Engineering Physics PDF Lecture Notes

ENGINEERING PHYSICS Muhammad Faisal, PhD Department of Science and Humanities ENGINEERING PHYSICS UE20PH101 ENGINEERING PHYSICS Electromagnetic wave equations Class #2 Maxwell’s equations in differential form Maxwell’s equations in free space Ideas of electric and magn...

ENGINEERING PHYSICS Muhammad Faisal, PhD Department of Science and Humanities ENGINEERING PHYSICS UE20PH101 ENGINEERING PHYSICS Electromagnetic wave equations Class #2 Maxwell’s equations in differential form Maxwell’s equations in free space Ideas of electric and magnetic waves EM wave as coupled E and B waves ENGINEERING PHYSICS Electromagnetic wave equations Suggested Reading 1. Fundamentals of Physics, Resnik and Halliday, Chapters 34 2. NCERT Physics Book I grade 12 Chapters 1,4,6 Reference Videos 1. https://nptel.ac.in/courses/108/106/108106073/ 2. Engineering physics class #1 MAXWELL’S EQUATIONS - Importance! Goes on….. ENGINEERING PHYSICS Maxwell’s equations - Integral form & Differential form Equations connecting the existing ideas of electric and magnetic fields and their inter-related phenomena………… 𝝆 𝜵. 𝑬 = 𝜺𝒐 𝛁. 𝑩 = 𝟎 𝝏𝑩 𝛁𝐱𝑬 = − 𝝏𝒕 𝝏𝑬 𝛁𝐱𝑩 = 𝝁𝒐 𝒋Ԧ + 𝝁𝒐 𝜺𝒐 𝝏𝒕 𝝏φ 𝝏𝑩.𝐀𝐫𝐞𝐚 𝝏𝑩 𝝏𝐄 Help note: = = , 𝒔𝒊𝒎𝒊𝒍𝒂𝒓𝒍𝒚 𝝏𝒕 𝝏𝒕 𝝏𝒕 𝝏𝒕 ENGINEERING PHYSICS Gauss’s law for electric and magnetic fields For Electric fields Divergence of the electric field is given by the charge density divided by 𝜺𝒐 𝝆 𝜵. 𝑬 = 𝜺𝒐 For Magnetic fields Divergence of the magnetic field is uniformly zero 𝛁. 𝑩 = 𝟎 This implies the absence of magnetic monopoles ENGINEERING PHYSICS Faraday’s law and Ampere-Maxwell’s law Faraday’s law The curl of the induced electric field in a closed loop is proportional to the rate of change of associated magnetic flux with the loop 𝝏𝑩 𝛁𝐱𝑬 = − 𝝏𝒕 Ampere-Maxwell’s law The curl of the magnetic field in a closed loop is equal to the sum of the current density and the displacement current due to the time varying electric field 𝝏𝑬 𝛁𝐱𝑩 = 𝝁𝒐 𝒋Ԧ + 𝝁𝒐 𝜺𝒐 𝝏𝒕 ENGINEERING PHYSICS Maxwell’s equations in free space Summarized by Maxwell (1860)……….. In free space (which does not have sources of charges and currents) 𝛁. 𝑬 = 𝟎 (1) 𝜵. 𝑩 = 𝟎 (2) 𝝏𝑩 𝜵𝒙𝑬 = − (3) 𝝏𝒕 𝝏𝑬 𝜵𝒙𝑩 = +𝝁𝒐 𝜺𝒐 (4) 𝝏𝒕 ENGINEERING PHYSICS Electric and Magnetic waves: Pre-requisites A general wave equation, 𝟏 𝛛𝟐 𝐀 𝛁𝟐 𝐀 = , with 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 = 𝐯 𝐯 𝟐 𝛛𝐭 𝟐 Laplacian operator 𝛛𝟐 𝛛𝟐 𝛛𝟐 𝛁𝟐 = 𝛁. 𝛁 = + + 𝛛𝐱 𝟐 𝛛𝐲 𝟐 𝛛𝐳 𝟐 Vector identity 𝛁 × 𝛁 × 𝐀 = 𝛁 𝛁. 𝐀 − 𝛁𝟐 𝐀 ENGINEERING PHYSICS Wave equation for E vector: Electric waves in free space Taking the curl of Maxwell’s equation 3 𝛛𝐁 𝛁× 𝛁×𝐄 =𝛁× − 𝛛𝐭 𝛛𝛁×𝐁 this reduces to, 𝛁(𝛁. 𝐄) − 𝛁𝟐 𝐄 = − 𝛛𝐭 For free space, 𝛁. 𝐄 = 𝟎 (Maxwell’s equation 1), 𝛛𝛁×𝐁 Thus, −𝛁𝟐 𝐄 = − 𝛛𝐭 Substituting for curl of B (Maxwell’s equation 4) 𝛛 𝟐𝐄 𝛁𝟐 𝐄 = 𝛍𝐨 𝛆𝐨 𝛛𝐭 𝟐 𝟏 with 𝛍𝐨 𝛆𝐨 = 𝟐 , wave equation for electric wave in free 𝐜 𝟏 𝛛𝟐 𝐄 space, 𝛁𝟐 𝐄 = 𝐜 𝟐 𝛛𝐭 𝟐 ENGINEERING PHYSICS Wave equation for B vector: Magnetic waves in free space Taking the curl of Maxwell’s equation 4 𝛛𝐄 𝛁× 𝛁×𝐁 =𝛁× 𝛍𝐨 𝛆𝐨 𝛛𝐭 𝟐 𝛛𝛁×𝐄 this reduces to, 𝛁(𝛁. 𝐁) − 𝛁 𝐁 = 𝛍𝐨 𝛆𝐨 𝛛𝐭 [As per Vector identity 𝛁 × 𝛁 × 𝐀 = 𝛁 𝛁. 𝐀 − 𝛁𝟐 𝐀 ] 𝛛𝐁 For free space, 𝛁. 𝐁 = 𝟎 and 𝛁𝐱𝐄 = − (Maxwell’s equation 3) 𝛛𝐭 𝛛𝟐 𝐁 Applying the above, 𝛁𝟐 𝐁 = 𝛍𝐨 𝛆𝐨 𝟐 𝛛𝐭 The general form of magnetic wave in free space at speed of light, 𝟏 𝛛𝟐 𝐁 𝟏 𝛁𝟐 𝐁 = , with 𝛍𝐨 𝛆𝐨 = 𝐜 𝟐 𝛛𝐭 𝟐 𝐜𝟐 ENGINEERING PHYSICS Electromagnetic wave equations Maxwell’s Conclusion: 1. Both electric and magnetic waves propagate with 𝟏 speed of light 𝐜 = 𝛍𝐨 𝛆𝐨 2. Light waves (radiation) as electromagnetic waves 3. Light waves are transverse waves and electromagnetic waves are transverse in nature 4. Electric and magnetic fields are mutually perpendicular and perpendicular to the direction of propagation ENGINEERING PHYSICS Propagation of a 1D transverse wave 𝑬𝒚 = 𝑬𝒐𝒚 𝒄𝒐𝒔 𝝎𝒕 + 𝒌𝒛 𝒐𝒓𝑬𝒐𝒚 𝒔𝒊𝒏 𝝎𝒕 + 𝒌𝒛 ENGINEERING PHYSICS Electromagnetic waves in free space Analysis: E and B are mutually perpendicular to each other Consider a 1D electric wave 𝑬𝒙 associated with EM radiation propagating in the Z direction as, 𝑬𝒙 = 𝑬𝒐𝒙 𝒄𝒐𝒔 𝝎𝒕 + 𝒌𝒛 𝒐𝒓𝑬𝒐𝒙 𝒔𝒊𝒏 𝝎𝒕 + 𝒌𝒛 The electric field vector has only x component and other two components Ey and Ez are zero ENGINEERING PHYSICS Electromagnetic waves in free space The associated magnetic component of the EM wave is evaluated as, 𝛛𝐁 Using Maxwell’s third equation, 𝛁𝐱𝐄 = − 𝛛𝐭 𝐢Ƹ 𝐣Ƹ መ 𝐤 𝛛 𝛛 𝛛 Evaluating curl of the electric field 𝛁𝐱𝐄 = 𝛛𝐱 𝛛𝐲 𝛛𝐳 𝐄𝐱 𝟎 𝟎 𝛛𝐄𝐱 መ ∗ 𝟎 = 𝐣Ƹ 𝛛 = 𝐢Ƹ × 𝟎 + 𝐣Ƹ ∗ +𝐤 𝑬𝒐𝒙 𝐜𝐨𝐬 𝛚𝐭 + 𝐤𝐳 𝛛𝐳 𝛛𝐳 = 𝐣Ƹ ∗ 𝐤 ∗ 𝑬𝒐𝒙 𝐬𝐢𝐧(𝛚𝐭 + 𝐤𝐳) 𝛛𝐁 Thus, − = 𝐣Ƹ ∗ 𝐤 ∗ 𝑬𝒐𝒙 𝐬𝐢𝐧(𝛚𝐭 + 𝐤𝐳) 𝛛𝐭 ENGINEERING PHYSICS Electromagnetic waves in free space 𝛛𝐁 Integrating − with respect to time gives magnetic component, 𝛛𝐭 𝟏 𝟏 𝛛𝐁 𝐁 = 𝐣Ƹ ∗ 𝛚 ∗ 𝑬𝒐𝒙 𝐜𝐨𝐬 𝛚𝐭 + 𝐤𝐳 = 𝐣.Ƹ 𝐄𝐱 ∗ − 𝛛𝐭 = 𝐣Ƹ ∗ 𝐤 ∗ 𝑬𝒐𝒙 𝐬𝐢𝐧(𝛚𝐭 + 𝐤𝐳) 𝐜 𝐤 𝛚 (𝐜 = , is the velocity of the radiation) ∫ sin x dx = -cos x 𝐤 Thus, Ƹ 𝒐𝒙 𝒄𝒐𝒔 𝝎𝒕 + 𝒌𝒛 𝑬𝒙 = 𝒊𝑬 𝟏 𝑩𝒚 = 𝒋Ƹ 𝛚 ∗ 𝑬𝒐𝒙 𝒄𝒐𝒔 𝝎𝒕 + 𝒌𝒛 𝐤 ENGINEERING PHYSICS Electromagnetic waves in free space Conclusion: Magnetic field (B) of the EM wave is Y component In phase with the E field variations 𝛚 Phase velocity of the wave, 𝐜 = 𝐤 𝟏 Magnitude of B wave is times the magnitude of the E wave 𝒄  ENGINEERING PHYSICS Electromagnetic waves in free space EM waves have coupled E and B field components which are mutually perpendicular Both E and B are perpendicular to the direction of propagation Image courtesy: ResearchGate  ENGINEERING PHYSICS Electromagnetic waves in free space Practical Observation: Heat from the sun can travel to the earth and humans can send any type of signal via radio waves ! Electric and Magnetic Fields in "Free Space" - a region without charges or currents (like air) - can travel with a single speed - c One of the greatest discovery, and one of the unique properties that the universe exhibit! ENGINEERING PHYSICS Class 2. Quiz … The concepts which are not correct are…. 1. Electric waves in free space are longitudinal 2. Magnetic waves in free space are transverse 3. The curl of a magnetic field is uniformly zero 4. The divergence of a magnetic field can be non zero 5. The curl of an electric field is always linked to a time varying magnetic field 6. The divergence of a vector field is a scalar 7. A plane electric wave propagating along z direction has both x and y components 8. Electric and magnetic waves are always out of phase THANK YOU Muhammad Faisal, Ph.D. Associate Professor, Department of Science and Humanities [email protected] +91 80 50829629 ENGINEERING PHYSICS Muhammad Faisal, PhD Department of Science and Humanities ENGINEERING PHYSICS Energy transported by EM waves Class #3 Energy in an electric field Energy in a magnetic field Energy transported by Electric and Magnetic waves Total Energy of the EM wave Poynting Vector and average energy transported Polarization of EM waves ENGINEERING PHYSICS Energy transported by EM waves Suggested Reading 1. Fundamentals of Physics, Resnik and Halliday, Chapters 34 2. NCERT Physics Book I grade 12 Chapters 8 Reference Videos 1. https://nptel.ac.in/courses/108/106/108106073/ 2. UE20PH101_week1_class 1,2 ENGINEERING PHYSICS Energy in an electric field 𝟏 Energy per unit volume = 𝛆𝟎 𝐄 𝟐 𝟐 The energy per unit