic world,
Introduction to Quantum Mechanics (Prelude to Nuclear Shell Model) Heisenberg Uncertainty Principle In the microscopic world, x p h π If you try to specify/measure the exact position of a particle you
More informationQuantum Theory. Thornton and Rex, Ch. 6
Quantum Theory Thornton and Rex, Ch. 6 Matter can behave like waves. 1) What is the wave equation? 2) How do we interpret the wave function y(x,t)? Light Waves Plane wave: y(x,t) = A cos(kx-wt) wave (w,k)
More informationCHEM 301: Homework assignment #5
CHEM 30: Homework assignment #5 Solutions. A point mass rotates in a circle with l =. Calculate the magnitude of its angular momentum and all possible projections of the angular momentum on the z-axis.
More informationWelcome back to PHY 3305
Welcome back to PHY 335 Today s Lecture: Hydrogen Atom Pt 1 Sad, quantum surfer, Alone, forlorn on the beach, His wave form collapsed. ThinkGeek.com via Ben Wise AnNouncements Reading Assignment for Nov
More informationChemistry 532 Practice Final Exam Fall 2012 Solutions
Chemistry 53 Practice Final Exam Fall Solutions x e ax dx π a 3/ ; π sin 3 xdx 4 3 π cos nx dx π; sin θ cos θ + K x n e ax dx n! a n+ ; r r r r ˆL h r ˆL z h i φ ˆL x i hsin φ + cot θ cos φ θ φ ) ˆLy i
More informationto the potential V to get V + V 0 0Ψ. Let Ψ ( x,t ) =ψ x dx 2
Physics 0 Homework # Spring 017 Due Wednesday, 4/1/17 1. Griffith s 1.8 We start with by adding V 0 to the potential V to get V + V 0. The Schrödinger equation reads: i! dψ dt =! d Ψ m dx + VΨ + V 0Ψ.
More informationCHAPTER 8 The Quantum Theory of Motion
I. Translational motion. CHAPTER 8 The Quantum Theory of Motion A. Single particle in free space, 1-D. 1. Schrodinger eqn H ψ = Eψ! 2 2m d 2 dx 2 ψ = Eψ ; no boundary conditions 2. General solution: ψ
More informationFinal Exam: Tuesday, May 8, 2012 Starting at 8:30 a.m., Hoyt Hall.
Final Exam: Tuesday, May 8, 2012 Starting at 8:30 a.m., Hoyt Hall. Chapter 38 Quantum Mechanics Units of Chapter 38 38-1 Quantum Mechanics A New Theory 37-2 The Wave Function and Its Interpretation; the
More information8 Wavefunctions - Schrödinger s Equation
8 Wavefunctions - Schrödinger s Equation So far we have considered only free particles - i.e. particles whose energy consists entirely of its kinetic energy. In general, however, a particle moves under
More informationFinal Exam. Tuesday, May 8, Starting at 8:30 a.m., Hoyt Hall.
Final Exam Tuesday, May 8, 2012 Starting at 8:30 a.m., Hoyt Hall. Summary of Chapter 38 In Quantum Mechanics particles are represented by wave functions Ψ. The absolute square of the wave function Ψ 2
More informationAtkins & de Paula: Atkins Physical Chemistry 9e Checklist of key ideas. Chapter 7: Quantum Theory: Introduction and Principles
Atkins & de Paula: Atkins Physical Chemistry 9e Checklist of key ideas Chapter 7: Quantum Theory: Introduction and Principles classical mechanics, the laws of motion introduced in the seventeenth century
More informationThe Hydrogen atom. Chapter The Schrödinger Equation. 2.2 Angular momentum
Chapter 2 The Hydrogen atom In the previous chapter we gave a quick overview of the Bohr model, which is only really valid in the semiclassical limit. cf. section 1.7.) We now begin our task in earnest
More informationPhysics 220. Exam #2. May 23 May 30, 2014
Physics 0 Exam # May 3 May 30, 014 Name Please read and follow these instructions carefully: Read all problems carefully before attempting to solve them. Your work must be legible, with clear organization,
More informationUnderstand the basic principles of spectroscopy using selection rules and the energy levels. Derive Hund s Rule from the symmetrization postulate.
