Answer on Question #51446, Physics, Solid State Physics
Write down the wave function ψ 211 \psi_{211} ψ 211
Solution:
The wave function ψ 211 1 \psi_{211} 1 ψ 211 1
ψ 211 = 1 2 6 a 0 3 ⋅ r a 0 exp [ − r 2 a 0 ] 3 8 π sin θ e i φ \psi_{211} = \frac{1}{2\sqrt{6a_0^3}} \cdot \frac{r}{a_0} \exp\left[-\frac{r}{2a_0}\right] \sqrt{\frac{3}{8\pi}} \sin \theta e^{i\varphi} ψ 211 = 2 6 a 0 3 1 ⋅ a 0 r exp [ − 2 a 0 r ] 8 π 3 sin θ e i φ
The probability current density for this state is given by Eq.(2).
j ˉ = i e ℏ 2 m 0 ( ψ ∇ ψ ∗ − ψ ∗ ∇ ψ ) \bar{j} = \frac{\mathrm{i} \mathrm{e} \hbar}{2 \mathrm{m}_0} \left(\psi \nabla \psi^* - \psi^* \nabla \psi\right) j ˉ = 2 m 0 ie ℏ ( ψ ∇ ψ ∗ − ψ ∗ ∇ ψ )
So,
j r = i e ℏ 2 m 0 ( ψ ∂ ψ ∗ ∂ r − ψ ∗ ∂ ψ ∂ r ) = 0 j_r = \frac{\mathrm{i} \mathrm{e} \hbar}{2 \mathrm{m}_0} \left(\psi \frac{\partial \psi^*}{\partial \mathrm{r}} - \psi^* \frac{\partial \psi}{\partial \mathrm{r}}\right) = 0 j r = 2 m 0 ie ℏ ( ψ ∂ r ∂ ψ ∗ − ψ ∗ ∂ r ∂ ψ ) = 0 j θ = i e ℏ 2 m 0 1 r ( ψ ∂ ψ ∗ ∂ θ − ψ ∗ ∂ ψ ∂ θ ) = 0 j_\theta = \frac{\mathrm{i} \mathrm{e} \hbar}{2 \mathrm{m}_0} \frac{1}{\mathrm{r}} \left(\psi \frac{\partial \psi^*}{\partial \theta} - \psi^* \frac{\partial \psi}{\partial \theta}\right) = 0 j θ = 2 m 0 ie ℏ r 1 ( ψ ∂ θ ∂ ψ ∗ − ψ ∗ ∂ θ ∂ ψ ) = 0 j φ = i e ℏ 2 m 0 1 r sin θ ( ψ ∂ ψ ∗ ∂ φ − ψ ∗ ∂ ψ ∂ φ ) = j_\varphi = \frac{\mathrm{i} \mathrm{e} \hbar}{2 \mathrm{m}_0} \frac{1}{\mathrm{r} \sin \theta} \left(\psi \frac{\partial \psi^*}{\partial \varphi} - \psi^* \frac{\partial \psi}{\partial \varphi}\right) = j φ = 2 m 0 ie ℏ r sin θ 1 ( ψ ∂ φ ∂ ψ ∗ − ψ ∗ ∂ φ ∂ ψ ) = i e ℏ 2 m 0 1 r sin θ 1 2 6 a 0 3 ⋅ r a 0 exp [ − r 2 a 0 ] 3 8 π sin θ ( − i e i φ e − i φ − i e − i φ e i φ ) = \frac{\mathrm{i} \mathrm{e} \hbar}{2 \mathrm{m}_0} \frac{1}{\mathrm{r} \sin \theta} \frac{1}{2\sqrt{6 \mathrm{a}_0^3}} \cdot \frac{\mathrm{r}}{\mathrm{a}_0} \exp\left[-\frac{\mathrm{r}}{2 \mathrm{a}_0}\right] \sqrt{\frac{3}{8\pi}} \sin \theta \left(-\mathrm{i} \mathrm{e}^{\mathrm{i} \varphi} \mathrm{e}^{-\mathrm{i} \varphi} - \mathrm{i} \mathrm{e}^{-\mathrm{i} \varphi} \mathrm{e}^{\mathrm{i} \varphi}\right) = 2 m 0 ie ℏ r sin θ 1 2 6 a 0 3 1 ⋅ a 0 r exp [ − 2 a 0 r ] 8 π 3 sin θ ( − i e i φ e − i φ − i e − i φ e i φ ) = = e ℏ 2 a 0 m 0 6 a 0 3 ⋅ exp [ − r 2 a 0 ] 3 8 π = \frac{\mathrm{e} \hbar}{2 \mathrm{a}_0 \mathrm{m}_0 \sqrt{6 \mathrm{a}_0^3}} \cdot \exp\left[-\frac{\mathrm{r}}{2 \mathrm{a}_0}\right] \sqrt{\frac{3}{8\pi}} = 2 a 0 m 0 6 a 0 3 e ℏ ⋅ exp [ − 2 a 0 r ] 8 π 3
Answer: e ℏ 2 a 0 m 0 6 a 0 3 ⋅ exp [ − r 2 a 0 ] 3 8 π \frac{\mathrm{e} \hbar}{2 \mathrm{a}_0 \mathrm{m}_0 \sqrt{6 \mathrm{a}_0^3}} \cdot \exp\left[-\frac{\mathrm{r}}{2 \mathrm{a}_0}\right] \sqrt{\frac{3}{8\pi}} 2 a 0 m 0 6 a 0 3 e ℏ ⋅ exp [ − 2 a 0 r ] 8 π 3
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#339914 on Dec 2023
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