volume in an electric field is dependent only on the strength of the field ! Help note: A capacitor stores energy in the form of electric field ENGINEERING PHYSICS Energy in a magnetic field This energy stored per unit volume, 𝟏 𝐁𝟐 = 𝟐 𝛍𝟎 The energy per unit volume in a magnetic field is also dependent only on the strength of the field ! Help note: An inductor stores energy in the form of magnetic field ENGINEERING PHYSICS Energy of EM waves Energy content of the electric component 𝟏 𝟏 = 𝛆𝐨 𝐄𝐱 = 𝛆𝐨 𝐄𝐨𝐱 𝟐 𝐜𝐨𝐬 𝟐 (𝛚𝐭 + 𝐤𝐳) 𝟐 𝟐 𝟐 𝟐 𝟏 𝐁𝐲 Energy content of the magnetic component = 𝟐 𝛍𝐨 𝟐 𝟏 𝟐 𝟏 𝐁𝐲 Total energy content of the EM wave = 𝛆 𝐄 + 𝟐 𝐨 𝐱 𝟐 𝛍𝐨 1 𝟏 𝐄𝐱 𝟐 𝟏 𝟏 𝟏 = 𝛆𝐨 𝐄𝐱 𝟐 + 𝟐 [Since, 𝐁𝒚 = 𝐄𝐱 ∗ and 𝐜 = 𝒐𝒓 𝝁𝒐 = ] 2 𝟐 𝐜 𝛍𝐨 𝐜 𝛍𝐨 𝛆𝐨 𝒄𝟐 𝜺𝒐 = 𝛆𝐨 𝐄𝐱 𝟐 , transported in the z-direction Important: Classically the energy of waves is equivalent to it’s intensity (square of the amplitude)! ENGINEERING PHYSICS Average energy of EM waves The average energy of the EM wave transmitted 𝐜𝛆𝐨 Energy transported in one cycle, Total energy/cycle = 𝐄𝐱 𝟐 𝐓 𝐜𝛆𝐨 𝐓 𝟐 < 𝑬𝒏𝒆𝒓𝒈𝒚 > = න 𝐄𝐱 𝐝𝐭 𝐓 𝟎 𝐜𝛆𝐨 𝐓 𝟐 = න 𝐄𝐨𝐱 𝐜𝐨𝐬 𝟐 𝛚𝐭 + 𝐤𝐳 𝐝𝐭 𝟏 𝟏 𝟏 𝐓 𝟎 [Since, 𝐁𝒚 = 𝐄𝐱 ∗ and 𝐜 = 𝒐𝒓 𝝁𝒐 = ] 𝐜 𝛍𝐨 𝛆𝐨 𝒄𝟐 𝜺𝒐 𝟐 𝟏 𝟐 𝟏 𝐁 𝐨𝐲 𝟏 𝐄𝐨𝐱 𝐁𝐨𝐲 = 𝛆𝐨 𝐜𝐄𝐨𝐱 = 𝐜 = 𝟐 𝟐 𝛍𝐨 𝟐 𝛍𝐨 Total energy contained in a box = 𝛆𝐨 𝐄𝐱 𝟐 ∗ 𝒅𝑨 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒃𝒐𝒙 𝒂𝒓𝒆𝒂 𝒙 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 = 𝒅𝑨 𝒙 𝒄. 𝒅𝒕 = 𝐜 𝛆𝐨 𝐄𝐱 𝟐 𝒄. 𝒅𝒕  ENGINEERING PHYSICS Poynting vector Ԧ describes the EM energy Poynting vector (𝐒) transported per unit time per unit volume 𝟏 𝐒Ԧ = 𝐄 × 𝐁 = 𝐜 𝟐 𝛆𝐨 𝐄 × 𝐁 𝛍𝐨 𝟏 𝟏 Ԧ is the direction of 𝐜= 𝒐𝒓 𝝁𝒐 = 𝟐 Direction of Poynting vector (𝐒) 𝛍𝐨 𝛆 𝐨 𝒄 𝜺𝒐 propagation of EM waves Its time dependent (magnitude varies in time) ENGINEERING PHYSICS Polarization of electromagnetic waves Natural light is generally unpolarized, all planes of propagation being equally probable  ENGINEERING PHYSICS Polarization of electromagnetic waves A plane wave is called linearly polarized. The addition of a horizontally and vertically linearly polarized waves of the same amplitude in the same phase also result in a linearly polarized at a 45o angle If light is composed of two plane waves of equal amplitude but differing in phase by 90°, then the light is said to be circularly polarized If two plane waves of differing amplitude are related in phase by 90°, or if the relative phase is other than 90° then the light is said to be Visualization of Circular elliptically polarized and Elliptical ENGINEERING PHYSICS Class 3. Quiz ……. The concepts which apply to electromagnetic waves…. 1. Energy of electric wave is proportional to the amplitude the wave 2. Energy of magnetic longitudinal wave is proportional to the square of the amplitude 3. Total energy of the EM wave is dependent only on the electric wave 4. Total energy of the EM wave cannot be indicated in terms of magnetic field 5. Average energy of EM wave is equal to the energy transported in one cycle 6. Direction of Poynting vector is along the amplitude variation of the electric wave 7. A linearly polarized wave can only be a plane wave with restricted Y component 8. Two waves out of phase by 90o and unequal amplitude form a circularly polarized wave THANK YOU Muhammad Faisal, Ph.D. Associate Professor, Department of Science and Humanities [email protected] +91 80 50829629 ENGINEERING PHYSICS Department of Science and Humanities A numerical to review the whole concept of EM waves The electric field associated with an EM radiation (light) is given by, 𝑬 𝒙, 𝒕 = 𝟏𝟎𝟑 𝐜𝐨𝐬(𝝎𝒕 − 𝝅𝒙𝟑𝒙𝟏𝟎𝟔 𝒛) Evaluate 1. Speed of the Electric vector 2. Wavelength 3. Frequency 4. Period of the wave 5. Magnetic field associated with the wave 6. Direction of propagation of the magnetic transverse wave 7. Amplitude of the electric field vector 8. Amplitude and direction of the transverse magnetic wave 9. Peak energy of the electric wave 10. Maximum instantaneous energy of the magnetic wave 11. Maximum energy of the EM wave 12. Average energy of the EM wave 13. What are the expected changes in the above parameters if the electric wave vector changes to 𝑬 𝐱, 𝒕 = 𝟏𝟎𝟑 𝐬𝐢𝐧 (𝝎𝒕 − 𝝅. 𝟑𝒙𝟏𝟎𝟔 𝒛) 14. What are the expected changes if the electric wave vector changes to 𝑬 𝒚, 𝒕 = 𝟏𝟎𝟑 𝐬𝐢𝐧 (𝝎𝒕 − 𝝅. 𝟑𝒙𝟏𝟎𝟔 𝐱) A numerical to review the whole concept of EM waves The electric field associated with an EM radiation (light) is given by, 𝑬 𝒙, 𝒕 = 𝟏𝟎𝟑 𝐜𝐨𝐬(𝝎𝒕 − 𝝅𝒙𝟑𝒙𝟏𝟎𝟔 𝒛) 1. Speed of the Electric vector Speed of the electric vector = speed of light = 3x108 m/s 2. Wavelength Wavelength = 𝜆 = 2𝜋 2𝜋 , 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑎𝑔𝑎𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦, 𝑘 = 𝑘 𝜆 2𝜋 2𝑥3.14 On substitution, 𝜆 = = = 6666 𝐴𝑜 𝑘 3𝑥106 𝑥𝜋 3. Frequency 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 3𝑥108 𝑚/𝑠 Frequency, 𝜈 = 𝑏𝑎𝑠𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 = = 4.5 𝑥 1014 𝑠 −1 𝑜𝑟 𝐻𝑧 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ 6666 𝑥10−10 𝑚 4. Period of the wave 1 Period of the wave, 𝑇 = = 2.2 𝑥 10−15 𝑠 𝜈  A numerical to review the whole concept of EM waves The electric field associated with an EM radiation (light) is given by, 𝑬 𝒙, 𝒕 = 𝟏𝟎𝟑 𝐜𝐨𝐬(𝝎𝒕 − 𝝅𝒙𝟑𝒙𝟏𝟎𝟔 𝒛) 5. Magnetic field associated with the wave Magnetic field associated with the wave, 𝟏 From the analysis, 𝐁𝒚 = 𝐄𝐱 ∗ (For an EM wave with E-component along x axis , 𝐜 and propagating along z axis, the corresponding magnetic wave must be along y axis with same direction of propagation of E-wave) Thus, Bx=0 and Bz=o (x-axis for E wave, z-axis for common direction of propagation) B=By propagating along z axis 𝐸(𝑥, 𝑡) 103 cos(ωt − 𝜋𝑥3𝑥106 𝑧) −6 cos(ωt − 𝜋𝑥3𝑥106 𝑧) 𝐵 𝑦, 𝑡 = = = 3.33𝑥10 𝑐 3𝑥108 6. Direction of propagation of the magnetic transverse wave Same as direction of propagation of E-wave  A numerical to review the whole concept of EM waves The electric field associated with an EM radiation (light) is given by, 𝑬 𝒙, 𝒕 = 𝟏𝟎𝟑 𝐜𝐨𝐬(𝝎𝒕 − 𝝅𝒙𝟑𝒙𝟏𝟎𝟔 𝒛) 7. Amplitude of the electric field vector WKT, general representation of an E or B field vector as, 𝐸 𝑥, 𝑡 = 𝐸0𝑥 cos (ωt ± kz) or 𝐵 𝑥, 𝑡 = 𝐵0𝑥 sin(ωt ± kz), where E0x or Box is the amplitude Thus, for E wave𝐸 𝑥, 𝑡 = 103 cos(ωt − 𝜋𝑥3𝑥106 𝑧), Amplitude is 103 8. Amplitude and direction of the transverse magnetic wave For magnetic wave 𝐵 𝑦, 𝑡 = 3.33𝑥10−6 cos(ωt − 𝜋𝑥3𝑥106 𝑧) Amplitude is −6 3 3.33𝑥10 and direction of propagation is along z-axis (same as for E wave)  A numerical to review the whole concept of EM waves The electric field associated with an EM radiation (light) is given by, 𝑬 𝒙, 𝒕 = 𝟏𝟎𝟑 𝐜𝐨𝐬(𝝎𝒕 − 𝝅𝒙𝟑𝒙𝟏𝟎𝟔 𝒛) 9. Peak energy of the electric wave Energy content of the electric component 𝟏 = 𝛆𝐨 𝐄 𝐱 𝟐 , for peak energy Ex = Amplitude = 103 𝟐 𝛆𝐨 = 𝟖. 𝟖𝟓𝒙𝟏𝟎−𝟏𝟐 𝐀𝐧𝐬𝐰𝐞𝐫 = 𝟒. 𝟒𝟐𝟓𝐱𝟏𝟎−𝟔 10. Maximum instantaneous energy of the magnetic wave Maximum instantaneous energy of the magnetic wave Maximum instantaneous energy corresponding that at the amplitude 𝟐 𝟏 𝐁𝐲 Energy content of the magnetic component = , where By=Boy=3.33𝑥10−6 𝟐 𝛍𝐨 𝛍𝐨 = 𝟒𝝅𝒙𝟏𝟎−𝟕 , 𝐀𝐧𝐬𝐰𝐞𝐫 = 𝐬𝐚𝐦𝐞 𝐚𝐬 𝐟𝐨𝐫 𝐄 𝐰𝐚𝐯𝐞! 11. Maximum energy of the EM wave Maximum energy of the EM wave will be sum of the maximum energy of E wave and maximum 𝟐 𝟏 𝟐 𝟏 𝐁𝐲 energy of B wave, Total energy content of the EM wave = 𝛆 𝐄 + 𝟐 𝐨 𝐱 𝟐 𝛍𝐨 A numerical to review the whole concept of EM waves The electric field associated with an EM radiation (light) is given by, 𝑬 𝒙, 𝒕 = 𝟏𝟎𝟑 𝐜𝐨𝐬(𝝎𝒕 − 𝝅𝒙𝟑𝒙𝟏𝟎𝟔 𝒛) 12. Average energy of the EM wave Average energy of the EM wave 𝟏 𝟐 𝟏 𝐁𝐨𝐲 𝟐 𝟏 𝐄𝐨𝐱 𝐁𝐨𝐲 = 𝛆𝐨 𝐜𝐄𝐨𝐱 = 𝐜 = 𝟐 𝟐 𝛍𝐨 𝟐 𝛍𝐨 13. What are the expected changes in the above parameters if the electric wave vector changes to 𝑬 𝐱, 𝒕 = 𝟏𝟎𝟑 𝐬𝐢𝐧 (𝝎𝒕 − 𝝅. 