Atomic Structure & Quantum Numbers (PDF)

Part 3B 1 (https://saintschemistry10.weebly.com/history-of-the-atom.html) Outline 3.3 Schrodinger Equation for energy of Hydrogen Atom 3.4 Quantum Numbers 3.5 Orbital Magnetic moment (𝜇! )...

Part 3B 1 (https://saintschemistry10.weebly.com/history-of-the-atom.html) Outline 3.3 Schrodinger Equation for energy of Hydrogen Atom 3.4 Quantum Numbers 3.5 Orbital Magnetic moment (𝜇! ) 2  (%)' ! 3.3 Schrodinger Equation for energy of Hydrogen Atom Hydrogen-like ions, 𝑈 𝑟 = − ()*"+ 𝐻𝑒 , , 𝐿𝑖 ", , 𝐵𝑒 -, Schrodinger’s model (1D) In Schrodinger’s model of the hydrogen atom, the electron (-e) is in a potential energy trap due to its electrical attraction to the proton (+e) at the center of the atom. From 𝑞! 𝑞" (−𝑒)(+𝑒) 𝑈= = 4𝜋𝜖# 𝑟 4𝜋𝜖# 𝑟 Note: in reality 1. Potential well is 3D and finite (similar finite potential well). 2. Difficult to sharply define walls. 3. The walls vary in depth with radial distance r. ℏ! 𝑑 ! 𝜓 𝑥 − + 𝑈 𝑥 𝜓 𝑥 = 𝐸𝜓 𝑥 2𝑚 𝑑𝑥 ! 3 3.3 Schrodinger Equation for energy of Hydrogen Atom ไม่ ออกสอบ Wave function (3D) : Spherical polar coordinates R(r)= radial function Θ = polar function Φ = azimuthal function Spherical polar This procedure gives three coordination of H atom differential equations, each of a 𝑚𝑒 3 1 single variable (r, θ, φ). 𝐸2 = − 5 5 ) 5 8𝜀4 ℎ 𝑛 4 3.3 Schrodinger Equation for energy of Hydrogen Atom Schrodinger’s model From Schrodinger Eq. (Spherical coordinates.) Hydrogen-like atom./.1. 𝐸- = − 𝑒𝑉 𝐻𝑒 , , 𝐿𝑖 ", , 𝐵𝑒 -, -!./.1. 𝐸- = − (Z ) 2 -! 𝑒𝑉 Bohr Model 𝑚𝑒 3 1 (Z= atomic number) 𝐸2 = − 5 5 ) 5 8𝜀4 ℎ 𝑛./.1. 𝐸- = − 𝑒𝑉 -! 5 3.3 Schrodinger Equation for energy of Hydrogen Atom Example 1: Calculate the potential energy for a hydrogen atom at n=1 (1.602 𝑥 10.!/ 𝐶) E (−1.602 𝑥 10.!/ 𝐶) 𝑈= (4𝜋) E (8.85 𝑥 10.!" 𝐶 " /𝑁 E 𝑚" ) E (52.92 𝑥 10.!" 𝑚) Soln 𝑞! 𝑞" (−𝑒)(+𝑒) − (2.566 𝑥 10.-0 𝐶 " ) 𝑈= = = 4𝜋𝜀# 𝑟 4𝜋𝜀# 𝑟 (12.56) E (468.34 𝑥 10."( 𝐶 " /𝑁 E 𝑚) Permittivity of free space 𝜀# = 8.85 𝑥 10.!" 𝐶 " /𝑁 E 𝑚" = − 0.000436 𝑥 10.!( 𝑁 E 𝑚 Charge of proton e = 1.602 𝑥 10.!/ 𝐶 = − 4. 36 𝑥 10.!0 𝐽 Charge of electron −e = −1.602 𝑥 10.!/ 𝐶 𝑛 = 1 ⇒ 𝑟 = 𝑎# = 52.92 𝑝𝑚 = − 𝟒. 𝟑𝟔 𝒙 𝟏𝟎.𝟏𝟖 𝑱 − 4.4 𝑥 10.!0 = 𝑒𝑉 1.602 𝑥 10.!/ 𝑈 = −27. 216 𝑒𝑉 6 3.3 Schrodinger Equation for energy of Hydrogen Atom Example 1.1: According to previous problem, calculate the kinetic energy for a hydrogen atom at n=1. From Part 2(A) Slides 10-11 𝐸. = − 13.61 𝑒𝑉 𝐸 =𝐾+𝑈 (1) From previous slide 1 1 𝑒 ! = 𝑚𝑣 ! + (− 2 : ) 4𝜋𝜀" 𝑟 𝑈 = −27. 216 𝑒𝑉 (3) 3 −→ 1 , 𝑛 = 1./.1. − 13.61 = 𝐾 + (−27. 216) 𝐸- = − -! 𝑒𝑉 (2) 𝐾 ≈ 13.61 𝑒𝑉 7 3.3 Schrodinger Equation for energy of Hydrogen Atom Example 2: (1.602 𝑥 10.!