CHEM 5314: Advanced Physical Chemistry Overall Goals: Use quantum mechanics to understand that molecules have quantized translational, rotational, vibrational, and electronic energy levels. In a large
More informationCollection of formulae Quantum mechanics. Basic Formulas Division of Material Science Hans Weber. Operators
Basic Formulas 17-1-1 Division of Material Science Hans Weer The de Broglie wave length λ = h p The Schrödinger equation Hψr,t = i h t ψr,t Stationary states Hψr,t = Eψr,t Collection of formulae Quantum
More informationModern Physics. Unit 6: Hydrogen Atom - Radiation Lecture 6.3: Vector Model of Angular Momentum
Modern Physics Unit 6: Hydrogen Atom - Radiation ecture 6.3: Vector Model of Angular Momentum Ron Reifenberger Professor of Physics Purdue University 1 Summary of Important Points from ast ecture The magnitude
More informationUNIVERSITY OF SURREY FACULTY OF ENGINEERING AND PHYSICAL SCIENCES DEPARTMENT OF PHYSICS. BSc and MPhys Undergraduate Programmes in Physics LEVEL HE2
Phys/Level /1/9/Semester, 009-10 (1 handout) UNIVERSITY OF SURREY FACULTY OF ENGINEERING AND PHYSICAL SCIENCES DEPARTMENT OF PHYSICS BSc and MPhys Undergraduate Programmes in Physics LEVEL HE PAPER 1 MATHEMATICAL,
More informationLecture 6. Four postulates of quantum mechanics. The eigenvalue equation. Momentum and energy operators. Dirac delta function. Expectation values
Lecture 6 Four postulates of quantum mechanics The eigenvalue equation Momentum and energy operators Dirac delta function Expectation values Objectives Learn about eigenvalue equations and operators. Learn
More informationSolution 01. Sut 25268
Solution. Since this is an estimate, more than one solution is possible depending on the approximations made. One solution is given The object of this section is to estimate the the size of an atom. While
More information11 Quantum theory: introduction and principles
Part 2: Structure Quantum theory: introduction and principles Solutions to exercises E.b E.2b E.3b E.4b E.5b E.6b Discussion questions A successful theory of black-body radiation must be able to explain
More informationQUANTUM PHYSICS II. Challenging MCQ questions by The Physics Cafe. Compiled and selected by The Physics Cafe
QUANTUM PHYSICS II Challenging MCQ questions by The Physics Cafe Compiled and selected by The Physics Cafe 1 Suppose Fuzzy, a quantum-mechanical duck of mass 2.00 kg, lives in a world in which h, the Planck
More informationMATH325 - QUANTUM MECHANICS - SOLUTION SHEET 11
MATH35 - QUANTUM MECHANICS - SOLUTION SHEET. The Hamiltonian for a particle of mass m moving in three dimensions under the influence of a three-dimensional harmonic oscillator potential is Ĥ = h m + mω
More informationPhysics 342: Modern Physics
Physics 342: Modern Physics Final Exam (Practice) Relativity: 1) Two LEDs at each end of a meter stick oriented along the x -axis flash simultaneously in their rest frame A. The meter stick is traveling
More informationPhysics 2203, 2011: Equation sheet for second midterm. General properties of Schrödinger s Equation: Quantum Mechanics. Ψ + UΨ = i t.
General properties of Schrödinger s Equation: Quantum Mechanics Schrödinger Equation (time dependent) m Standing wave Ψ(x,t) = Ψ(x)e iωt Schrödinger Equation (time independent) Ψ x m Ψ x Ψ + UΨ = i t +UΨ
More informationChemistry 432 Problem Set 4 Spring 2018 Solutions
Chemistry 4 Problem Set 4 Spring 18 Solutions 1. V I II III a b c A one-dimensional particle of mass m is confined to move under the influence of the potential x a V V (x) = a < x b b x c elsewhere and
More informationElectron in a Box. A wave packet in a square well (an electron in a box) changing with time.