𝟑𝒙𝟏𝟎𝟔 𝒛) No change! 14. What are the expected changes if the electric wave vector changes to 𝑬 𝒚, 𝒕 = 𝟏𝟎𝟑 𝐬𝐢𝐧 (𝝎𝒕 − 𝝅. 𝟑𝒙𝟏𝟎𝟔 𝐱) All the magnitude remains the same, and only the direction changes! ENGINEERING PHYSICS Black-body radiation Class #4 Overview of failure of EM wave theory Black body radiation Cavity Oscillators Classical estimation of energy density Max Planck’s estimation of energy density  ENGINEERING PHYSICS Overview of failure of classical EM wave theory EM Radiation (e.g. Radio waves, microwaves, infrared, visible light, ultraviolet, x-rays and gamma radiation) - Described as mutually perpendicular sinusoidal electric and magnetic fields and perpendicular to the direction of propagation of the waves Classical wave theory - Assumed that energy content of the wave is proportional to the square of the amplitude of the waves (wavelength/frequency independence on energy!) Wave theory successfully explains the phenomena of reflection, refraction, interference, diffraction and polarization of light  ENGINEERING PHYSICS Overview of failure of classical EM wave theory Classical wave theory could not explain many observed phenomena 1. Photo-electric Effect 2. Spectrum of Hydrogen Emissions (Atomic Spectra) 3. Black-Body Radiation Spectrum 4. Compton Scattering Resulted in the birth and rise of Quantum Mechanics! Our focus: Black-Body Radiation and Compton Scattering  ENGINEERING PHYSICS Black-body radiation Classically the interaction of radiation with matter (by absorption and emission) gives the color of the material Gustav Robert Kirchhoff found materials which absorb all incident rays Such a material on heating would emit all wavelengths of radiation absorbed Black-body (not necessarily black!) Absorbs all radiations falling on it Emits all wavelengths (frequencies) as it absorbed Emissivity is unity  ENGINEERING PHYSICS Black-body radiation spectrum ▪ Radiation depends only on the temperature of the object, and not on what it is made of (material independent) ▪ all emit the same blackbody spectrum if their temperatures are the same ▪ As the temperature increases, emits more energy (Area under curve) ▪ As the temperature of object increases, the peak wavelength becomes shorter (higher frequency) (Blue stars are hotter than red stars!) ENGINEERING PHYSICS Black-body radiation spectrum Radiation depends only on the temperature of the object, and not on what it is made of (a metal block, a ceramic vase, and a piece of charcoal, etc. all emit the same blackbody spectrum if their temperatures are the same.) ENGINEERING PHYSICS Black-body radiation spectrum (Frequency vs intensity) Blackbody Radiation – Interesting reality! Related to Our experience on seeing objects- interaction of light with objects, color of objects, color of emitting objects (sun, lava, metal bar, star, etc) Generally, any object with a temperature above absolute zero emits light. (Human Body in the IR region!) If the object is perfectly black (it doesn't reflect any light), then the light that comes from it is called blackbody radiation. The energy of blackbody radiation is not shared evenly by all wavelengths of light.  ENGINEERING PHYSICS Black-body model (Cavity oscillators) ▪ Practically modeled as a cavity - not allowing any incident radiation to escape due to multiple reflections inside ▪ This cavity when heated, emit radiation of every possible frequency and rate of emission increases with temperature Multiple reflections of EM energy inside the cavity Image courtesy: https://chem.libretexts.org ENGINEERING PHYSICS Black-body radiation spectrum How to understand this spectrum? Analysis by Rayleigh-Jeans To study the energy density of radiation, 𝝆 𝝂 𝒅𝝂 = 𝑬 𝒅𝑵 , between 𝝂 𝒂𝒏𝒅 𝝂 + 𝜹𝝂  ENGINEERING PHYSICS Classical estimation of energy density Analysis by Rayleigh-Jeans To understand the energy density of radiation - Assuming black-body as cavity oscillators (trapped oscillations of EM energy) The number of oscillators with frequencies between 𝝂 𝒂𝒏𝒅 𝝂 + 𝜹𝝂 𝟖𝝅 𝟐 is calculated as 𝒅𝑵 = 𝝂 𝒅𝝂 𝒄𝟑 Rayleigh and Jeans showed that the number of modes was proportional to 𝝂𝟐 ENGINEERING PHYSICS Classical estimation of energy density Rayleigh and Jeans considered the average energy of the oscillators as per Maxwell-Boltzmann distribution law 𝐚𝐬 𝑬 = 𝒌𝑩 𝑻 Thus, expression for the energy density (energy per unit volume) of radiations with frequencies between 𝝂 𝒂𝒏𝒅 𝝂 + 𝜹𝝂 as 𝟖𝝅 𝟐 𝝆 𝝂 𝒅𝝂 = 𝑬 𝒅𝑵 = 𝟑 𝝂 𝒅𝝂𝑘𝐵 𝑇 𝒄 This is the Rayleigh Jeans law which is in contradiction with the experimental observations  ENGINEERING PHYSICS Failure of Rayleigh-Jeans’ law Treating EM waves as classical oscillators failed to explain the experimental observations (intensity of radiations were found decrease with increase in frequency - termed as ultra-violet catastrophe) Failure in high frequency region (shorter wavelength) Image courtesy: hyperphysics  ENGINEERING PHYSICS Max Planck’s analysis – Quantum theory of radiation Solution to the failure of classical approach! Max Planck (quantum theory of radiation, 1900) This theory proposed that the energy of the oscillator model of a black body (cavity oscillator) are restricted to multiples of a fundamental natural frequency 𝝂 times a constant (𝒉 = 𝟔. 𝟔𝒙𝟏𝟎−𝟑𝟒 𝑱𝒔) ie.,𝑬 = 𝒏𝒉𝝂 Thus black body radiations are from a collection of harmonic oscillators of different frequencies and the energy of the radiations has to be packets of 𝒉𝝂 With this concept, the average energy of the oscillators were −𝒉𝒗ൗ 𝒉𝝂∗𝒆 𝒌𝑻 𝒉𝝂 evaluated as, 𝑬 = −𝒉𝒗ൗ = 𝒉𝒗 𝟏− 𝒆 𝒌𝑻 𝒆 ൗ𝒌𝑻 −𝟏  ENGINEERING PHYSICS Max Planck’s analysis – Quantum theory of radiation Thus, the energy density of radiations 𝝆 𝝂 𝒅𝝂 = 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒅𝒆𝒔 𝒙 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝒆𝒏𝒆𝒓𝒈𝒚 𝟖𝝅 𝟐 𝒉𝝂 𝟖𝝅𝒉𝝂𝟑 𝟏 𝑬 𝒅𝑵 = 𝟑 𝝂 𝒅𝝂 𝒉𝒗 = 𝟑 𝒉𝒗 𝒅𝝂 𝒄 𝒆 ൗ𝒌𝑻 − 𝟏 𝒄 𝒆 ൗ𝒌𝑻 − 𝟏 Intensity decrease at higher frequencies (smaller wavelengths - UV catastrophe) since the excitation of the oscillators to the higher energy states is less probable at lower temperatures At higher temperatures the thermal energy 𝒌𝑻 enables oscillations at higher frequency 𝑛𝝂 Planck’s expression gives excellent co-relation with experimental results The foundation stone- for era of quantum physics!  