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Unformatted text preview: HOMEWORK ASSIGNMENT 5 PHYS852 Quantum Mechanics I, Spring 2008 Topics covered: Fine structure, hyperfine structure, timedependent perturbation theory 1. Consider a hydrogen atom in the ground state n = 1, ℓ = 0. Using the classical kinetic energy formula T = Mv 2 2 , at what radius does the electrons velocity begin to exceed 10% of the speed of light? Using the ground state orbital R 10 ( r ), compute the probability of finding the electron inside this radius. Do the same for the 2S and 2P levels. If you were to generalize these results, would you expect relativistic corrections to increase or decrease with n ? What about ℓ ? Answer : We start from E = T + V , which gives − V = T − E . Setting v = c// 10 gives e 2 4 πǫ r = Mc 2 200 + planckover2pi1 2 2 Ma 2 n 2 Solving for r gives r n = 50 a 2 e 2 Mn 2 (100 planckover2pi1 2 + a 2 M 2 c 2 n 2 ) πǫ where a = 5 . 29177 × 10 − 11 m e = 1 . 60218 × 10 − 19 C M = 9 . 10442 × 10 − 31 kg (note this is the reduced mass M = M e M p / ( M e + M p )) c = 299792458m/s This gives r 1 = 5 . 60605 × 10 − 13 . r 2 = 5 . 62841 × 10 − 13 . We note the these two radii differ by less than a percent. The probability to be within this radius is P ( n,ℓ ) = integraldisplay r n dr r 2  R nℓ ( r )  2 = parenleftbigg 2 na parenrightbigg 3 ( n − ℓ − 1)! 2 n ( n + ℓ )! integraldisplay r n dr r 2 e − 2 r/na parenleftbigg 2 r na parenrightbigg 2 ℓ vextendsingle vextendsingle vextendsingle vextendsingle L 2 ℓ +1 n − ℓ − 1 parenleftbigg 2 r na parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle 2 = ( n − ℓ − 1)! 2 n ( n + ℓ )! integraldisplay 2 rn na duu 2+2 ℓ e − u vextendsingle vextendsingle vextendsingle L 2 ℓ +1 n − ℓ − 1 ( u ) vextendsingle vextendsingle vextendsingle 2 Integration gives P (1 , 0) = 1 . 56031 × 10 − 6 P (2 , 0) = 1 . 97366 × 10 − 7 P (3 , 2) = 1 . 12433 × 10 − 2 From this we can surmise that relativistic effects should decrease with n , and decrease very strongly with ℓ . The reason for the decrease with n is that as n increases, the classical turning point increases, giving the electron access to a larger nonrelativistic region. For ℓ it is clear that nonzero angular momentum strongly suppresses access to small radii, due to the repulsive angular momentum term in the radial equation. 1 2. Compute the complete firstorder fine structure shift of the 1S state. One can separate the spinorbit term into radial and angular parts giving ( vector L · vector S R 3 ) = ( nℓ  R − 3  nℓ )( ℓm ℓ m 2  vector L · vector S  ℓm ℓ m s ) The angular part vanishes due to an exact symmetry, while the radial part is infinite due to the divergence of the contribution from very small radius. At such small radius, particularly for S orbitals, relativistic corrections beyond the P 4 term come into effect. So we can assume that additional relativistic effects will constrain the radial term to a large but finite value. Thus the spinorbit termrelativistic effects will constrain the radial term to a large but finite value....
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This note was uploaded on 11/26/2010 for the course PHYSICS PHYS 852 taught by Professor Michaelmoore during the Spring '10 term at Michigan State University.
 Spring '10
 MichaelMoore
 mechanics, Energy, Kinetic Energy, Work

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