Electron in a Box A wave packet in a square well (an electron in a box) changing with time. Last Time: Light Wave model: Interference pattern is in terms of wave intensity Photon model: Interference in
More informationThe wavefunction ψ for an electron confined to move within a box of linear size L = m, is a standing wave as shown.
1. This question is about quantum aspects of the electron. The wavefunction ψ for an electron confined to move within a box of linear size L = 1.0 10 10 m, is a standing wave as shown. State what is meant
More informationSample Quantum Chemistry Exam 2 Solutions
Chemistry 46 Fall 7 Dr. Jean M. Standard Name SAMPE EXAM Sample Quantum Chemistry Exam Solutions.) ( points) Answer the following questions by selecting the correct answer from the choices provided. a.)
More informationNotes for Special Relativity, Quantum Mechanics, and Nuclear Physics
Notes for Special Relativity, Quantum Mechanics, and Nuclear Physics 1. More on special relativity Normally, when two objects are moving with velocity v and u with respect to the stationary observer, the
More informationChemistry 3502/4502. Exam I. February 6, ) Circle the correct answer on multiple-choice problems.
D Chemistry 3502/4502 Exam I February 6, 2006 1) Circle the correct answer on multiple-choice problems. 2) There is one correct answer to every multiple-choice problem. There is no partial credit. On the
More informationIf electrons moved in simple orbits, p and x could be determined, but this violates the Heisenberg Uncertainty Principle.
CHEM 2060 Lecture 18: Particle in a Box L18-1 Atomic Orbitals If electrons moved in simple orbits, p and x could be determined, but this violates the Heisenberg Uncertainty Principle. We can only talk
More informationλ = ( nm) n2 m 2 n 2 m 2, n > m.
CHAPTER 11 The Need for Quantum Mechanics SECTION 11.1 11.1 Equation (11.3) states λ = (91.176 nm) n m n m, n > m. The energy of a photon of this wavelength is E = hν = hc/λ. Substituting the expression
More informationElectronic Structure of Atoms. Chapter 6
Electronic Structure of Atoms Chapter 6 Electronic Structure of Atoms 1. The Wave Nature of Light All waves have: a) characteristic wavelength, λ b) amplitude, A Electronic Structure of Atoms 1. The Wave
More informationPreliminary Examination - Day 1 Thursday, August 9, 2018
UNL - Department of Physics and Astronomy Preliminary Examination - Day Thursday, August 9, 8 This test covers the topics of Thermodynamics and Statistical Mechanics (Topic ) and Quantum Mechanics (Topic
More informationChemistry 3502/4502. Exam I Key. September 19, ) This is a multiple choice exam. Circle the correct answer.
D Chemistry 350/450 Exam I Key September 19, 003 1) This is a multiple choice exam. Circle the correct answer. ) There is one correct answer to every problem. There is no partial credit. 3) A table of
More informationIntroduction to Quantum Physics and Models of Hydrogen Atom
Introduction to Quantum Physics and Models of Hydrogen Atom Tien-Tsan Shieh Department of Applied Math National Chiao-Tung University November 7, 2012 Physics and Models of Hydrogen November Atom 7, 2012
More informationChapter 6. Quantum Theory of the Hydrogen Atom
Chapter 6 Quantum Theory of the Hydrogen Atom 1 6.1 Schrodinger s Equation for the Hydrogen Atom Symmetry suggests spherical polar coordinates Fig. 6.1 (a) Spherical polar coordinates. (b) A line of constant
More informationFun With Carbon Monoxide. p. 1/2
Fun With Carbon Monoxide p. 1/2 p. 1/2 Fun With Carbon Monoxide E = 0.25 ± 0.05 ev Electron beam results p. 1/2 Fun With Carbon Monoxide E = 0.25 ± 0.05 ev Electron beam results C V (J/K-mole) 35 30 25
More informationIndicate if the statement is True (T) or False (F) by circling the letter (1 pt each):
Indicate if the statement is (T) or False (F) by circling the letter (1 pt each): False 1. In order to ensure that all observables are real valued, the eigenfunctions for an operator must also be real
More information8.05 Quantum Physics II, Fall 2011 FINAL EXAM Thursday December 22, 9:00 am -12:00 You have 3 hours.