ENGINEERING PHYSICS Black-body radiation : Summary of classical & quantum Image courtesy: hyperphysics  ENGINEERING PHYSICS Black-body radiation : Summary of classical & quantum Image courtesy: hyperphysics ENGINEERING PHYSICS Black-body radiation : Points of Relevance The trademark of modern physics - Planck's constant, h = 6.63x10-34 Js The failure of classical physics to explain blackbody radiation, the photoelectric effect, and the atomic spectra demolished the foundations of classical physics. Planck's constant is very tiny, only about 6 x 10-34, so in our everyday world, quantum effects makes difference in the 34th decimal place Large objects obey Newton's laws (the average behavior of their component atoms) ENGINEERING PHYSICS Class 4. Quiz … The black body radiation concepts which are correct … 1. Rayleigh and Jeans could explain the radiation curves for the higher wavelengths and not the lower wave lengths 2. Classically the average energy of the oscillators cannot be found 3. Max Planck suggested that the average energy of oscillators have to evaluated using a summation of energies and probabilities ENGINEERING PHYSICS Class 4. Numerical … The frequency of harmonic oscillator at 50oc is 6.2x1012 per sec. Estimate the average energy of the oscillator as per Planck's idea of cavity oscillator, also compare the same with classical average energy and average energy by R-J law. 1. Average energy of the oscillator as per Planck's idea −34 𝒉𝝂 𝒉 = 6.63𝒙10 𝑬 = 𝒉𝒗ൗ k = 1.38 x 10 -23 𝒆 𝒌𝑻 −𝟏 B 2. Average energy of the oscillator as per Classical analysis 𝑬 = 𝒌𝑩 𝑻 3. Partially successful classical analysis is R-J law 𝑬 = 𝒌𝑩 𝑻 THANK YOU Muhammad Faisal, Ph.D. Associate Professor, Department of Science and Humanities [email protected] +91 80 50829629 ENGINEERING PHYSICS Department of Science and Humanities ENGINEERING PHYSICS Unit I : Review of concepts leading to Quantum Mechanics Week #2 Class #7 Atomic Spectra Photo Electric effect Compton effect Compton shift Dual nature of radiation  ENGINEERING PHYSICS Atomic spectra Atoms of different elements have distinct spectra Atomic spectroscopy allows the identification of a sample's elemental composition (An important tool for material characterization) Atomic absorption lines are observed in the solar spectrum, referred to as Fraunhofer lines Robert Bunsen and Gustav Kirchhoff discovered new elements by observing their emission spectra The existence of discrete line - Emission spectra Absence of discrete lines -Absorption spectra  ENGINEERING PHYSICS Atomic spectra analysis – Classical Classical physics - orbiting electron is constantly changing direction and emit electromagnetic radiation As a result, the electron should be continually losing energy! The electron should lose all of its energy and spiral down into the proton In other words, atoms should not exist! Hydrogen spectra https://physics.weber.edu/carroll/honors/failures.htm  ENGINEERING PHYSICS Atomic spectra analysis – Quantum explanation Based on Max Planck's idea that energy comes in quanta, energy can be absorbed or emitted in terms of quanta Particle-Particle interaction leading to absorption and emission spectra of atoms! The explanation of the line spectrum of atoms: in terms of transition between quantized energy states of an atom https://physics.weber.edu/carroll/honors/failures.htm ENGINEERING PHYSICS Photoelectric effect Electron emission from metals under irradiation - Photo electric effect Instantaneous emission of electrons with kinetic energy dependent on wavelength of radiation Energy of photo electrons independent of intensity of radiation Failure of EM wave theory to explain observed results https://notes.tyrocity.com/photoelectric-effect/  ENGINEERING PHYSICS Photoelectric effect – Quantum explanation Quantum phenomenon Einstein’s concepts of photons Low energy electron-photon interaction (Particle-Particle interaction!) Transfer of energy and momentum to the photo electron 𝒉𝝂 = 𝑾 + 𝑲𝑬𝒆 Waves can have dual nature – depending on the nature of interaction with matter ! Single photon-electron interaction Image courtesy: hyperphysics  ENGINEERING PHYSICS Scattering of X-rays by target materials – Compton effect Scattering of X Rays by different target materials Observation: Scattered X rays have a higher wavelength than the incident X rays Wavelength of scattered X rays depend on the angle of scattering Scattering of EM waves with electrons do not explain the observed change in wavelength-Classical explanation fails! Image courtesy: hyperphysics  ENGINEERING PHYSICS Compton Effect Arthur H Compton proposed a high energy photon -electron interaction (particle-particle) X ray energies are in the range of 100KeV Electrons can gain energy and can be emitted with relativistic velocities Image courtesy: hyperphysics ENGINEERING PHYSICS Compton shift Change in wavelength (Compton Shift) was calculated as 𝒉 𝝀𝒇 − 𝝀𝒊 = 𝜟𝝀 = 𝟏 − 𝒄𝒐𝒔 𝜽 𝒎𝒆 𝒄 Compton shift 𝜟𝝀 is independent of the incident wavelength 𝜟𝝀 depends only on the scattering angle. 𝒉 = 𝝀𝒄 is termed as the Compton wavelength 𝒎𝒆 𝒄 For electrons, 𝝀𝒄 =2.42 x 10-12 m Maximum value of Compton shift for 𝒉 the angle 180o is (2x ) 𝒎𝒆 𝒄 ENGINEERING PHYSICS Compton shift: Dependency on scattering angle Change in wavelength (Compton Shift) 𝒉 𝝀𝒇 − 𝝀𝒊 = 𝜟𝝀 = 𝟏 − 𝒄𝒐𝒔 𝜽 𝒎𝒆 𝒄 Energy lost by photon = Energy gained by electron hν−hν′= KE of electron If instead of electron if the X-ray photon interacts with proton? Compton shift- Dependence on angle ENGINEERING PHYSICS Relativistic concepts of momentum and energy Pre-requisites to derive Compton shift Rest mass energy of a particle given by E= moc2. the kinetic energy of a particle with momentum p is given by pc (no specific knowledge on mass, e.g: photon) The total energy of the particle is given by 𝑬= 𝒑𝟐 𝒄𝟐 + 𝒎𝒐 𝟐 𝒄𝟒 ENGINEERING PHYSICS Compton shift derivation: Conservation of momentum and energy Momentum conservation along the incident direction Pi + 0 = Pf cosθ + Pe cosφ Momentum conservation in the perpendicular direction 0 = Pf sinθ - Pe sinφ ENGINEERING PHYSICS Conservation of momentum in X ray scattering Momentum conservation along the incident direction - 𝒑𝒊 + 𝟎 = 𝒑𝒇 𝒄𝒐𝒔𝜽 + 𝒑𝒆 𝒄𝒐𝒔𝝓. Momentum conservation in a perpendicular direction - 𝟎 = 𝒑𝒇 𝒔𝒊𝒏𝜽 − 𝒑𝒆 𝒔𝒊𝒏𝝓 Conservation of momentum before and after collision 𝒑𝒆 𝟐 = 𝒑𝒊 𝟐 + 𝒑𝒇 𝟐 − 𝟐𝒑𝒊 𝒑𝒇 𝒄𝒐𝒔𝜽 … 1. ENGINEERING PHYSICS Compton shift derivation: Conservation of energy Conservation of energy before and after collision 𝒑𝒊 𝒄 + 𝒎𝒐 𝒄𝟐 = 𝒑𝒇 𝒄 + 𝒑𝒆 𝟐 𝒄𝟐 + 𝒎𝒐 𝟐 𝒄𝟒 𝒑𝒆 𝟐 = 𝒑𝒊 𝟐 + 𝒑𝒇 𝟐 − 𝟐𝒑𝒊 𝒑𝒇 + 𝟐𝒎𝒐 𝒄 𝒑𝒊 − 𝒑𝒇 --- 2 Comparing equations 1 & 2 𝒑𝒆 𝟐 = 𝒑𝒊 𝟐 + 𝒑𝒇 𝟐 − 𝟐𝒑𝒊 𝒑𝒇 𝒄𝒐𝒔𝜽 … 1. (From conservation of momentum) 𝒑𝒆 𝟐 = 𝒑𝒊 𝟐 + 𝒑𝒇 𝟐 − 𝟐𝒑𝒊 𝒑𝒇 + 𝟐𝒎𝒐 𝒄 𝒑𝒊 − 𝒑𝒇 --- 2 (From conservation of energy) −𝟐𝒑𝒊 𝒑𝒇 + 𝟐𝒎𝒐 𝒄 𝒑𝒊 − 𝒑𝒇 = −𝟐𝒑𝒊 𝒑𝒇 𝒄𝒐𝒔𝜽 ---- 3. ENGINEERING PHYSICS Compton Shift derivation 𝒉 𝒉 With 𝒑𝒊 = and 𝒑𝒇 = equation 3. simplifies to 𝝀𝒊 𝝀𝒇 𝒉 𝝀𝒇 − 𝝀𝒊 = 𝚫𝝀 = 𝟏 − 𝐜𝐨𝐬 𝜽 𝒎𝒆 𝒄 ENGINEERING PHYSICS Compton effect: Conclusion Compton Effect- proved the particle nature of EM radiation Interaction of radiation with matter at sub-atomic matter requires radiation to be treated as particles - Photons Wave-Particle duality is a reality (radiation can behave like a particle at times and show the normal wave characteristics at other times) ENGINEERING PHYSICS Class 7. Quiz … The concepts which are incorrect …. 