/ 𝐶) E (−1.602 𝑥 10.!/ 𝐶) 𝑈= (4𝜋) E (8.85 𝑥 10.!" 𝐶 " /𝑁 E 𝑚" ) E (211.68 𝑥 10.!" 𝑚) Calculate the potential energy for a hydrogen atom at n=2 Soln − (2.566 𝑥 10.-0 𝐶 " ) = (12.56) E (1873.37 𝑥 10."( 𝐶 " /𝑁 E 𝑚) 𝑞! 𝑞" (−𝑒)(+𝑒) 𝑈= = 4𝜋𝜀# 𝑟 4𝜋𝜀# 𝑟 = − 0.000109 𝑥 10.!( 𝑁 E 𝑚 Permittivity of free space 𝜀# = 8.85 𝑥 10.!" 𝐶 " /𝑁 E 𝑚" = −1. 09 𝑥 10.!0 𝐽 Charge of proton e = 1.602 𝑥 10.!/ 𝐶 Charge of electron −e = −1.602 𝑥 10.!/ 𝐶 = −𝟏. 𝟎𝟗 𝒙 𝟏𝟎.𝟏𝟖 𝑱 𝑟 = (𝑎4)𝑛5 = − 1.09 𝑥 10.!0 𝑒𝑉 1.602 𝑥 10.!/ 𝑟 = 52.92 𝑝𝑚 ) 4 𝑈 = −6. 804 𝑒𝑉 𝑟 = 211.68 𝑝𝑚 8 3.3 Schrodinger Equation for energy of Hydrogen Atom Example 2.1: According to previous problem, calculate the kinetic energy for a hydrogen atom at n=2. From Part 2(A) Slides 10-11 𝐸5 = − 3.403 𝑒𝑉 𝐸 =𝐾+𝑈 (1) From previous slide 1 1 𝑒 ! 𝑈 = −6. 804 𝑒𝑉 (3) = 𝑚𝑣 ! + (− : ) 2 4𝜋𝜀" 𝑟 3 −→ 1 , 𝑛 = 2 −3.403 = 𝐾 + (−6. 804)./.1. 𝐾 ≈ 3.40 𝑒𝑉 𝐸- = − 𝑒𝑉 (2) -! 9 3.3 Schrodinger Equation for energy of Hydrogen Atom Schrodinger’s model (1D) 𝑞! 𝑞" (−𝑒)(+𝑒) 𝑈= = 4𝜋𝜖# 𝑟 4𝜋𝜖# 𝑟./.1. 𝐸- = − 𝑒𝑉 -! https://www.researchgate.net/figure/Coulomb-potential-in-the- Hydrogen-atom-showing-the-discrete-energy-states-that- 10 correspond_fig6_307416983 3.3 Schrodinger Equation for energy of Hydrogen Atom Schrodinger’s model (1D)./.1. 𝐸- = − 𝑒𝑉 -! 1) The lowest level, for n = 1, is the ground state of hydrogen. 2) Higher levels correspond to excited states. Note: 1) Energy levels have negative values. 2) The greatest value of n (𝒏 = ∞) is 𝑬# = 𝟎. 3) Any energy greater than 𝐸# (zero), the electron and proton are not bound together (no hydrogen atom) 4) At the E > 0 region, it is nonquantized region An energy-level diagram for the hydrogen11 atom. 3.4 Quantum Numbers (Orbital quantum number) 12 3.4 Quantum Numbers Quantum Number Principal 𝑛 Energy level. n = 1,2,3,… è size of the electron (shell) cloud è the energy level of the electron. Orbital Angular 𝑙 Shape of the orbital 𝑙 = 0, 1, 2, 3, … 𝑛 − 1 Determines the shape Momentum (subshell) of the orbital (e.g., l=0 (Azimuthal) for s, l=1 for p, l=2 for d, l=3 for f). Magnetic 𝑚X Orientation of the orbital 𝑙 = 0, 𝑚3 = 0 Specifies the orientation in space 𝑙 = 1, 𝑚3 = −1, 0, 1 of the orbital around 𝑙 = 2, 𝑚3 = −2, −1, 0, 1, 2 the nucleus. Spin 𝑚Y Spin of the electron 1 Direction of the ± 2 electron’s spin, creating magnetic properties 13 3.4 Quantum Numbers 14 3.4 Quantum Numbers 𝐴𝑡 𝑛 = 2 𝐴𝑡 𝑛 = 2 𝑙 = 0 (suborbital : s) 𝑙 = 1(suborbital : p) 𝑚X = −1, 0, 1 (2, 0, 0) (2, 1, 0) (2, 1, ±1) " ( 𝜓4,3,6# ) " " 15 ( 𝜓4,3,6# ) ( 𝜓4,3,6# ) 16 3.4 Quantum Numbers Quantum Numbers for the Hydrogen Atom 1. Energy of the hydrogen atom Each set of quantum numbers (n, l, ml) identifies à need principle quantum number n. the wave function of a particular quantum state. %&.