111 年 國立臺灣聯合大學系統(清華、政治、陽明交通、中央四校聯招)研究所電機類《近代物理》
第 1 題
For a particle with an energy of tunneling through a finite barrier height of and width of , solve the Schrodinger wave equation to calculate the tunneling probability and select the correct answer corresponding to various , , and for different particles.
(A) For a neutron with meV incident on a barrier of meV and nm, the tunneling probability is ppb
(B) For a neutron with meV incident on a barrier of meV and nm, the tunneling probability is ppb
(C) For an electron with eV incident on a barrier with eV and nm, the tunneling probability is ppm
(D) For an electron with eV incident on a barrier with eV and nm, the tunneling probability is ppb
(E) For an electron with eV incident on a barrier with eV and nm, the tunneling probability is ppb
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考慮一維矩形勢壘,粒子能量 時,薛丁格方程在勢壘內的解為
對厚度 的勢壘,穿透係數(近似)
必須常數
(A) 中子
(B) 同上,只改
第 2 題
For a nuclear magnetic dipole of Hydrogen placed in a magnetic field (T: Tesla), calculate the Zeeman effect to be observed in a spectral line of various wavelengths and the resolution , select the correct solution ()
(A) T, nm, m pm
(B) T, nm, m fm
(C) T, nm, m fm
(D) T, nm, m fm
(E) T, nm, m fm
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(a) 本題考查的核心概念
本題屬於 正常 Zeeman 效應(Normal Zeeman effect),核心在於:
- 磁偶極矩與外加磁場的相互作用
對於氫原子的核磁矩(質子磁矩),其中
為 核磁子,。
- 能級位移與光譜線波長變化的關係
合併得
(此式已把 代入,)。
(b) 完整解題步驟
- 列出常數
| 常數 | 符號 | 數值 (SI) |
|---|---|---|
| 基本電荷 | ||
| 質子質量 | ||
| 光速 | ||
| 核子 g 因子 | ||
| 圓周率 |
把常數代入式子,先求出 係數
- 計算每一選項的 Zeeman 位移(把波長換成公尺)
| 選項 | (T) | (nm) | () | 計算式 | (m) | (pm) |
|---|
第 3 題
Consider a p-n junction diode. Which of the following statements is wrong?
(A) In the absence of any bias, there is no drift current in the junction.
(B) A forward bias will result in a rapid increase in the current.
(C) A reverse bias will turn off the diode.
(D) A diffusion current exists in both forward and reverse bias.
(E) The current-voltage curve is temperature sensitive.
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解說
- 平衡時:在零偏壓下,pn 接面內部有由內建電場產生的漂移電流,與濃度梯度引起的擴散電流相等且方向相反,形成電流平衡。故說「不存在漂移電流」不正確。
- 其餘選項皆符合 pn 接面的特性:
- 前向偏壓降低勢壘,使載流子快速注入,電流指數上升。
第 4 題
The atomic radii in terms of the lattice constant , for the structure of
(A) simple cubic is , face-centered cubic is , body-centered cubic is , and diamond is , respectively.
(B) simple cubic is , face-centered cubic is , body-centered cubic is , and diamond is , respectively.
(C) simple cubic is , face-centered cubic is , body-centered cubic is , and diamond is , respectively.
(D) simple cubic is , face-centered cubic is , body-centered cubic is , and diamond is , respectively.
(E) simple cubic is , face-centered cubic is , body-centered cubic is , and diamond is , respectively.
