115年 - 115 台灣聯合大學系統_碩士班招生考試_電機類:近代物理#139674

科目:研究所、轉學考(插大)-近代物理學 | 年份:115年 | 選擇題數:20 | 申論題數:0

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所屬科目:研究所、轉學考(插大)-近代物理學

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16. We usually need to use the concept of "effective mass" to analyze and predict the properties of electronic and optoelectronic devices. It can be calculated by $\frac{1}{m^{*}} = \frac{1}{h^{2}}\frac{\partial^{2}E(k)}{\partial k^{2}}$, where $m^{*}$ is the effective mass, and $E(k)$ is the energy of the carrier as a function of the wave vector $k$. The effective mass of carriers in a material can be measured using the cyclotron resonance method. In short, this method is performed by applying a magnetic field to the material and irradiating it with electromagnetic waves, assuming that the absorption frequency of the sample matches the cyclotron frequency of the carriers in the magnetic field. Hint: You may start the derivation from the classical equation of motion of a charged particle in a magnetic field. The magnetic force is $qvB$, where $q$ is the carrier charge, $v$ is its velocity, and $B$ is the magnitude of the magnetic field, and it provides the centripetal force for circular motion ($F = \frac{m^{*}v^{2}}{r}$). Please choose the correct answer. (A) The idea of effective mass is only needed for semiconductors. For metals, the free electron model is sufficient. (B) Effective mass can be less than $m_e$, where $m_e$ is the mass of an electron. (C) The effective mass is determined by the curvature of the energy band E(k). (D) When the magnetic field is $0.15\mathrm{T}$ and the absorption frequency $= 2.1\times 10^{10}\mathrm{Hz}$, the effective mass of an electron is $0.3m_e$. (E) If the absorption frequency is $1.4 \times 10^{10} \mathrm{~Hz}$ and the radius is $3.4 \times 10^{-7} \mathrm{~m}$, the velocity of the electron should be $3 \times 10^{4} \mathrm{~m/s}$.

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