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EXAMPLE 6.3 Verify that (6.75) is the ground-state energy eigenfunction for the hydrogen atom with the appropriate ground-state energy. SOLUTION Since the ground-state has 7 = 0, the radial Schrödinger equation (6.61) is given by or ħ² 2 d [-(+²) - 2m dr.² r dr 4π for ħ2² 2m 2 (ar² + ² dr) dr.2 r e² E =- e² 4πeor e-r/ao = Ee-r/ao where we have divided out the normalization constant from both sides of the equa- tion and have set Z = 1. Taking the derivatives, we obtain ħ2² 1 2 e² e-r/ao = Ee-r/ao 2m a aor 4π for Notice that the 1/r terms multiplying the wave function on the left-hand side must cancel so we end up with a constant. Therefore, ħ² 2ma R1,0 = ER1,0 ħ² e² mao 4+€0 consistent with (6.76). Consequently, the remaining constant on the left-hand side must equal the energy E: me4 2(4π €0)²h²

EXAMPLE 6.3 Verify that (6.75) is the ground-state energy eigenfunction for the hydrogen atom with the appropriate ground-state energy. SOLUTION Since the ground-state has 7 = 0, the radial Schrödinger equation (6.61) is given by or ħ² 2 d [-(+²) - 2m dr.² r dr 4π for ħ2² 2m 2 (ar² + ² dr) dr.2 r e² E =- e² 4πeor e-r/ao = Ee-r/ao where we have divided out the normalization constant from both sides of the equa- tion and have set Z = 1. Taking the derivatives, we obtain ħ2² 1 2 e² e-r/ao = Ee-r/ao 2m a aor 4π for Notice that the 1/r terms multiplying the wave function on the left-hand side must cancel so we end up with a constant. Therefore, ħ² 2ma R1,0 = ER1,0 ħ² e² mao 4+€0 consistent with (6.76). Consequently, the remaining constant on the left-hand side must equal the energy E: me4 2(4π €0)²h²

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Prove that the equation provided in the first image is an energy eigenfunction for the hydrogen atom with the appropriate eigenvalue.

(Hint: See the provided Example 6.3)

R₂,1(r) = √32a0. 3/2 Zr ² Zre-Zr/2a0 do
Transcribed Image Text:R₂,1(r) = √32a0. 3/2 Zr ² Zre-Zr/2a0 do
R1,0(r) = 2 or Z 3/2 e ao EXAMPLE 6.3 Verify that (6.75) is the ground-state energy eigenfunction for the hydrogen atom with the appropriate ground-state energy. -Zr/ao SOLUTION Since the ground-state has / = 0, the radial Schrödinger equation (6.61) is given by ħ² 2m dr2 ħ² d² 2 d e² [ - 200 (1 + r) - Ter] RID = ERLA R1,0 ER1,0 2m dr2 r dr 4π d² 2 d + r dr (6.75) e² 4лor e-r/ao = Ee-r/ao where we have divided out the normalization constant from both sides of the equa- tion and have set Z = 1. Taking the derivatives, we obtain 2 e² [-(-3) 2 2m af aor E -r/ao = Notice that the 1/r terms multiplying the wave function on the left-hand side must cancel so we end up with a constant. Therefore, = Ee-r/ao ħ² e² mao 4л€0 consistent with (6.76). Consequently, the remaining constant on the left-hand side must equal the energy E: me4 ħ² 2ma 2(4π €0)²h² consistent with (6.72). Notice how solving the Schrödinger equation determines both the energy and the size (through ao) of the atom.
Transcribed Image Text:R1,0(r) = 2 or Z 3/2 e ao EXAMPLE 6.3 Verify that (6.75) is the ground-state energy eigenfunction for the hydrogen atom with the appropriate ground-state energy. -Zr/ao SOLUTION Since the ground-state has / = 0, the radial Schrödinger equation (6.61) is given by ħ² 2m dr2 ħ² d² 2 d e² [ - 200 (1 + r) - Ter] RID = ERLA R1,0 ER1,0 2m dr2 r dr 4π d² 2 d + r dr (6.75) e² 4лor e-r/ao = Ee-r/ao where we have divided out the normalization constant from both sides of the equa- tion and have set Z = 1. Taking the derivatives, we obtain 2 e² [-(-3) 2 2m af aor E -r/ao = Notice that the 1/r terms multiplying the wave function on the left-hand side must cancel so we end up with a constant. Therefore, = Ee-r/ao ħ² e² mao 4л€0 consistent with (6.76). Consequently, the remaining constant on the left-hand side must equal the energy E: me4 ħ² 2ma 2(4π €0)²h² consistent with (6.72). Notice how solving the Schrödinger equation determines both the energy and the size (through ao) of the atom.
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