1. EM waves can explain discrete spectral lines 2. Compton effect cannot be observed in the visible region 3. Compton shift for protons are higher than that for electrons 4. The maximum shift in the wavelength is 4.84 pm 5. Maximum momentum transfer to electron happens when the angle of scattering is 180o ENGINEERING PHYSICS Class 7. Numericals X-rays of wavelength 0.112 nm is scattered from a carbon target. Calculate the wavelength of X-rays scattered at an angle 90o with respect to the original direction. What is the energy lost by the X-ray photons? What is the energy gained by the electrons? If the incident x-ray retraces back what will be the shift? 𝒉 𝑪𝒐𝒎𝒑𝒕𝒐𝒏 𝒔𝒉𝒊𝒇𝒕, 𝜟𝝀 = 𝟏 − 𝒄𝒐𝒔 𝜽 𝒎𝒆 𝒄 𝑼𝒑𝒐𝒏 𝒔𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒊𝒐𝒏, 𝜟𝝀 = 𝟎. 𝟎𝟐𝟒 𝑨𝒐 𝑾𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒄𝒂𝒕𝒕𝒆𝒓𝒆𝒅 𝑿 − 𝒓𝒂𝒚𝒔, 𝝀′ = 𝝀 + 𝜟𝝀 = 𝟏. 𝟏𝟐 + 𝟎. 𝟎𝟐𝟒 = 𝟏. 𝟏𝟒𝟒 𝑨𝒐 Energy lost by the X-ray photons = Energy gained by the electron = Kinetic energy gained by the electron 𝒄 𝒄 𝐄𝐧𝐞𝐫𝐠𝐲 𝐥𝐨𝐬𝐭 𝐛𝐲 𝐭𝐡𝐞 𝑿 − 𝐫𝐚𝐲 𝐩𝐡𝐨𝐭𝐨𝐧𝐬 = 𝒉𝝂 − 𝐡𝝂′ = 𝐡( ) − 𝒉( ) = Energy gained by the 𝝀 𝝀′ electron 1.77x10-15 – 1.74x10-15=3.14x10-17 J =196 eV Here, θ = 180o maximum shift, head on collision ENGINEERING PHYSICS Class 7. Numericals X-rays of wavelength 1 𝑨𝒐 is scattered from a proton target. Calculate the wavelength of X-rays scattered at an angle 45o with respect to the original direction. What is the energy lost by the X-ray photons? What is the energy gained by the proton? 𝒉 𝑪𝒐𝒎𝒑𝒕𝒐𝒏 𝒔𝒉𝒊𝒇𝒕, 𝜟𝝀 = 𝟏 − 𝒄𝒐𝒔 𝜽 =1.32x10-15x0.3 = 3.87x10-16 𝒎𝒑𝒓𝒐𝒕𝒐𝒏 𝒄 𝑐 𝑐 Energy lost by the X − ray photons = hν − hν′ = h( ) − ℎ( ) = 𝜆 𝜆′ Energy gained by the proton Evaluate the maximum Compton shift when an X-ray photon interacts with a neutron. (Wavelength of X-rays 0.112 nm) 𝒉 Neutron Compton scattering 𝑪𝒐𝒎𝒑𝒕𝒐𝒏 𝒔𝒉𝒊𝒇𝒕, 𝜟𝝀 = 𝟏 − 𝒄𝒐𝒔 𝜽 Proton Compton scattering 𝒎𝒏𝒆𝒖𝒕𝒓𝒐𝒏 𝒄 Compton scattering (with electrons) THANK YOU Muhammad Faisal, Ph.D. Associate Professor, Department of Science and Humanities [email protected] +91 80 50829629 ENGINEERING PHYSICS Department of Science and Humanities ENGINEERING PHYSICS Dual nature of radiation EM radiation Wave nature Particle nature Successfully explains Successfully explains ✓ Reflection ✓ Black body radiation spectra particle-particle ✓ Refraction ✓ Atomic spectra interaction ✓ Interference ✓ Photoelectric effect ✓ Diffraction ✓ Compton effect ✓ Polarization Energy α intensity (square of Amplitude) Energy is quantized in terms of hν Wavelength and frequency – no effect on energy Wavelength and frequency - effect on energy DUAL NATURE OF RADIATION! ENGINEERING PHYSICS Unit I : Review of concepts leading to Quantum Mechanics Week #2 Class #9 Double slit experiment de Broglie hypothesis Dual nature of matter Concept of matter waves  ENGINEERING PHYSICS Young’s double slit experiment Young’s double slit experiment on interference and diffraction of radiations Characteristic wave experiment Well defined experiment to demonstrate wave nature of light (EM radiation) Image courtesy: hyperphysics ENGINEERING PHYSICS de Broglie hypothesis Based on the analysis of dual nature of radiation, de Broglie hypothesis Moving matter (form of energy) should also exhibit wave characteristics 𝒉 Wavelength of this associated waves, 𝝀 = (p = mv, 𝒑 momentum of the particle) Wavelengths of macro particles are extremely small to be measured Wavelengths of moving sub atomic particles are in the measurable range (𝝀 ~𝟏𝟎−𝟏𝟎𝒎) – relevance to microscopic scale  ENGINEERING PHYSICS Dual nature of matter – Experimental verification Experimental verification of de Broglie’s hypothesis Davisson and Germer’s experiment (electron scattering by Ni crystals) 𝒉 𝒉 𝒉 de Broglie wavelength 𝝀 = = = 𝒑 𝟐𝒎𝑬 𝟐𝒎𝒆𝑽 Electron diffraction confirmed at particular settings (54 V, angle of scattering 50o) Satisfied Bragg’s law 𝝀 = 𝟐𝒅 sin 𝜽, by ‘electron waves’! Conclusion: Dual nature of matter - matter and matter waves! Image courtesy: hyperphysics Davisson-Germer experiment  ENGINEERING PHYSICS Double slit experiment with electrons Diffraction is characteristic wave phenomenon Double slit experiment with a particle (single electrons or photons – one at a time) show wave nature – Particle diffraction! Building up of the diffraction pattern of electrons scattered from a crystal surface Image courtesy: hyperphysics  If electrons were classical particles they would be detected in two places, directly in line with each slit. We do not see two maxima in front of the slits. Instead, we see an interference pattern that is characteristic of two waves. https://education.pasco.com/epub/PhysicsNGSS/BookInd-1235.html  A single electron is detected at one place on the screen each time the electron gun fires. As the number of electrons increases ( e.g. 10,000 electrons one at a time). We observe the same interference pattern similar to electrons at once! Conclusion: Electron must have a wave nature too! Our classical concept of an electron as a particle is not adequate to explain the double slit experiment. If the electron is a wave, we can explain the two-slit diffraction pattern by constructive and destructive interference. ENGINEERING PHYSICS Concept of matter waves Need a mathematical concept to describe matter waves Any representative wave should be able to give information about the position and momentum of the system Simple sine or cosine waves fall short (Momentum can be inferred from wavelengths 𝒑 = 𝒉/𝝀 but Position is not well defined) ENGINEERING PHYSICS Class 9. Quiz … The following concepts are not true of matter waves …. 1. de Broglie wavelength of moving particles cannot be measured 2. Double slit experiments cannot be performed with particles 3. Single photon experiments can be performed 4. Sine or cosine waves can describe particle motion accurately 5. The momentum of a particle is independent of the position of the particle ENGINEERING PHYSICS Class 9. Numericals… Find the de Broglie wavelength of electrons moving with a speed of 107 m/s (Ans: 7.28 x 10-11 m) 𝒉 𝒅𝑬 − 𝑩𝒓𝒐𝒈𝒍𝒊𝒆 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉, 𝝀 = 𝒑 𝑷𝒍𝒂𝒏𝒄𝒌′ 𝒔𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕, 𝒉 = 𝟔. 𝟔𝟑𝒙𝟏𝟎−𝟑𝟒 𝑱𝒔 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏, 𝒑 = 𝒎𝒂𝒔𝒔 𝒙 𝒗𝒆𝒍𝒐𝒄𝒕𝒚 An alpha particle is accelerated through a potential difference of 1 kV. Find its de Broglie wavelength. 𝒉 𝒉 𝒉 𝒉 𝒅𝒆 − 𝑩𝒓𝒐𝒈𝒍𝒊𝒆 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚, 𝝀 = = = = 𝟐𝒎𝑬 𝟐𝒎𝑲𝑬 𝟐𝒎𝒆𝑽 𝟑 𝟐𝒎 𝑲𝑩 𝑻 𝟐 𝒉 𝒎 = 𝟒 𝒎𝒑 = 𝟒 𝒎𝒏 = 𝟔. 𝟔𝟖 × 𝟏𝟎−𝟐𝟕 𝒌𝒈 𝝀= 𝟐𝒎𝒒𝑽 𝒒 = 𝟐 × 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 = 𝟑. 𝟐 × 𝟏𝟎−𝟏𝟗 𝑪 𝝀 = 3.21 x 10-13 m  ENGINEERING PHYSICS Class 9. Numericals… Compare the momenta and energy of an electron and photon whose de Broglie wavelength is 650nm (Ans: Ratio of momenta =1; ratio of energy of electron to 𝒉 energy of photon = = 𝟏. 