Modern Physics

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PHY251 Homework Set 8 Reading: Chapter 9 Homework: Chapter 9, Question 2, Problems 11,14,15,19; (extra credit: 23) Hints and Solutions Question IX.2 (10 points) In the quantum theory it is generally not possible to specify all three components of the angular momentum simultaneously. Is the l =0 angular momentum state an exception to this rule? Hints: No hints Solution: Yes, for l =0 the magnitude of the angular momentum is zero, and hence all three components are trivially zero as well. However, this state does not violate the Heisenberg uncertainty relationship because the position and the momentum of the electron in a l =0 state are still uncertain as required by the rule p x · x h /2. Problem IX.11 (15 points) Find r ² for the ground state of hydrogen. Compare your result with r ² for the same state. Hints: See example 9-6. Solution: r ² = r ²| ψ 100 ( r )| 2 dV = r ²| R 10 ( r ) Y 00 | 2 dV = ∫∫∫ r ²| R 10 ( r ) Y 00 | 2 r ² dr sin θ d θ d φ = = r ²|2 a 0 - 3/2 e - r / a 0 ( 1 / 4 π )| 2 r ² dr sin θ d θ d φ = [ π - 1 a 0 - 3 r 4 e -2 r / a 0 dr ] 4 π = = 4 a 0 - 3 r 4 e -2 r / a 0 dr = 4 a 0 - 3 4!/(2/ a 0 ) 5 = (96/32) a 0 ² = 3 a 0 ². The value r : r = r | ψ 100 ( r )| 2 dV = r | R 10 ( r ) Y 00 | 2 dV = ∫∫∫ r | R 10 ( r ) Y 00 | 2 r ² dr sin θ d θ d φ = = r |2 a 0 - 3/2 e - r / a 0 ( 1 / 4 π )| 2 r ² dr sin θ d θ d φ = [ π - 1 a 0 - 3 r ³ e -2 r / a 0 dr ] 4 π = phy251_hw08_solutions file:///C|/DOCUME~1/PHY_CO~1/PHY251/PHY251_hw08_solutions.html (1 of 4) [4/14/2001 4:19:26 AM]
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= 4 a 0 - 3 r ³ e -2 r / a 0 dr = 4 a 0 - 3 3!/(2/ a 0 ) 4 = (24/16) a 0 = 3 / 2 a 0 . Hence: r ² = 9 / 4 a 0 ² < r ² In the above we have used: 0 x n e - ax dx = n !/ a n +1 Problem IX.14 (15 points) Show that in the state n =2, l =1, and m =0, r = 5 a 0 / Z . Set Z
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