(% 1) Principal quantum number (n) à Energy of the 𝐸$ = − 𝑒𝑉 $! state 2) Orbital Angular Momentum Quantum Number 2. Wave functions (l) à Magnitude of the angular momentum. (shape) à need 3 quantum numbers, corresponding to the three dimensions in which the electron can move. 3) Magnetic quantum number (ml) à Orientation in space of this angular momentum vector. 4) Spin Quantum Number (mₛ): Spin of the electron: +1/2 or -1/2. 17 3.4 Quantum Numbers Quantum Numbers for the Hydrogen Atom 1) Principal quantum number (n) 1. The energy of an electron and the size of the orbital in an atom. 2. n is the main shell of electron, n = 1(K), 2(L), 3(M), … 3. Maximum number of electrons in each shell =2𝑛! e.g. n=2, number of electron = 2(4) 𝐾 𝐿 M 18 3.4 Quantum Numbers 2) Orbital Angular Momentum Quantum Number (l) 𝑙 = 0, 1, 2, 3, … 𝑛 − 1 𝑽𝒂𝒍𝒖𝒆 𝒐𝒇 𝒍 𝑳𝒆𝒕𝒕𝒆𝒓 à Determining the angular distribution of the electron's probability density around the nucleus. 𝟎 𝒔 à The shape of the electron's orbital (subshell). 𝟏 𝒑 𝟐 𝒅 𝟑 𝒇 19 3.4 Quantum Numbers subshell 3) Magnetic quantum number (ml) 𝒍 - Orientation in space of this angular momentum vector. 𝟎 𝒔 -There are (𝟐𝒍 + 𝟏) magnetic quantum numbers containing orbitals present in each subshell. 𝑙 = 0; 𝑚3 = 0 𝟏 𝒑 𝑙 = 1; 𝑚3 = 0, ±1 𝟐 𝒅 𝑙 = 2; 𝑚3 = 0, ±1, ±2 𝟑 𝒇 𝑙 = 3; 𝑚3 = 0, ±1, ±2, ±3 20 3.4 Quantum Numbers 3) Magnetic quantum number (ml) ml : describes how the orbital behaves in the presence of a external magnetic field. No external magnetic field: Orbitals with the same principal (n) and azimuthal (l) quantum numbers but different magnetic quantum numbers (ml) have the same energy. With external magnetic field: The orbitals split into different energy levels according to the value of ml. (splitting due to the interaction between the magnetic field and the magnetic moment.) Normal Zeeman effect 21 3.4 Quantum Numbers Angular Momentum (L) & Angular momentum in the z-direction (𝑳𝒛) Bohr condition (n=1,2,3,…) (magnitude) Here, Angular momentum is (magnitude) 22 3.4 Quantum Numbers Angular Momentum (L) & Angular momentum in the z-direction (𝑳𝒛) (magnitude) 23 3.4 Quantum Numbers Angular Momentum (L) & Angular momentum in the z-direction (𝑳𝒛) For example, 𝐿m = ? , 𝜃 = ? at 𝑙 = 1 Soln. 𝑳 = 1(1 + 1)ℏ = 2ℏ For each 𝑙, there are 2𝑙 + 1 of 𝑚). 