𝟖𝟔𝟕𝒙𝟏𝟎−𝟔 ) 𝟐𝒎𝝀𝒄 𝒑𝟐 𝑬𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 𝒘𝒊𝒕𝒉 𝒌𝒏𝒐𝒘𝒏 𝒓𝒆𝒔𝒕 𝒎𝒂𝒔𝒔 = 𝑲𝒊𝒏𝒆𝒕𝒊𝒄 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝟐𝒎 𝒉 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒐𝒇 𝒂 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒅𝒆 − 𝑩𝒓𝒐𝒈𝒍𝒊𝒆 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉, 𝒑 = 𝝀 𝒉𝟐 𝒑𝟐 𝒉𝟐 𝒔𝒊𝒏𝒄𝒆, 𝒑𝟐 = , 𝑬𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 = = 𝝀𝟐 𝟐𝒎 𝟐𝒎𝝀𝟐 𝑬𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒑𝒉𝒐𝒕𝒐𝒏 𝒎𝒂𝒔𝒔 𝒏𝒐𝒕 𝒅𝒆𝒕𝒆𝒄𝒕𝒆𝒅 𝒃𝒖𝒕 𝒎𝒂𝒔𝒔 𝒆𝒇𝒇𝒆𝒄𝒕 𝒊𝒔 𝒅𝒆𝒕𝒆𝒄𝒕𝒆𝒅! = 𝒉𝝂 = 𝒉𝒄 𝒄 , 𝒔𝒊𝒏𝒄𝒆, 𝝂 = 𝝀 𝝀 𝒉𝟐 𝒆𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 𝟐𝒎𝝀𝟐 𝒉 𝑪𝒐𝒎𝒑𝒂𝒓𝒆 𝒆𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 𝒕𝒐 𝒑𝒉𝒐𝒕𝒐𝒏 = = 𝒉𝒄 = 𝒆𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒑𝒉𝒕𝒐𝒏 𝟐𝒎𝝀𝒄 𝝀  ENGINEERING PHYSICS Class 9. Numericals… Calculate the de Broglie wavelength of electrons and protons if their kinetic energies are i) 1% and ii) 5% of their rest mass energies. (Rest mass energy of electron = 8.19x 10-14 J; rest mass energy of protons = 1.503 x 10-10J) 𝒉 𝒅𝑬 − 𝑩𝒓𝒐𝒈𝒍𝒊𝒆 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉, 𝝀 = 𝒑 𝒉 𝒉 𝒉 𝒉 𝒅𝒆 − 𝑩𝒓𝒐𝒈𝒍𝒊𝒆 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚, 𝝀 = = = = 𝟐𝒎𝑬 𝟐𝒎𝑲𝑬 𝟐𝒎𝒆𝑽 𝟑 𝟐𝒎 𝑲𝑩 𝑻 𝟐 𝑹𝒆𝒔𝒕 𝒎𝒂𝒔𝒔 𝒆𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒂 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆 𝒐𝒇 𝒎𝒂𝒔𝒔 𝒎, 𝑬 = 𝒎𝒐 𝒄𝟐 𝑭𝒐𝒓 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 𝒓𝒆𝒔𝒕 𝒎𝒂𝒔𝒔, 𝒎 = 𝟗. 𝟏𝟏𝒙𝟏𝟎−𝟑𝟏 𝒌𝒈 𝒂𝒏𝒅 𝒇𝒐𝒓 𝒑𝒓𝒐𝒕𝒐𝒏, 𝒎𝒑 = 𝟏. 𝟔𝟕𝒙𝟏𝟎−𝟐𝟕 𝒌𝒈 Electron 1% , 𝝀 = 𝟏. 𝟕𝟐 𝒙𝟏𝟎−𝟏𝟏 𝒎 Proton 1% , 𝝀 = 𝟗. 𝟑𝟔 𝒙 𝟏𝟎−𝟏𝟔 𝒎 Electron 5% , 𝝀 = 𝟕. 𝟔𝟖 𝒙𝟏𝟎−𝟏𝟐 𝒎 Proton 5%, 𝝀 = 𝟒. 𝟏𝟖 𝒙𝟏𝟎−𝟏𝟓 𝒎 ENGINEERING PHYSICS Class 9. Numericals… An electron and a photon have a wavelength of 2.0 Ao. Calculate their momenta and total energies. (Eelectron=8.2x10-14 Ephoton =10x10-16J, Pelectron=Pphoton=3.32x10-24) 𝒉 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒐𝒇 𝒂 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒅𝒆 − 𝑩𝒓𝒐𝒈𝒍𝒊𝒆 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉, 𝒑 = 𝝀 𝑻𝒐𝒕𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒐𝒏 𝒘𝒊𝒕𝒉 𝒌𝒏𝒐𝒘𝒏 𝒎𝒂𝒔𝒔, 𝑬 𝒑𝟐 = 𝑹𝒆𝒔𝒕 𝒎𝒂𝒔𝒔 𝒆𝒏𝒆𝒓𝒈𝒚 + 𝑲𝒊𝒏𝒆𝒕𝒊𝒄 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝒎𝒐 𝒄𝟐 + 𝟐𝒎 𝒉𝒄 𝑻𝒐𝒕𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒑𝒉𝒐𝒕𝒐𝒏, 𝑬 = 𝒉𝝂 = 𝝀 ENGINEERING PHYSICS Class 9. Numericals… What is the wavelength of an hydrogen atom moving with a mean velocity corresponding to the average kinetic energy of hydrogen atoms 𝒉 under thermal equilibrium at 293K? ( 𝝀 = = 𝟏. 𝟒𝟕𝒙𝟏𝟎−𝟏𝟎 𝒎 ) 𝟑𝒎𝒌𝑻 Mass of hydrogen atom = 1.67 x 10-27 kg 𝒅𝒆 − 𝑩𝒓𝒐𝒈𝒍𝒊𝒆 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚, 𝒉 𝒉 𝒉 𝒉 𝝀= = = = 𝟐𝒎𝑬 𝟐𝒎𝑲𝑬 𝟐𝒎𝒆𝑽 𝟑 𝟐𝒎 𝑲𝑩 𝑻 𝟐 THANK YOU Muhammad Faisal, Ph.D. Associate Professor, Department of Science and Humanities [email protected] +91 80 50829629 ENGINEERING PHYSICS Department of Science and Humanities ENGINEERING PHYSICS Unit I : Review of concepts leading to Quantum Mechanics DUALITY OF RADIATION & MATTER PARTICLE WAVE NATURE NATURE MATTER WAVES PHOTON EM WAVES E = hν MATTER (e.g. Davison- E & B waves Germer expt: (BlackBodyRadiation (Mass) (e.g: Reflection, electron diffraction, , Compton effect, (elephant, electron, interference, etc.) double slit expt) Photoelectric effect, cricket ball) atomic spectra) How to represent? ENGINEERING PHYSICS Unit I : Review of concepts leading to Quantum Mechanics Week #3 Class #10(7,8) Superposition of waves Phase and group velocities Group velocity relations ENGINEERING PHYSICS Concept of matter waves - superposition of waves Wave packets describe matter waves - with a defined wavelength and an amplitude maximum (for both momentum and position) unlike sine or cos waves How to represent a wave packet: Superposition of two waves  ENGINEERING PHYSICS Wave packets as matter waves - Mathematical analysis Two sinusoidal waves: Wave 1: y1 (frequency ω and propagation constant k) Wave 2: y2 (frequency 𝝎 + 𝜟𝝎 and propagation constant k+ Δk) 𝒚𝟏 = 𝑨𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙) 𝒚𝟐 = 𝑨 𝐬𝐢𝐧{ 𝝎 + 𝜟𝝎 𝒕 + 𝒌 + 𝜟𝒌 𝒙} 𝜟𝒘𝒕 + 𝜟𝒌𝒙 Superposition gives a wave packet 𝒚 = 𝒚𝟏 + 𝒚𝟐 = 𝟐𝑨𝒔𝒊𝒏 𝒘𝒕 + 𝒌𝒙. 𝐜𝐨𝐬{ 𝟐 }  ENGINEERING PHYSICS Wave packets as matter waves – Detailed mathematical analysis Two sinusoidal waves: 𝒚𝟏 = 𝑨𝒔𝒊𝒏(𝝎𝒕 + 𝒌𝒙) 𝒚𝟐 = 𝑨 𝐬𝐢𝐧{ 𝝎 + 𝜟𝝎 𝒕 + 𝒌 + 𝜟𝒌 𝒙} 𝒚 = 𝒚𝟏 + 𝒚𝟐 𝒂−𝒃 𝒂+𝒃 𝒘𝒆 𝒌𝒏𝒐𝒘 𝒕𝒉𝒂𝒕, 𝒔𝒊𝒏 𝒂 + 𝒔𝒊𝒏 𝒃 = 𝟐 𝒄𝒐𝒔 𝒔𝒊𝒏 𝟐 𝟐 𝑺𝒊𝒎𝒊𝒍𝒂𝒓𝒍𝒚 𝒉𝒆𝒓𝒆, 𝒚𝟏 + 𝒚𝟐 𝝎𝒕 − 𝒌𝒙 − (𝝎 + ∆𝝎)𝒕 − (𝒌 + ∆𝒌)𝒙 𝝎𝒕 − 𝒌𝒙 + (𝝎 + ∆𝝎)𝒕 − (𝒌 + ∆𝒌)𝒙 = 𝟐𝑨 𝒄𝒐𝒔. 𝒔𝒊𝒏 𝟐 𝟐 ∆𝝎)𝒕 + (∆𝒌)𝒙 𝟐𝝎𝒕 − 𝟐𝒌𝒙 + (∆𝝎)𝒕 − (∆𝒌)𝒙 = 𝟐𝑨 𝒄𝒐𝒔 −. 𝒔𝒊𝒏 𝟐 𝟐 𝑻𝒉𝒊𝒔 𝒓𝒆𝒅𝒖𝒄𝒆𝒔 𝒕𝒐, 𝒓𝒆𝒔𝒖𝒍𝒕𝒂𝒏𝒕 𝒚 = 𝒚𝟏 + 𝒚𝟐 ∆𝝎)𝒕 + (∆𝒌)𝒙 = 𝟐𝑨 𝒄𝒐𝒔. 𝒔𝒊𝒏 𝝎𝒕 − 𝒌𝒙 𝟐 𝑺𝒊𝒏𝒄𝒆, 𝒄𝒐𝒔 − 𝒙 = 𝒄𝒐𝒔 𝒙 , 𝑨𝒔𝒍𝒐 (𝟐𝝎𝒕 + ∆𝝎)𝒕 ≈ 𝟐𝝎𝒕 𝒂𝒏𝒅 𝟐𝒌𝒙 − ∆𝒌 𝒙 ≈ 𝟐𝒌𝒙 ENGINEERING PHYSICS Wave packets as Matter waves k defines λ - which defines momentum 𝟐𝝅 𝒉 (𝑷𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒊𝒐𝒏 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕, 𝒌 = 𝒂𝒏𝒅 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒂𝒔 𝒑𝒆𝒓 𝒅𝒆 − 𝑩𝒓𝒐𝒈𝒍𝒊𝒆′ 𝒔𝒉𝒚𝒑𝒐𝒕𝒉𝒆𝒔𝒊𝒔, 𝒑 = 𝝀) 𝝀 Spread around the central maximum can be the approximate position of the particle (fulfill both the requirements – position and momentum)  ENGINEERING PHYSICS Phase and group velocities: Velocities associated with matter waves 𝜟𝒘𝒕+𝜟𝒌𝒙 Wave packet, 𝒚 = 𝒚𝟏 + 𝒚𝟐 = 𝟐𝑨𝐜𝐨𝐬. 𝒔𝒊𝒏 𝒘𝒕 + 𝒌𝒙 𝟐 The phase velocity of the wave packet is the velocity of a representative point on 𝝎 the wave packet, 𝒗𝒑 = 𝒌 The group velocity of the wave packet is the velocity of common velocity of the 𝒅𝝎 superposed wave group, 𝒗𝒈 = 𝒅𝒌 𝒅𝝎 𝒗𝒈 = 𝒅𝒌 𝝎 𝒗𝒑 = 𝒌 ENGINEERING PHYSICS Relation between phase and group velocity Group velocity, 𝒅𝝎 𝒅 𝒅𝒗𝒑 𝒗𝒈 = = 𝒗𝒑. 𝒌 = 𝒗𝒑 + 𝒌 𝒅𝒌 𝒅𝒌 𝒅𝒌 𝒅𝒗𝒑 𝒅𝒗𝒑 𝒅𝝀 Here, =. 𝒅𝒌 𝒅𝝀 𝒅𝒌 𝒅𝝀 𝟐𝝅 2𝜋 2𝜋 And = − 𝟐 (since, 𝑘 = 𝑜𝑟 𝜆= ) 𝒅𝒌 𝒌 𝜆 𝑘 𝟐𝝅 𝒅𝒗𝒑 𝒅𝒗𝒑 Hence, 𝒗𝒈 = 𝒗𝒑𝒉 − = 𝒗𝒑𝒉 −𝝀 𝒌 𝒅𝝀 𝒅𝝀 Conclusion: Vg is dependent on Vp and also on the phase velocity change with wavelength ENGINEERING PHYSICS Phase and group velocity relation Group velocity = Phase velocity! Is it possible? In a non-dispersive medium (where velocity of the waves independent of the 𝒅𝒗𝒑 wavelength), Vg = Vp (𝝀 =0, since phase velocity does not change with 𝝀) 𝒅𝝀 In a dispersive medium (where the velocity of the waves depends on the wavelength) Vg can be Vp Interesting relations of Group velocity & Phase velocity – Case 1 Vg - half the phase velocity (case of Vg < Vp) 𝒗𝒈 = 𝒗𝒑 /𝟐 𝒅𝒗𝒑 Group velocity of a wave packet is given by 𝒗𝒈 = 𝒗𝒑 − 𝝀 𝒅𝝀 𝒗𝒑 𝒅𝒗𝒑 𝒗𝒑 𝑨𝒔 𝒗𝒈 = , 𝒘𝒆 𝒈𝒆𝒕 𝝀 = 𝟐 𝒅𝝀 𝟐 𝒅𝒗𝒑 𝟏 𝒅𝝀 Thus, = This on integration yields 𝒍𝒏 𝒗𝒑 ∞ 𝒍𝒏 𝝀 or 𝒗𝒑 𝟐 𝝀 𝒗𝒑 ∞ 𝝀 This implies that the phase velocity is proportional to the square root of the wavelength ENGINEERING PHYSICS Phase and group velocity relation Interesting relations of Group velocity & Phase velocity – Case 2 Vg - twice the phase velocity (case of Vg > Vp) 𝒗𝒈 = 𝟐𝒗𝒑 𝒅𝒗𝒑 Group velocity of a wave packet is given by 𝒗 𝒈 = 𝒗𝒑 − 𝝀 𝒅𝝀 𝒅𝒗𝒑 𝒅𝝀 = − 𝒗𝒑 𝝀 𝟏 This on integration yields 𝒍𝒏 𝒗𝒑 ∞ 𝒍𝒏 or 𝒗𝒑 ∞ 𝝀−𝟏 𝝀 This implies that the phase velocity is inversely proportional to the wavelength ENGINEERING PHYSICS Relation between group and particle velocities 𝒅𝝎 Group velocity, 𝒗𝒈 = 𝒅𝒌 𝑬 The angular frequency 𝝎 = where E is the energy of the wave and ℏ 𝒅𝑬 hence 𝒅𝝎 = ℏ 𝒑 𝒅𝒑 The wave vector 𝒌 = where p is the momentum and hence 𝒅𝒌 = ℏ ℏ 𝒉 (Propagation constant is transformed to momentum, 𝒑 = ℏ𝒌 & ℏ= 𝒊𝒔 𝒕𝒉𝒆 𝒓𝒆𝒅𝒖𝒄𝒆𝒅 𝑷𝒍𝒂𝒏𝒄𝒌′ 𝒔𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕) 𝟐𝝅 𝒅𝑬 𝒅𝝎 ℏ 𝒅𝑬 Therefore the group velocity 𝒗𝒈 = = 𝒅𝒑 = As group velocity is the 𝒅𝒌 𝒅𝒑 ℏ 𝒑𝟐 velocity of the wave Since 𝑬 = 𝟐𝒎 𝒎𝒂𝒕𝒕𝒆𝒓 𝒘𝒂𝒗𝒆 − 𝒆𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒐𝒗𝒊𝒏𝒈 𝒎𝒂𝒕𝒕𝒆𝒓 𝒊𝒔 𝒌𝒊𝒏𝒆𝒕𝒊𝒄 𝒆𝒏𝒆𝒓𝒈𝒚 , packet representing a 𝒑𝟐 𝒅𝑬 𝒅( ) 𝟏 𝒅𝒑𝟐 𝒑 particle, group velocity 𝟐𝒎 Group velocity 𝒗𝒈 = = = = = 𝒗 where v is the should be the same as 𝒅𝒑 𝒅𝒑 𝟐𝒎 𝒅𝒑 𝒎 particle velocity! particle velocity, vparticle  ENGINEERING PHYSICS Group and particle velocities As group velocity is the velocity of the wave packet representing a particle, group velocity should be the same as particle velocity! 