8.05 Quantum Physics II, Fall 0 FINAL EXAM Thursday December, 9:00 am -:00 You have 3 hours. Answer all problems in the white books provided. Write YOUR NAME and YOUR SECTION on your white books. There
More information1 r 2 sin 2 θ. This must be the case as we can see by the following argument + L2
PHYS 4 3. The momentum operator in three dimensions is p = i Therefore the momentum-squared operator is [ p 2 = 2 2 = 2 r 2 ) + r 2 r r r 2 sin θ We notice that this can be written as sin θ ) + θ θ r 2
More informationThe 3 dimensional Schrödinger Equation
Chapter 6 The 3 dimensional Schrödinger Equation 6.1 Angular Momentum To study how angular momentum is represented in quantum mechanics we start by reviewing the classical vector of orbital angular momentum
More informationChemistry 3502/4502. Exam I. September 19, ) This is a multiple choice exam. Circle the correct answer.
D Chemistry 350/450 Exam I September 9, 003 ) This is a multiple choice exam. Circle the correct answer. ) There is one correct answer to every problem. There is no partial credit. 3) A table of useful
More informationSTRUCTURE OF MATTER, VIBRATIONS AND WAVES, AND QUANTUM PHYSICS
Imperial College London BSc/MSci EXAMINATION June 2008 This paper is also taken for the relevant Examination for the Associateship STRUCTURE OF MATTER, VIBRATIONS AND WAVES, AND QUANTUM PHYSICS For 1st-Year
More informationChapter 12: Phenomena
Chapter 12: Phenomena K Fe Phenomena: Different wavelengths of electromagnetic radiation were directed onto two different metal sample (see picture). Scientists then recorded if any particles were ejected
More informationProbability and Normalization
Probability and Normalization Although we don t know exactly where the particle might be inside the box, we know that it has to be in the box. This means that, ψ ( x) dx = 1 (normalization condition) L
More informationChapter 38 Quantum Mechanics
Chapter 38 Quantum Mechanics Units of Chapter 38 38-1 Quantum Mechanics A New Theory 37-2 The Wave Function and Its Interpretation; the Double-Slit Experiment 38-3 The Heisenberg Uncertainty Principle
More informationProblem Set 5: Solutions
University of Alabama Department of Physics and Astronomy PH 53 / eclair Spring 1 Problem Set 5: Solutions 1. Solve one of the exam problems that you did not choose.. The Thompson model of the atom. Show
More informationLast Name or Student ID
12/05/18, Chem433 Final Exam answers Last Name or Student ID 1. (2 pts) 12. (3 pts) 2. (6 pts) 13. (3 pts) 3. (3 pts) 14. (2 pts) 4. (3 pts) 15. (3 pts) 5. (4 pts) 16. (3 pts) 6. (2 pts) 17. (15 pts) 7.
More informationPhysics 443, Solutions to PS 1 1
Physics 443, Solutions to PS. Griffiths.9 For Φ(x, t A exp[ a( mx + it], we need that + h Φ(x, t dx. Using the known result of a Gaussian intergral + exp[ ax ]dx /a, we find that: am A h. ( The Schrödinger
More informationThe Schrödinger Equation
Chapter 13 The Schrödinger Equation 13.1 Where we are so far We have focused primarily on electron spin so far because it s a simple quantum system (there are only two basis states!), and yet it still
More informationExamination Radiation Physics - 8N120, 2 November
Examination Radiation Physics - 8N0, November 0-4.00-7.00 Four general remarks: This exam consists of 6 assignments on a total of pages. There is a table on page listing the maximum number of that can
More informationModern Physics for Scientists and Engineers Instructors Solutions. Joseph N. Burchett
Modern Physics for Scientists and Engineers Instructors Solutions Joseph N. Burchett January 1, 011 Introduction - Solutions 1. We are given a spherical distribution of charge and asked to calculate the
More informationTopics Covered: Motion in a central potential, spherical harmonic oscillator, hydrogen atom, orbital electric and magnetic dipole moments
PHYS85 Quantum Mechanics I, Fall 9 HOMEWORK ASSIGNMENT Topics Covered: Motion in a central potential, spherical harmonic oscillator, hydrogen atom, orbital electric and magnetic dipole moments. [ pts]
More informationIV. Electronic Spectroscopy, Angular Momentum, and Magnetic Resonance
IV. Electronic Spectroscopy, Angular Momentum, and Magnetic Resonance The foundation of electronic spectroscopy is the exact solution of the time-independent Schrodinger equation for the hydrogen atom.
