思考题和习题解答 解:(1)v2da 4IN20r2e-2a dr=4IN2. 2! 4N22 Wy=常数v ar rasin 0 de sine o 4cr 10r2e|-a) uLe ' A (2)本征值 I - uao 6626×10-Js) 8n2×9104×10-kg×5292×102m) =-2.18×10-8J=-136eV·162· 思考题和习题解答 解：(1) ψ τ 2 d 1 ∫ = ( ) ( ) 1 π 2 2 4π 2 2! 4π e d 4π 3 2 3 2 3 2 0 2 2 2 ∫ = ⋅ = = = ∞ − α α α α N N r r N N r ∴ π 3 α N = H$ψ ψ = 常数 ( ) ψ ε α α μ ψ ε α μ ψ ε ψ θ θ φ θ μ θ θ α ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ − ⎠ ⎞ ⎜ ⎝ ⎛ = − − − − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ = − − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − ⋅ − r e r r e r r r N r e r r r r r r r 0 2 2 2 0 2 2 2 2 0 2 2 2 2 2 2 2 2 2 4π 2 2 4π e 1 2 sin 4π 1 sin sin 1 1 2 h h h ∴ 0 4π 0 2 2 − = r e μr ε αh 0 2 0 2 π 1 h a e = = ε μ α 3 π 0 1 a N = (2) 2 0 2 2 2 2 8π 2 a h μ μ α = − = − h 本征值 ( ) ( )2 2 31 -12 2 34 8π 9.104 10 kg 52.92 10 m 6.626 10 J s × × × × × ⋅ = − − − 2.18 10 J 13.6 eV 18 = − × = − −