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核心觀念
本題考查晶體結構中「原子半徑 與晶格常數 的關係」。基本假設是:
- 原子視為彼此接觸的硬球。
- 沿著最近鄰原子排列方向,兩個原子中心距離等於 。
- 由晶胞幾何結構求出最近鄰原子中心距離,再令其等於 。
四種結構的最近鄰接觸方向不同:
| 結構 | 最近鄰接觸方向 |
|---|---|
| Simple cubic, SC | 立方體邊 |
| Face-centered cubic, FCC | 面對角線 |
| Body-centered cubic, BCC | 體對角線 |
| Diamond | 晶胞體對角線上的特定位置 |
解題方法
1. Simple cubic
簡單立方結構中,最近鄰原子沿著立方體邊方向接觸。
立方體邊長為 ,因此:
得到:
2. Face-centered cubic
面心立方結構中,角落原子與面心原子沿著面對角線接觸。
面對角線長度為:
在一條面對角線上,排列為:
其總長度包含四個原子半徑:
因此:
3. Body-centered cubic
體心立方結構中,角落原子、體心原子與對角角落原子沿著體對角線接觸。
立方體體對角線長度為:
沿體對角線共有四個原子半徑:
所以:
4. Diamond structure
鑽石結構可視為 FCC 晶格加上位移為
的基底原子。
最近鄰原子間的位移向量為:
因此最近鄰距離為:
因為最近鄰原子彼此接觸:
所以:
第 5 題
Which of the following statements is correct?
(A) An LED is constructed from a p-n junction based on a certain semi-conducting material with energy gap of 1.55 eV. The wavelength of the emitted light is 80 nm.
(B) of the available volume is occupied by hard spheres in contact in a simple cubic arrangement.
(C) If the Debye temperature for iron is known to be 360 K, the maximum frequency is 75 THz.
(D) The Fermi energy in gold is 5.54 eV. The average energy of the free electrons in gold at 0 K is right the Fermi energy and the corresponding speed of free electrons is m/s.
(E) The density of states function for electrons in a metal is given by . The Fermi level at a temperature few degrees above absolute zero for copper which has electrons per cubic meter is about 9.35 eV.
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解析
(A) 能隙 時,發射光子能量 ,故 ,遠大於 80 nm,敘述錯誤。
(B) 簡單立方(SC)堆積中每個晶胞僅有 1 個完整球體,體積佔比
故可用體積正好為 ,敘述正確。
(C) 德拜溫度 ,最大頻率
第 6 題
Which of the following principles of classical physics is violated in the Bohr model?
(A) Newton's
(B) Coulomb's law
(C) Accelerating charges radiate energy
(D) Particles always have a well-defined position and momentum
(E) More than one of the above
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參考書等級:大學一年級普通物理或電磁學教材(如《大學物理(上)》(楊振寧))
- Bohr 模型假設電子於原子核周圍的固定圓形軌道上運動,儘管在圓軌道上電子做向心加速,卻不會因加速而發射電磁波,亦即不會失去能量。
第 7 題
Consider a particle in the relativistic regime, with the energy given by the expression ( = energy, = momentum, = mass, = light speed). What is the leading-order relativistic correction to the kinetic energy?
(A)
(B)
(C)
(D)
(E) None of the above
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核心觀念
題目考查相對論性能量與動能的低速展開。
相對論性能量為
動能定義為總能量扣除靜止能量:
當動量滿足 時,可用二項式展開,找出相對於經典動能 的第一個修正項。
解題方法
先將總能量整理成適合展開的形式:
利用二項式展開公式
令
因此
展開後得到
扣除靜止能量 :
其中第一項
是經典動能;相對於經典結果的 leading-order relativistic correction 為下一項:
選項分析
第 8 題
You found an interesting looking piece of metal. You want to find out what kind of metal that is and so you decide to replace the cathode in your photoelectric effect apparatus with that metal piece. You still have the laser diode from the blue-ray player (405 nm wavelength). In this setup you find a stopping potential of 0.16 eV. What kind of metal is it?
🖼️【此處有附圖,請對照原卷】
(A) Sodium (work function eV)
(B) Calcium (work function eV)
(C) Silver (work function eV)
(D) Lead (work function eV)
(E) Some other material
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核心觀念
本題考查光電效應方程式:
其中:
- :入射光子能量
- :金屬功函數
- :光電子最大動能
停止電位 與最大動能的關係為:
若以電子伏特表示,數值上可直接寫成:
解題方法
藍光雷射波長為 ,光子能量為:
利用 :
由光電效應方程式:
代入數值:
因此該金屬的功函數約為 ,對應 Calcium(鈣)。
選項分析
第 9 題
A radio transmitter of 1 kW operates at a frequency of 880 kHz. How many photons per second does it emit?