𝒅𝝎 Group velocity, 𝒗𝒈 = = 𝑷𝒂𝒓𝒕𝒊𝒄𝒍𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒅𝒌 ENGINEERING PHYSICS Class 10. Quiz … The concepts which true of matter waves …. 1. Wave packet is a cosine wave 2. The outline connecting the peaks of the wave packet is a low frequency wave 3. Wave packets are longitudinal 4. In a non dispersive medium the group velocity is equal to the phase velocity ENGINEERING PHYSICS Class 10. Numericals 1. How does group velocity signify for a quantum particle? Group velocity is the particle velocity of the quantum particle 1 2. The ω-k relation for a certain set of waves is 𝜔 = 2𝑘. Find the phase 3 and group velocities. 𝝎 𝑷𝒉𝒂𝒔𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝒗𝒑 = 𝒌 𝒅𝝎 𝑮𝒓𝒐𝒖𝒑 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝒗𝒈 = 𝒅𝒌 𝟏 𝝎 𝟐𝒌𝟑−𝟐 𝑷𝒉𝒂𝒔𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝒗𝒑 = = = 𝟐𝒌 𝟑 𝒌 𝒌 𝟏 𝒅𝝎 𝒅(𝟐𝒌𝟑 ቁ 𝟏 −𝟐 𝟐 −𝟐 𝑮𝒓𝒐𝒖𝒑 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝒗𝒈 = = = 𝟐.. 𝒌 𝟑 =. 𝒌 𝟑 𝒅𝒌 𝒅𝒌 𝟑 𝟑 ENGINEERING PHYSICS Class 10. Numericals 3. A wave packet is represented as, 𝒚 = 𝟏𝟎 𝒔𝒊𝒏 𝟑𝟎𝒕 − 𝟒𝟎𝒙. 𝒄𝒐𝒔 𝟎. 𝟑𝒕 − 𝟎. 𝟓 𝒙 Find the phase and group velocities 𝝎 𝑷𝒉𝒂𝒔𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝒗𝒑 = 𝒌 𝒅𝝎 𝑮𝒓𝒐𝒖𝒑 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝒗𝒈 = 𝒅𝒌 𝑯𝒆𝒓𝒆, 𝝎 = 𝟑𝟎, 𝒌 = 𝟒𝟎, ∆𝝎 = 𝟎. 𝟑, ∆𝒌 = 𝟎. 𝟓 THANK YOU Muhammad Faisal, Ph.D. Associate Professor, Department of Science and Humanities [email protected] +91 80 50829629 ENGINEERING PHYSICS Department of Science and Humanities ENGINEERING PHYSICS Unit I : Review of concepts leading to Quantum Mechanics Week #2 Class #8 Analysis of wave packet Heisenberg’s Uncertainty Principle Two fundamental ideas which Applications of Uncertainty Principle cannot be violated in any theory of quantum systems ----- 1) Electron’s non-existence inside nucleus De Broglie hypothesis and 2) Gamma Ray microscope Heisenberg’s Uncertainty principle ENGINEERING PHYSICS Heisenberg’s Uncertainty Principle 1. Position momentum uncertainty: The position and momentum of a particle cannot be determined simultaneously with unlimited precision 𝒉 ∆𝒙. ∆𝒑 ≥ ℏ/𝟐 ≥ , 𝒘𝒉𝒆𝒓𝒆, ∆𝒙 ∆𝒑 - uncertainty in the position & momentum 𝟒𝝅 (determined simultaneously) Uncertainty relation is valid for any conjugate pairs 2. Energy Time uncertainty: The energy and life time of a particle in a state cannot be determined simultaneously 𝒉 with unlimited precision, ∆𝑬. ∆𝒕 ≥ ℏ/𝟐 ≥ 𝟒𝝅 ∆𝑬 & ∆𝒕 - the uncertainty in energy and life time of the particle ENGINEERING PHYSICS Heisenberg’s Uncertainty Principle 3. Uncertainty relation for circular motion: The angular position and angular momentum of a particle in a circular motion cannot be determined simultaneously with unlimited precision 𝒉 ∆𝜽. ∆𝑳 ≥ ℏ/𝟐 ≥ 𝟒𝝅 ∆𝜽 is the uncertainty in the angular position ∆L is the uncertainty in the angular momentum (determined simultaneously)  ENGINEERING PHYSICS Heisenberg’s analysis of wave packets Wave packets describe matter waves Wave packets have inherent components of uncertainties Spread in the estimation of position (say, along x axis, ∆𝒙) and propagation constant (∆𝒌) of the wave is intrinsically related Product of the deviations, Moving particle ∆𝒙. ∆𝒌 ≥ 𝟏/𝟐 Standard form of the uncertainty principle, 𝒉 ∆𝒙. ∆𝒑 ≥ ℏ/𝟐 ≥ 𝟒𝝅 (propagation constant in terms of momentum, 𝒑 = ℏ𝒌, thus ∆𝒑 = ∆ℏ𝒌 = ℏ∆𝒌 ) Image courtesy: 𝒉 𝟐𝝅 ibphysics.org (ℏ = 𝒊𝒔 𝒓𝒆𝒅𝒖𝒄𝒆𝒅 𝑷𝒍𝒂𝒏𝒄𝒌′ 𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 & 𝒌 = ) 𝟐𝝅 𝝀  ENGINEERING PHYSICS Applications of uncertainty principle: 1Non-existence of electrons inside nuclei Electrons cannot exist inside nucleus, but in 𝜷 decay an electron is emitted from the nucleus with energies of the order of 8 MeV! Why? Assuming electron to be inside the nucleus (confined to nuclear diameter), estimate the energy of the electron using uncertainty principle Then, uncertainty in the position ∆𝒙 ≈ 𝟏𝟎−𝟏𝟒 𝒎 (≈ nuclear diameter) Image courtesy: wikipedia  ENGINEERING PHYSICS Non-existence of electrons inside nuclei Corresponding uncertainty in the momentum of the electron using uncertainty relation ∆𝒙. ∆𝒑 ≥ ℏ/𝟐 ℏ ∆𝒑 = = 𝟓. 𝟐𝟖 × 𝟏𝟎−𝟐𝟏 𝒌𝒈𝒎𝒔−𝟏 (minimum possible as ∆𝒙 ≈ 𝟏𝟎−𝟏𝟒 𝒎 ≈ nuclear 𝟐.∆𝒙 diameter, 𝒙 ) Hence the momentum of the electron p cannot be lesser than ∆𝒑, (p ≈ ∆𝒑 ) Kinetic energy of the electron, 𝟐 Thus, energy of the electron should 𝒑𝟐 ∆𝒑𝟐 𝟏 ℏ be quite high to be an integral 𝑬= = = ≈ 𝟗𝟔 𝑴𝒆𝑽 𝟐𝒎 𝟐𝒎 𝟐𝒎 𝟐. ∆𝒙 member of the nuclei! ✓ The actual energies of electron emitted by radioactive nuclei are very less compared to the above estimate ✓ Conclude that the electron cannot be a permanent part of the nuclei, thus illustrating the power of the uncertainty principle www.assignmentpoint.com  ENGINEERING PHYSICS Applications of uncertainty principle: 2. Gamma ray microscope:- A thought experiment! Experiment to “measure” the position of an electron to “observe” electron (wavelength to be comparable to size of electron - so gamma rays of wavelength ≈ 𝟏𝟎−𝟏𝟐 𝒎 Resolution of the microscope comparable to 𝝀 the position uncertainty 𝜟𝒙 ≈ 𝒔𝒊𝒏 𝜽 Resolution of the microscope: Minimum distance at which two distinct points of a specimen can still be seen High energy 𝜸-rays impart momentum to Image courtesy: the electrons (following the principles of wikipedia Compton Effect)  ENGINEERING PHYSICS Gamma ray microscope – A thought experiment! 𝒉 Momentum gained by the electron in the x direction, 𝒑𝒙 ≈ ± 𝒔𝒊𝒏 𝜽 𝝀 𝒉 Momentum uncertainty, ∆𝒑𝒙 ≈ 𝟐 𝒔𝒊𝒏 𝜽 (cannot be greater than Px) 𝝀 Product of the uncertainties 𝝀 𝒉 ∆𝒙. ∆𝒑𝒙 ≈ ∗ 𝟐 𝒔𝒊𝒏 𝜽 ≈ 𝟐𝒉 𝒔𝒊𝒏 𝜽 𝝀 𝒉 Greater than ! x component of momentum, 𝟒𝝅 𝒉 Conforms to the uncertainty principle 𝒑𝒙 ≈ ± 𝒔𝒊𝒏 𝜽 𝝀 Conclusion: Simultaneous determination of the position and momentum results in an inherent uncertainty  ENGINEERING PHYSICS Heisenberg’s Uncertainty Principle - Conclusion Two fundamental ideas which cannot be violated in any theory of quantum systems ----- De Broglie hypothesis and Heisenberg’s Uncertainty principle Ref: Slide share This consequence is called Heisenberg’s Uncertainty Principle! ENGINEERING PHYSICS Class 10. Quiz … The concepts which true of the uncertainty principle …. 1. Uncertainty principle is based on the measurement accuracies of equipments used 2. The position of a particle cannot be determined accurately 3. The momentum of a particle always has an uncertainty which is related to the uncertainty in the position of the particle 4. Electrons cannot be confined to a nucleus as it is energetically not feasible 5. Electrons cannot be confined to a nucleus as it’s size is bigger the nuclear diameter  ENGINEERING PHYSICS Other forms of uncertainty relations Other forms of Uncertainty Relations 1 𝑷𝒐𝒔𝒊𝒕𝒊𝒐𝒏 − 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏, ∆𝒙. ∆𝝀 𝒉 𝟏 𝟏 𝑨𝒔, ∆𝒑 =. ∆ = 𝒉. ∆ = 𝒉. (− 𝟐. ∆𝝀ቇ 𝝀 𝝀 𝝀 𝑻𝒉𝒖𝒔 𝒑𝒐𝒔𝒊𝒕𝒊𝐨𝐧 − 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒄𝒂𝒏 𝒂𝒍𝒔𝒐 𝒃𝒆 𝒘𝒓𝒊𝒕𝒕𝒆𝒏 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒂𝒏𝒅 𝒘𝒂𝒗𝒆𝒍𝒆𝒈𝒕𝒉 𝟏 ∆𝒙. 