More informationBasic Quantum Mechanics
Frederick Lanni 10feb'12 Basic Quantum Mechanics Part I. Where Schrodinger's equation comes from. A. Planck's quantum hypothesis, formulated in 1900, was that exchange of energy between an electromagnetic
More informationSCIENCE VISION INSTITUTE For CSIR NET/JRF, GATE, JEST, TIFR & IIT-JAM Web:
Test Series: CSIR NET/JRF Exam Physical Sciences Test Paper: Quantum Mechanics-I Instructions: 1. Attempt all Questions. Max Marks: 185 2. There is a negative marking of 1/4 for each wrong answer. 3. Each
More informationQuantum Physics Lecture 6
Quantum Physics Lecture 6 Bohr model of hydrogen atom (cont.) Line spectra formula Correspondence principle Quantum Mechanics formalism General properties of waves Expectation values Free particle wavefunction
More informationPhysics 401: Quantum Mechanics I Chapter 4
Physics 401: Quantum Mechanics I Chapter 4 Are you here today? A. Yes B. No C. After than midterm? 3-D Schroedinger Equation The ground state energy of the particle in a 3D box is ( 1 2 +1 2 +1 2 ) π2
More information( ) in the interaction picture arises only
Physics 606, Quantum Mechanics, Final Exam NAME 1 Atomic transitions due to time-dependent electric field Consider a hydrogen atom which is in its ground state for t < 0 For t > 0 it is subjected to a
More informationOpinions on quantum mechanics. CHAPTER 6 Quantum Mechanics II. 6.1: The Schrödinger Wave Equation. Normalization and Probability
CHAPTER 6 Quantum Mechanics II 6.1 The Schrödinger Wave Equation 6. Expectation Values 6.3 Infinite Square-Well Potential 6.4 Finite Square-Well Potential 6.5 Three-Dimensional Infinite- 6.6 Simple Harmonic
More informationCHAPTER 6 Quantum Mechanics II
CHAPTER 6 Quantum Mechanics II 6.1 The Schrödinger Wave Equation 6.2 Expectation Values 6.3 Infinite Square-Well Potential 6.4 Finite Square-Well Potential 6.5 Three-Dimensional Infinite-Potential Well
More informationSemiconductor Physics and Devices
EE321 Fall 2015 September 28, 2015 Semiconductor Physics and Devices Weiwen Zou ( 邹卫文 ) Ph.D., Associate Prof. State Key Lab of advanced optical communication systems and networks, Dept. of Electronic
More informationWaves and the Schroedinger Equation
Waves and the Schroedinger Equation 5 april 010 1 The Wave Equation We have seen from previous discussions that the wave-particle duality of matter requires we describe entities through some wave-form
More informationAngular Momentum. Classically the orbital angular momentum with respect to a fixed origin is. L = r p. = yp z. L x. zp y L y. = zp x. xpz L z.
Angular momentum is an important concept in quantum theory, necessary for analyzing motion in 3D as well as intrinsic properties such as spin Classically the orbital angular momentum with respect to a
More informationIntent. a little geometry of eigenvectors, two-dimensional example. classical analog of QM, briefly
Intent a little geometry of eigenvectors, two-dimensional example classical analog of QM, briefly review of the underpinnings of QM, emphasizing Hilbert space operators with some simplifying assumptions???
More informationPhysics 606, Quantum Mechanics, Final Exam NAME ( ) ( ) + V ( x). ( ) and p( t) be the corresponding operators in ( ) and x( t) : ( ) / dt =...