𝑚) = −1, 0, +1 cos 𝜃 = −1/ 2, 0, +1/ 2 𝜃 = 135°, 90°, 45° 𝐿* = −ℏ, 0, +ℏ 24 3.4 Quantum Numbers Angular Momentum (L) & Angular momentum in the z-direction (𝑳𝒛) 𝑳 =? , 𝑎𝑡 𝑙 = 1 𝑎𝑛𝑑 𝑙 = 2 𝑭𝒓𝒐𝒎 𝑳 = 𝑙(𝑙 + 1)ℏ A𝑡 𝑙 = 1 𝑳 = 1(1 + 1)ℏ = 2ℏ A𝑡 𝑙 = 2 𝑳 = 2(2 + 1)ℏ = 6ℏ 25  3.3 Quantum Numbers Angular Momentum (L) & Angular momentum in the z-direction (𝑳𝒛) 3 𝒂) 𝒍 = 3 à 𝑳𝒛 = ? Homework 𝑚) = −3, −2, −1, 0, 1, 2, 3 b) The length of the vector 𝐿 =? From, c) 𝜃 value of each z-component =? d) Drawing the orientations in space and the z components of vector 𝐿. 𝐿7 = −3ℏ, −2ℏ, −1ℏ, 0, ℏ, 2ℏ, 3ℏ 26 3.4 Quantum Numbers Quantum Numbers for the Hydrogen Atom 4) Spin quantum number (𝒎𝒔 ) a fundamental property that describes subatomic particles , e.g. electrons. 1 1 𝑚8 = , − 2 2 u u "spin-up" = 5 , "spin-down" = −5 27 3.5 Orbital Magnetic moment (𝜇! ) Magnetic Moments Classical physics view (Object with intrinsic magnetic properties ) Magnetic moment (𝝁) is used to measure the tendency of an object to interact with an external magnetic field (𝑩). Object: molecule, atom, nucleus, or subatomic particle. Magnetic moment (𝝁) in bar Magnetic moment in a magnet with north and south poles current loop (right-hand rule) (dipoles) may call the magnetic dipole moment. 28 (magnitude) 3.5 Orbital Magnetic moment (𝜇! ) Orbital Magnetic Moments (𝜇z ) Classical physics view (charged particle orbits around the nucleus) If vector 𝜇⃑ is exposure to external magnetic No external magnetic filed With external magnetic filed (𝐵) filed 𝐵, it will experience the torque 𝜏⃑ (direction: upward) (twisting force), making the alignment of 𝜇. ⃑ 𝜏⃑ = 𝜇⃑ 𝑥𝐵 𝜏⃑ = 𝜇z 𝑥𝐵 𝜏 = 𝜇z 𝐵𝑠𝑖𝑛𝜃 The magnetic moment vector The angle 𝜃 is the angle between the vector 𝜇⃑ (𝜇) ⃑ attempts to align with the and external field 𝐵. magnetic filed (𝐵) 29  3.5 Orbital Magnetic moment (𝜇! ) From, Orbital Magnetic moment (𝜇z ) 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 From, 𝒕= 1. Circulating electron as a circular loop à current 𝒗 𝜇 = 𝐼𝐴 𝒒 𝟐𝝅𝒓 𝑰= (1) 𝒕= (𝟑) 𝑞 𝒕 𝒗 =( )(𝜋𝑟 " ) q = charge of the electron 2𝜋𝑟𝑚 t = time for one cycle (also period) 𝐸𝑞. 2 −→ 𝐸𝑞. (3) 𝑝 𝑞 𝒑 𝟐𝝅𝒓 𝟐𝝅𝒓𝒎 =( )(𝜋𝑟 " ) 𝒗= (𝟐) 𝑡= = 2𝜋𝑟𝑚 𝒎 (𝒑/𝒎) 𝒑 𝑝 𝑞 2. If the electron moves with speed 𝒗 around the 𝟐𝝅𝒓𝒎 = 𝑟𝑝 loop radius ‘r’ 𝑡= (𝟒) 2𝑚 𝒑 𝑞 𝜇= E𝐿 2𝑚 (magnitude of the magnetic moment) For electron with vector form, −𝑒 𝝁𝑳 = )𝑳 30 2𝑚 3.5 Orbital Magnetic moment (𝜇! ) Orbital Magnetic moment (𝜇z ) −𝑒 𝝁𝑳 = )𝑳 2𝑚 For electron with vector form, (negative sign tells the direction, see the direction of 𝝁𝑳 and 𝑳 ) −𝑒 𝜇= -𝐿 2𝑚 31 3.5 Orbital Magnetic moment (𝜇! ) Bohr Magneton 𝑒 𝜇z = − )𝐿 2𝑚 𝑒 =− 𝑙 𝑙+1 ℏ 2𝑚 𝒆ℏ =− 𝑙 𝑙+1 𝟐𝒎 Bohr Magneton A reference for expressing the 𝜇z = −𝝁~ 𝑙 𝑙 + 1 magnetic moments of 𝑒ℏ electrons due to their orbital (negative sign tells the direction) 𝜇~ = or spin motion. 2𝑚 𝜇~ = 9.27𝑥10 53 𝐽/𝑇. (T=tesla) 32