3 Explanation Given Information: The angles between S and μ s and between L and μ L are 180 o . Formula Used: We know, magnitudes of orbital angular momentum L is given by- L = l ( l + 1 ) ℏ Also, magnitudes of spin angular momentum S is given by- S = s ( s + 1 ) ℏ ........... ( 1 ) Andmagnitudes of total angular momentum J is given by- J = j ( j + 1 ) ℏ Where ℏ represents Planck’s reduced constant and l , s and j represent orbital angular momentumquantum number,spin angular momentumquantum number and total angular momentumquantum number respectively Total angular momentum vector J is given by- J = L + S Magnetic dipole vector μ J will be- μ J = − e 2 m e ( L + 2 S ) ........ ( 2 ) Here L represents orbital angular momentum vector, S representsspin angular momentum vector, e representsmagnitude charge and m e representsmass of electron. Dot product of vectors A and B is: A ⋅ B = | A | | B | cos θ Here | A | and | B | represent magnitudes of vectors A and B respectively, and θ is the angle between the vectors. Calculation: To find the angle between J and μ the dot product is used and solved for the angle. J ⋅ μ J = | J | | μ J | cos θ cos θ = J ⋅ μ J | J | | μ J | θ = cos − 1 ( J ⋅ μ J | J | | μ J | ) That is then rewritten in terms of L and S using equations ( 2 ) . θ = cos − 1 ( J ⋅ μ J | J | | μ J | ) = cos − 1 [ ( L + S ) ⋅ { − e 2 m e ( L + 2 S ) } | J | | μ J | ] = cos − 1 [ − e 2 m e ( L + S ) ⋅ ( L + 2 S ) | J | | μ J | ] = cos − 1 [ − e 2 m e L ⋅ L + 2 S ⋅ S + 3 L ⋅ s | J | | μ J | ] Since the dot product of a vector with itself is just ¡its magnitude squared ( A ⋅ A = A 2 ) , the equation can be simplified slightly: θ = cos − 1 [ − e 2 m e L ⋅ L + 2 S ⋅ S + 3 L ⋅ S | J | | μ J | ] = cos − 1 [ − e 2 m e L 2 + 2 S 2 + 3 L ⋅ S J μ J ⋅ μ J ] The | J | was able to just be written as J since we already have an expression for that from equation ( 1 ) . That can be rewritten again with equation ( 2 ) , and simplified: θ = cos − 1 [ − e 2 m e L 2 + 2 S 2 + 3 L ⋅ S | J | | μ J | ] = cos − 1 [ − e 2 m e L 2 + 2 S 2 + 3 L ⋅ S J [ − e 2 m e ( L + 2 S ) ] . [ − e 2 m e ( L + 2 S ) ] ] = cos − 1 [ − e 2 m e L 2 + 2 S 2 + 3 L ⋅ S J ( e 2 m e ) 2 ( L ⋅ L + 4 S ⋅ S + 4 L ⋅ S ) ] That the dot product of a vector with itself is just its magnitude squared ( A ⋅ A = A 2 ) is used again. θ = cos − 1 [ − e 2 m e L 2 + 2 S 2 + 3 L ⋅ S J ( e 2 m e ) 2 ( L ⋅ L + 4 S ⋅ S + 4 L ⋅ S ) ] = cos − 1 [ − e 2 m e L 2 + 2 S 2 + 3 L ⋅ S J ( e 2 m e ) ( L ⋅ L + 4 S ⋅ S + 4 L ⋅ S ) ] = cos − 1 [ − L 2 + 2 S 2 + 3 L ⋅ S J ( L 2 + 4 S 2 + 4 L ⋅ S ) ] ......... ( 3 ) In order to proceed further, you need an expression for the spin-orbit coupling L . S . That can be gotten from the dot product of J with itself, using equation ( 2 ) : J = L + S J ⋅ J = ( L + S ) ⋅ ( L + S ) J ⋅ J = L ⋅ L + S ⋅ S + 2 L ⋅ S 2 L ⋅ S = J ⋅ J − L ⋅ L − S ⋅ S The self-dot products can then just be rewritten as the magnitudes squared again, and simplified 2 L ⋅ S = J ⋅ J − L ⋅ L − S ⋅ S 2 L ⋅ S = J 2 − L 2 − S 2 L ⋅ S = J 2 − L 2 − S 2 2 That can then be inserted into equation ( 3 ) for L ⋅ S θ = cos − 1 [ − L 2 + 2 S 2 + 3 L ⋅ S J L 2 + 4 S 2 + 4 L ⋅ S ] = cos − 1 [ − L 2 + 2 S 2 + 3 ( J 2 − L 2 − S 2 2 ) J L 2 + 4 S 2 + 4 ( J 2 − L 2 − S 2 2 ) ] = cos − 1 [ − L 2 + 2 S 2 + 3 2 ( J 2 − L 2 − S 2 ) J L 2 + 4 S 2 + 2 ( J 2 − L 2 − S 2 ) ] It’s then distributed and simplified

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Author: Randy Harris

Publisher: Addison Wesley

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The Bohr model

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Chapter 8, Problem 73E

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Angles between J and μL for 2p32 state of hydrogen.

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Chapter 8, Problem 72E

Chapter 8, Problem 74E

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Chapter 8 Solutions

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Angles between J and μ L for 2 p 3 2 state of hydrogen.

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Angles between J and μL for 2p32 state of hydrogen.

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