Physics 606, Quantum Mechanics, Final Exam NAME Please show all your work. (You are graded on your work, with partial credit where it is deserved.) All problems are, of course, nonrelativistic. 1. Consider
More informationChem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 2006 Christopher J. Cramer. Lecture 8, February 3, 2006 & L "
Chem 352/452 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 26 Christopher J. Cramer Lecture 8, February 3, 26 Solved Homework (Homework for grading is also due today) Evaluate
More informationQuantum Mechanics Solutions. λ i λ j v j v j v i v i.
Quantum Mechanics Solutions 1. (a) If H has an orthonormal basis consisting of the eigenvectors { v i } of A with eigenvalues λ i C, then A can be written in terms of its spectral decomposition as A =
More informationAtoms 2012 update -- start with single electron: H-atom
Atoms 2012 update -- start with single electron: H-atom x z φ θ e -1 y 3-D problem - free move in x, y, z - easier if change coord. systems: Cartesian Spherical Coordinate (x, y, z) (r, θ, φ) Reason: V(r)
More information[variable] = units (or dimension) of variable.
Dimensional Analysis Zoe Wyatt wyatt.zoe@gmail.com with help from Emanuel Malek Understanding units usually makes physics much easier to understand. It also gives a good method of checking if an answer
More informationBohr s Correspondence Principle
Bohr s Correspondence Principle In limit that n, quantum mechanics must agree with classical physics E photon = 13.6 ev 1 n f n 1 i = hf photon In this limit, n i n f, and then f photon electron s frequency
More informationLast Name or Student ID
12/05/18, Chem433 Final Exam Last Name or Student ID 1. (2 pts) 12. (3 pts) 2. (6 pts) 13. (3 pts) 3. (3 pts) 14. (2 pts) 4. (3 pts) 15. (3 pts) 5. (4 pts) 16. (3 pts) 6. (2 pts) 17. (15 pts) 7. (9 pts)
More informationH!!!! = E! Lecture 7 - Atomic Structure. Chem 103, Section F0F Unit II - Quantum Theory and Atomic Structure Lecture 7. Lecture 7 - Introduction
Chem 103, Section F0F Unit II - Quantum Theory and Atomic Structure Lecture 7 Lecture 7 - Atomic Structure Reading in Silberberg - Chapter 7, Section 4 The Qunatum-Mechanical Model of the Atom The Quantum
More informationIf you cannot solve the whole problem, write down all relevant equations and explain how you will approach the solution. Show steps clearly.
Letter ID Comprehensive Exam Session I Modern Physics (Including Stat.Mech) Physics Department- Proctor: Dr. Chris Butenhoff (Sat. Jan. 11 th, 2014) (3 hours long 9:00 to 12:00 AM) If you cannot solve
More informationQuiz 6: Modern Physics Solution
Quiz 6: Modern Physics Solution Name: Attempt all questions. Some universal constants: Roll no: h = Planck s constant = 6.63 10 34 Js = Reduced Planck s constant = 1.06 10 34 Js 1eV = 1.6 10 19 J d 2 TDSE
More informationCh. 1: Atoms: The Quantum World
Ch. 1: Atoms: The Quantum World CHEM 4A: General Chemistry with Quantitative Analysis Fall 2009 Instructor: Dr. Orlando E. Raola Santa Rosa Junior College Overview 1.1The nuclear atom 1.2 Characteristics
More informationReading: Mathchapters F and G, MQ - Ch. 7-8, Lecture notes on hydrogen atom.
Chemistry 356 017: Problem set No. 6; Reading: Mathchapters F and G, MQ - Ch. 7-8, Lecture notes on hydrogen atom. The H atom involves spherical coordinates and angular momentum, which leads to the shapes
More information(Refer Slide Time: 1:20) (Refer Slide Time: 1:24 min)
Engineering Chemistry - 1 Prof. K. Mangala Sunder Department of Chemistry Indian Institute of Technology, Madras Lecture - 5 Module 1: Atoms and Molecules Harmonic Oscillator (Continued) (Refer Slide Time:
More informationChemistry 3502/4502. Exam I. February 6, ) Circle the correct answer on multiple-choice problems.
A Chemistry 3502/4502 Exam I February 6, 2006 1) Circle the correct answer on multiple-choice problems. 2) There is one correct answer to every multiple-choice problem. There is no partial credit. On the
More information