(A) photons/s
(B) photons/s
(C) photons/s
(D) photons/s
(E) None of the above
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核心觀念
電磁波由光子組成,每個光子的能量為
其中 為普朗克常數, 為電磁波頻率。若發射功率為 ,則每秒發射的光子數 滿足
因此光子發射率為
解題方法
已知
取普朗克常數
單一光子的能量為
每秒發射的光子數為
故正確選項為 (D)。
選項分析
第 10 題
A galaxy in the constellation Ursa Major is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the corresponding wavelength measured by astronomers on the earth?
(A) 578 nm
(B) 940 nm
(C) 340 nm
(D) 1.15 µm
(E) 405 nm
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核心觀念
星系遠離地球時,所發出的光會產生紅移,也就是地球觀測到的波長變長。
本題速度為
相對於光速
有
由於 ,可採用近似的多普勒紅移公式:
其中:
- :光源在靜止時發出的波長
- :地球上的天文學家量測到的波長
- :光源遠離觀測者,造成紅移
解題方法
已知發射波長為
代入公式:
因此地球上的天文學家量測到的波長約為
若採用狹義相對論的精確縱向多普勒公式:
則
第 11 題
Consider the conduction band in the Na metal at T = 0 K (T = temperature). Which of the following statements are correct?
(A) The band derives primarily from the 3s orbital of Na atom.
(B) The band is completely filled.
(C) The Fermi energy lies inside the conduction band.
(D) Since conduction band electrons move quite freely in the metal, they do not bind the atoms together at all.
(E) All of the above.
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參考書等級:大學二年級固態物理課本或金屬物理教材。
關鍵概念
- 鈉原子外層為 ,金屬固態時 軌道疊合形成導帶,故導帶主要來源為 軌道 → (A) 正確。
- 金屬的導帶必須是 部分填充 才能提供自由電子;
第 12 題
Let 'v' and 's' both be binary electron degrees of freedom, with or , and or .
Below, we will use the following convention: a) A notation such as represents an one-electron state with , and ; b) A product such as denotes a two-electron state with the 1st electron in the state and the 2nd in . Which of the following states are consistent with the Pauli exclusion principle?
(A)
(B)
(C)
(D)
(E) None of the above
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先確認兩電子的總態必須對交換算符 反對稱:
對於每一選項,將交換 後與原式比較。
(A)
交換後得到
不符合反對稱。
(B)
交換後
第 13 題
and are both non-trivial solutions (i.e., are not zero everywhere) of the time-independent 1D Schrodinger equation. Which of the following is not a solution to that Schrodinger equation? (Note: A, B, and are arbitrary, non-zero constants. This question only asks for 'mathematical solution' to the Schrodinger equation. You don't need to normalize the wave functions nor fulfill any boundary conditions.)
(A)
(B)
(C)
(D)
(E) None of the above is solution to that Schrodinger equation.
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核心觀念
定態一維薛丁格方程可寫成
其中
移項後為
這是一個線性齊次微分方程。因此:
- 任一解乘上常數,仍是解。
- 多個解的線性組合,仍是解。
- 解若乘上一個非常數函數,通常不再是解。
解題方法
令
則已知
利用線性性,
其中 為常數。
因此,判斷重點在於:選項是否只是已知解的常數倍或線性組合。
選項分析
(A)
因為 是常數,
所以 仍是薛丁格方程的解。
(B)
由線性齊次性,
所以此選項是解。
(C)
先注意到
是兩個解的線性組合,因此為解。再乘上常數 ,仍然是解:
即使 ,所得的零函數也滿足齊次薛丁格方程;題目並未要求答案本身必須是非零解。因此此選項仍屬數學解。
第 14 題
Assuming there are particles, whose de Broglie waves shown in the figure below. (A: amplitude of the de Broglie waves, : wavelength of the de Broglie waves. All three particles have the same mass.)
Which of the following relations about the velocity of particles are correct?