𝒉. (− 𝟐. ∆𝝀) ≥ 𝒉 𝝀𝟐 𝝀𝟐 𝝀 𝟒𝝅 𝑻𝒉𝒖𝒔, ∆𝒙. ∆𝝀 ≥ − ≥ 𝟒𝝅 𝟒𝝅 2 𝑼𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒂𝒏𝒅 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, ∆𝒙. ∆𝒗 ∆𝐩 = ∆(𝒎𝒗) 𝒉 ∆𝒙. ∆𝒗 ≥ 𝟒𝝅. 𝒎  ENGINEERING PHYSICS Other forms of uncertainty relations Other forms of Uncertainty Relations 𝟏 3 𝑷𝒐𝒔𝒊𝒕𝒊𝒐𝒏 − 𝒑𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒊𝒐𝒏 𝒗𝒆𝒄𝒕𝒐𝒓 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏, ∆𝒙. ∆𝒌 ≥ 𝟐 (𝑭𝒓𝒐𝒎 𝒕𝒉𝒆 𝒄𝒐𝒏𝒄𝒆𝒑𝒕 𝒐𝒇 𝒘𝒂𝒗𝒆 𝒑𝒂𝒄𝒌𝒆𝒕) 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒊𝒏 𝒐𝒏𝒆 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒔 𝒕𝒐 4 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒖𝒏𝒄𝒆𝒓𝒂𝒕𝒊𝒏𝒕𝒚 𝒐𝒇 𝒐𝒕𝒉𝒆𝒓 𝒉 ∆𝒙𝒎𝒊𝒏. ∆𝒑𝒎𝒂𝒙 ≥ 𝟒𝝅 5 𝑨𝒄𝒄𝒖𝒓𝒂𝒄𝒚 𝒑𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒆𝒓𝒓𝒐𝒓 𝒂𝒏𝒅 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 E.g: The speed of an electron is measured to be 1 km/s with an accuracy of 0.005%, then uncertainty in velocity will be? 𝑼𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒊𝒏 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, ∆𝒗 = 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒙 𝒂𝒄𝒄𝒖𝒓𝒂𝒄𝒚 𝒐𝒓 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒙 𝒑𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒆𝒓𝒓𝒐𝒓 𝟎. 𝟎𝟎𝟓 ∆𝒗 = 𝟏𝟎𝟎𝟎𝒙 = 𝟎. 𝟎𝟓 𝒎Τ𝒔 𝟏𝟎𝟎  ENGINEERING PHYSICS Other forms of uncertainty relations Other forms of Uncertainty Relations 𝒉 6 𝑬𝒏𝒆𝒓𝒈𝒚 − 𝒕𝒊𝒎𝒆 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏, ∆𝐄. ∆𝐭 ≥ 𝟒𝝅 𝒉𝒄 𝟏 𝟏 𝑾𝑲𝑻, 𝑬 = 𝒉𝝂 𝑨𝒔, ∆𝑬 =. ∆ = 𝒉𝒄. ∆ = 𝒉𝒄. (− 𝟐. ∆𝝀ቇ 𝝀 𝝀 𝝀 𝑻𝒉𝒖𝒔 𝒆𝒏𝒆𝒓𝒈𝒚 − 𝒕𝒊𝒎𝒆 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒄𝒂𝒏 𝒂𝒍𝒔𝒐 𝒃𝒆 𝒘𝒓𝒊𝒕𝒕𝒆𝒏 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒅 𝒕𝒊𝒎𝒆 𝟏 𝒉 ∆𝒕. 𝒉𝒄. (− 𝟐. ∆𝝀) ≥ 𝝀 𝟒𝝅 𝝀𝟐 𝝀𝟐 𝑻𝒉𝒖𝒔, ∆𝒕. ∆𝝀 ≥ − ≥ 𝟒𝝅𝒄 𝟒𝝅𝒄 7 𝑬𝒏𝒆𝒓𝒈𝒚 − 𝒕𝒊𝒎𝒆 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒄𝒂𝒏 𝒂𝒍𝒔𝒐 𝒃𝒆 𝒘𝒓𝒊𝒕𝒕𝒆𝒏 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝒂𝒏𝒅 𝒕𝒊𝒎𝒆 𝑬 = 𝒉𝝂 𝟏 ∆𝑬. ∆𝝂 ≥ ∆𝑬 = ∆𝒉𝝂 = 𝒉. ∆𝝂 𝟒𝝅 ENGINEERING PHYSICS Class 10. Numericals 1. The speed of an electron is measured to be 1 km/s with an accuracy of 0.005%. Estimate the uncertainty in the position of the particle. 𝒉 𝑼𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒂𝒏𝒅 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, ∆𝒙. ∆(𝒑 = 𝒎𝒗) ≥ 𝟒𝝅 𝒉 ∆𝒙. ∆𝒗 ≥ 𝟒𝝅. 𝒎 𝑼𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒊𝒏 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, ∆𝒗 = 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒙 𝒂𝒄𝒄𝒖𝒓𝒂𝒄𝒚 𝟎. 𝟎𝟎𝟓 ∆𝒗 = 𝟏𝟎𝟎𝟎𝒙 = 𝟎. 𝟎𝟓 𝒎Τ𝒔 𝟏𝟎𝟎 𝒉 ∆𝒙 = , 𝒉 = 𝟔. 𝟔𝟑𝒙𝟏𝟎−𝟑𝟒 , 𝒎 = 𝒎𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 = 𝟗. 𝟏𝟏𝒙𝟏𝟎−𝟑𝟏 𝒌𝒈 𝟒𝝅𝒎∆𝒗 ∆𝒙 = 𝟏. 𝟏𝟓𝟗𝒙𝟏𝟎−𝟑 𝒎 ENGINEERING PHYSICS Class 10. Numericals… 2. The spectral line of Hg green is 546.1 nm has a width of 10-5 nm. Evaluate the minimum time spent by the electrons in the upper state before de excitation to ℏ 𝝀𝟐 the lower state. (Ans: ∆𝒕 = = = 𝟕. 𝟗𝟏𝒙𝟏𝟎−𝟗 𝒔 ) 𝟐.∆𝑬 𝟒𝝅𝒄𝜟𝝀 𝑰𝑴𝑷: 𝑺𝒑𝒆𝒄𝒕𝒓𝒂𝒍 𝒍𝒊𝒏𝒆 𝒂𝒏𝒅 𝒊𝒔 𝒘𝒊𝒅𝒕𝒉, 𝒕𝒉𝒂𝒕 𝒊𝒔 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒅 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒔𝒑𝒓𝒆𝒂𝒅 𝒊𝒏 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉, ∆𝝀 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒕𝒊𝒎𝒆 = 𝒕𝒊𝒎𝒆 𝒔𝒑𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒖𝒑𝒑𝒆𝒓 𝒆𝒏𝒓𝒈𝒚 𝒔𝒕𝒂𝒕𝒆 𝒃𝒆𝒇𝒐𝒓𝒆 𝒅𝒆𝒆𝒙𝒄𝒊𝒕𝒂𝒕𝒊𝒐𝒏 = 𝒍𝒊𝒇𝒆 𝒕𝒊𝒎𝒆 = ∆𝒕 𝑬𝒏𝒆𝒓𝒈𝒚 − 𝒕𝒊𝒎𝒆 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒄𝒂𝒏 𝒂𝒍𝒔𝒐 𝒃𝒆 𝒘𝒓𝒊𝒕𝒕𝒆𝒏 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒅 𝒕𝒊𝒎𝒆 𝝀𝟐 𝝀𝟐 ∆𝒕. ∆𝝀 ≥ − ≥ 𝟒𝝅𝒄 𝟒𝝅𝒄 ENGINEERING PHYSICS Class 10. Numericals 3. The uncertainty in the location of a particle is equal to it's de Broglie wavelength. Show that the corresponding uncertainty in its velocity is approx one tenth of it's ℏ 𝒉 𝒑 𝒗 𝒗 𝒗 velocity. (Ans: ∆𝒑 = = = Hence ∆𝒗 = = ≈ ) 𝟐.∆𝒙 𝟒𝝅𝝀 𝟒𝝅 𝟒𝝅 𝟏𝟐.𝟓𝟔 𝟏𝟎 𝒉 𝑷𝒐𝒔𝒊𝒕𝒊𝒐𝒏 − 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏, ∆𝒙. ∆𝒑 ≥ 𝟒𝝅 𝒉 𝑼𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒊𝒏 𝒍𝒐𝒄𝒂𝒕𝒊𝒐𝒏 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 = ∆𝒙 = 𝝀 = 𝒑 𝒉 𝒉 𝒗 𝒗 𝒗. ∆𝒑 ≥ 𝒑 ∆𝒗 ≥ = ≈ 𝒑 𝟒𝝅 ∆𝒑 ≥ 𝟒𝝅 𝟏𝟐.𝟓𝟔 𝟏𝟎 𝟒𝝅 ENGINEERING PHYSICS Class 10. Numericals 4. A proton is confined to a box of length 2 nm. What is the minimum uncertainty in its velocity? 𝑨𝒔𝒔𝒖𝒎𝒊𝒏𝒈 𝒕𝒉𝒆 𝒍𝒆𝒏𝒈𝒕𝒉 𝒊𝒕𝒔𝒆𝒍𝒇 𝒕𝒐 𝒃𝒆 𝒕𝒉𝒆 𝒎𝒂𝒙. 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒊𝒏 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏, 𝒉 ∆𝒙𝒎𝒂𝒙. ∆𝒑𝒎𝒊𝒏 ≥ 𝟒𝝅 𝒉 𝑼𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒂𝒏𝒅 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, ∆𝒙. ∆(𝒑 = 𝒎𝒗) ≥ 𝟒𝝅 𝒉 ∆𝒙. ∆𝒗 ≥ 𝟒𝝅. 𝒎 𝒎 = 𝒎𝒑𝒓𝒐𝒕𝒐𝒏 = 𝟏. 𝟔𝟕𝒙𝟏𝟎−𝟐𝟕 𝒌𝒈 ENGINEERING PHYSICS Class 10. Numericals 4. Show that if a component of angular momentum of the electron in a hydrogen atom is known to be 2ℏ within 5%, its angular orbital position in the plane perpendicular to the component will be? 𝒉 ∆𝜽. ∆𝑳 ≥ ℏ/𝟐 ≥ 𝟒𝝅 ∆𝜽 is the uncertainty in the angular position ∆L is the uncertainty in the angular momentum ∆𝑳 = 𝑳 𝒙 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒂𝒄𝒄𝒖𝒓𝒂𝒄𝒚 = 𝑳𝒙𝟓% 𝒐𝒏 𝒔𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒊𝒐𝒏, ∆𝜽 = 𝟓 𝐝𝐞𝐠𝐫𝐞𝐞𝐬 THANK YOU Muhammad Faisal, Ph.D. Associate Professor, Department of Science and Humanities [email protected] +91 80 50829629 ENGINEERING PHYSICS Department of Science and Humanities  ENGINEERING PHYSICS Unit I : Review of concepts leading to Quantum Mechanics Quick Recap! Phase consequences Analysis on velocity, De-Broglie’s matter matter waves - Vphase Hypothesis waves wave packet Group particle velocity, velocity, Vgroup Vparticle Heisenberg’s Inherent uncertainty-simultaneous Uncertainty Why to measurement Principle understand?! Applications: Non-existence 𝟏 𝒉 𝒉 𝝀𝟐 𝝀𝟐 ∆𝒙. ∆𝒌 ≥ ∆𝒙. ∆𝒑 ≥ ∆𝒙. ∆𝒗 ≥ ∆𝒙. ∆𝝀 ≥ − ≥ of electrons inside nucleus, 𝟐 𝟒𝝅 𝟒𝝅. 𝒎 𝟒𝝅 𝟒𝝅 gamma ray microscope 𝒉 𝒉 𝝀𝟐 𝝀𝟐 𝟏 𝒉 ∆𝒙𝒎𝒊𝒏. ∆𝒑𝒎𝒂𝒙 ≥ ∆𝐄. ∆𝐭 ≥ ∆𝒕. ∆𝝀 ≥ − ≥ ∆𝒕. ∆𝝂 ≥ ∆𝜽. ∆𝑳 ≥ ℏ/𝟐 ≥ 𝟒𝝅 𝟒𝝅 𝟒𝝅𝒄 𝟒𝝅𝒄 𝟒𝝅 𝟒𝝅 ENGINEERING PHYSICS Unit I : Review of concepts leading to Quantum Mechanics Week #3 Class #12 (As per less pl. class 9) Well behaved wave function Normalization and Probability concepts Wave function as a state function Double slit experiment revisited Linear Superposition of wave functions  ENGINEERING PHYSICS Unit I : Review of concepts leading to Quantum Mechanics Suggested Reading 1. Concepts of Modern Physics, Arthur Beiser, Chapter 5 2. Learning material prepared by the Department of Physics Reference Videos Video lectures : MIT 8.04 Quantum Physics I  ENGINEERING PHYSICS Wave functions Matter waves of moving bodies ( based on de-Broglie hypothesis) can represented by a wave function  (state of system in motion) - function of position and time - 𝚿 (𝒙, 𝒚, 𝒛, 𝒕) Three dimensional wave function in cartesian coordinates 𝚿 𝒙, 𝒚, 𝒛, 𝒕 = 𝝍(𝒙). 𝝓(𝒚). 𝝌(𝒛). 𝝋(𝒕) In general 𝝍(𝒙), 𝝓(𝒚), 𝝌(𝒛) are orthogonal functions  ENGINEERING PHYSICS Wave function and Probability density Wave function (probability amplitude) can be positive, negative or complex valued and can change with time Square of absolute magnitude of 𝝍 is called probability density (Max Born’s Approximation) Probability density represents probability of finding the particle in unit volume of space ForProbability a complexdensity must be a- Probability wave function positive real quantity