logoChat01.ai

一维势阱能量计算

बनाया गया: 11 जनवरी 2025

का उपयोग करके उत्तर दिया गया O1 द्वारा Chat01

प्रश्न

设有一电子在宽为 0.20nm 的一维无限深的方势阱中(1)计算电子在最低能级的能量?

विचार

Recalling the energy formula

I’m thinking about the energy levels in a one-dimensional infinite potential well, specifically the ground state, using En=n2π2ℏ22mL2E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}.

Calculating energy

I’m noting the constants: electron mass, ℏ\hbar, and well width. Substituting these into E1=π2ℏ22mL2E_1 = \frac{\pi^2 \hbar^2}{2 m L^2} for the ground state energy.

Tackling the denominator

I’m resolving the denominator by multiplying the electron mass, well width squared, and factoring in π\pi, reaching 7.2872×10−507.2872 \times 10^{-50}.

Calculating energy

I’m working through the numerator and denominator, factoring in π2\pi^2. The energy calculation focuses on the correctness of constants and coefficients, resulting in approximately 1.5×10−181.5 \times 10^{-18} J.

Refining the expression

I’m honing the energy formula, ensuring precise constants and coefficients, and arriving at a simplified expression for E1E_1 in joules.

Crunching the numbers

I’m carefully multiplying and simplifying the expressions for the numerator and denominator, ensuring precision with constants and coefficients, and ultimately calculating the ground state energy E1E_1 in joules.

Converting energy

OK, let me see. I’m mapping the ground state energy from joules to electron volts by dividing by the electron charge, confirming the outcome is around 9.4 eV.

Converting energy

I’m verifying the calculated ground state energy of around 9.4 eV by cross-checking typical formulas for infinite wells and noting consistency with known constants and electron charge.

Thought about 电子在方势阱中的能量 for 28 seconds

उत्तर

问题背景
在一维无限深方势阱（又称无限深方势井）模型中，势阱的势能函数可写为

V(x)={0,0<x<L∞,其它区域V(x) = \begin{cases} 0, & 0 < x < L\\ \infty, & \text{其它区域} \end{cases}

其中 LL 是势阱的宽度。本题中，给定电子所处的一维无限深方势阱宽度为

L=0.20 nm=0.20×10−9 m.L = 0.20 \,\mathrm{nm} = 0.20\times 10^{-9}\,\mathrm{m}.

1. 能级公式

对于质量为 mm 的粒子（此处为电子，质量记为 mem_e）在一维无限深方势阱中的定态解，能量本征值为

En  =  n2π2ℏ22 m L2,n=1,2,3,…E_n \;=\; \frac{n^2 \pi^2 \hbar^2}{2\,m\,L^2}, \quad n = 1,2,3,\dots

其中

  • ℏ=h2π≈1.054 571 817×10−34 J⋅s\hbar = \frac{h}{2\pi} \approx 1.054\,571\,817\times 10^{-34}\,\mathrm{J\cdot s} 为约化普朗克常数，
  • mm 为电子质量 me≈9.109×10−31 kgm_e \approx 9.109\times 10^{-31}\,\mathrm{kg}，
  • LL 为势阱宽度。

本题只要求最低能级（基态，n=1n=1）时的能量，即

E1  =  π2ℏ22 m L2.E_1 \;=\; \frac{\pi^2 \hbar^2}{2\,m\,L^2}.

2. 数值代入与计算

  1. 常数与参数：

    • 电子质量 me≈9.109×10−31 kg,m_e \approx 9.109 \times 10^{-31}\,\mathrm{kg},
    • 约化普朗克常数 ℏ≈1.054 571 817×10−34 J⋅s,\hbar \approx 1.054\,571\,817\times 10^{-34}\,\mathrm{J\cdot s},
    • 势阱宽度 L=0.20×10−9 m=2.0×10−10 m.L = 0.20\times 10^{-9}\,\mathrm{m} = 2.0\times 10^{-10}\,\mathrm{m}.

  2. 逐步计算（在 SI 制中得到结果，单位：焦耳）

    E1=π2 ℏ22 me L2E_1 = \frac{\pi^2\,\hbar^2}{2\,m_e\,L^2}
    • ℏ2=(1.054 571 817×10−34)2≈1.112×10−68 J2⋅s2\hbar^2 = (1.054\,571\,817\times 10^{-34})^2 \approx 1.112\times 10^{-68}\,\mathrm{J^2\cdot s^2};
    • π2≈9.8696\pi^2 \approx 9.8696;
    • 分子部分 π2 ℏ2  ≈  9.8696×1.112×10−68  ≈  1.10×10−67 J2⋅s2;\pi^2\, \hbar^2 \;\approx\; 9.8696 \times 1.112\times 10^{-68} \;\approx\; 1.10\times 10^{-67}\,\mathrm{J^2\cdot s^2};
    • 分母部分 2 me L2  =  2×9.109×10−31 kg×(2.0×10−10 m)2.2 \, m_e \, L^2 \;=\; 2 \times 9.109\times 10^{-31}\,\mathrm{kg} \times (2.0\times 10^{-10}\,\mathrm{m})^2. (2.0×10−10 m)2=4.0×10−20 m2;(2.0\times 10^{-10}\,\mathrm{m})^2 = 4.0\times 10^{-20}\,\mathrm{m}^2; meL2=9.109×10−31 kg×4.0×10−20 m2=3.6436×10−50 kg⋅m2;m_e L^2 = 9.109\times 10^{-31}\,\mathrm{kg} \times 4.0\times 10^{-20}\,\mathrm{m^2} = 3.6436\times 10^{-50}\,\mathrm{kg\cdot m^2}; 2 me L2≈7.2872×10−50 kg⋅m2.2\,m_e\,L^2 \approx 7.2872\times 10^{-50}\,\mathrm{kg\cdot m^2}.
    • 因此 E1  ≈  1.10×10−677.2872×10−50  =  1.51×10−18 J.E_1 \;\approx\; \frac{1.10\times 10^{-67}}{7.2872\times 10^{-50}} \;=\; 1.51\times 10^{-18}\,\mathrm{J}.

  3. 结果转换为电子伏特（eV\mathrm{eV}）

    1 eV = 1.602×10−191.602\times 10^{-19} J。
    因此

    E1(eV)  =  1.51×10−18 J1.602×10−19 J/eV  ≈  9.4 eV.E_1 \text{(eV)} \;=\; \frac{1.51\times 10^{-18}\,\mathrm{J}}{1.602\times 10^{-19}\,\mathrm{J/eV}} \;\approx\; 9.4\,\mathrm{eV}.

3. 结论

  • 以焦耳计：

    E1≈1.51×10−18 J.E_1 \approx 1.51\times 10^{-18}\,\mathrm{J}.

  • 以电子伏特计：

    E1≈9.4 eV.E_1 \approx 9.4\,\mathrm{eV}.

因此，电子在宽度为 0.20 nm0.20\,\mathrm{nm} 的一维无限深方势阱中的最低能级能量约为 9.4 eV9.4\,\mathrm{eV}。

इस प्रश्नोत्तर को साझा करें