(A) I > II
(B) II = I
(C) II > I
(D) III > II
(E) III = I
🖼️【此處有附圖,請對照原卷】
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核心觀念
本題利用德布羅意關係式:
三個粒子的質量相同,因此 相同,且 為普朗克常數。故粒子速度與波長成反比:
德布羅意波的振幅 不決定粒子速度;只有波長 會影響速度。
解題方法
比較圖中三個波的波長:
- 波長較短者,動量 較大,速度 較大。
- 波長相同者,速度相同。
- 波長較長者,速度較小。
因此各選項的判斷條件為:
- (A) :只有在 時正確。
- (B) :只有在 時正確。
第 15 題
An electron has a matter wavelength of and a rest energy . Which of the following answers is correct?
(A) The electron energy is .
(B) The kinetic energy (KE) of the electron is .
(C) The electron velocity is about .
(D) The group velocity is about .
(E) The phase velocity is about .
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第 16 題
Assume the Fermi energy of silver is and the number of conduction electrons is . Which of the following statements are correct?
(A) The corresponding velocity of a conduction electron is .
(B) If the resistivity of silver at room temperature is , the average time between collisions is .
(C) The mean free path is .
(D) The corresponding velocity of a conduction electron is .
(E) The mean free path is .
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第 17 題
Which of the following statements are correct?
(A) The splitting of spectrum lines in a magnetic field is known as Zeeman effect.
(B) The energy spacing between the components of the ground state energy level of hydrogen when split by a magnetic field of is .
(C) The energy spacing between the components of the ground state energy level of hydrogen when split by a magnetic field of is .
(D) Splitting levels by equal amounts in the presence of a magnetic field is called normal Zeeman effect.
(E) Splitting levels by equal amounts in the presence of a magnetic field is called anomalous Zeeman effect.
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第 18 題
For an electron trapped in a quantum well with infinitely high barrier at the boundaries, and , derive the wave function and calculate the expectation values of specified terms. Herein, is the linear momentum operator given by .
(A) , .
(B) , .
(C) , .
(D) , .
(E) , .
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第 19 題
Which of the following statements are correct?
(A) All motion is relative and the speed of light in free space is the same for all observers.
(B) Although they lack rest mass, photons behave as though they have gravitational mass.
(C) The de Broglie wave group associated with a moving body travels with the same velocity as the body.
(D) The energies of electrons liberated by light depend on the frequency of the light.
(E) Only the quantum theory of light can explain the origin of blackbody radiation.
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第 20 題
Select the correct electron configurations and quantum numbers of the electrons considering LS coupling effect for various atoms.
(A) ; : , , .
(B) ; : , , .
(C) ; : , , , or .
(D) ; : , , , or .
(E) ; : , , , or , .
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Electron rest mass:
Light velocity:
Proton rest mass:
Electron charge:
Neutron rest mass:
Planck's constant:
Hydrogen atomic mass:
Permittivity of free space:
第 21 題10 分
(10%) Right figure is an x-ray tube.
🖼️【此處有附圖,請對照原卷】
(a) (5%) Describe the generation of x-ray.
(b) (5%) Determine the minimum wavelength , when voltage is .
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核心觀念
陰極放出的電子經電壓 加速後撞擊陽極靶材,會因急遽減速而產生制動輻射;若電子撞出靶材原子的內層電子,外層電子填補空缺時也會放出特徵 X 光。最短波長對應單一光子取得電子全部動能的極限,因此使用
解題方法
圖中是抽空的 X 光管,右側為陰極,左側為靶材;電子由陰極移向靶材,靶材放出 X 光。電子通過電位差 時獲得動能 。令這份動能全部轉換成一個 X 光光子的能量,即可求出最短波長:
第 22 題10 分
(10%) Modern Metal-Oxide-Semiconductor Field-Effect Transistor (MOSFET), its device structure is as following figure.
🖼️【此處有附圖,請對照原卷】
(a) (5%) Determine electron matter wavelength , when electron with velocity and mass .
(b) (3%) Electron transport from source (I) to drain (III), electron will meet a potential barrier (II), which established by gate. Explain electron tunneling effect briefly.
(c) (2%) Give a reasonable minimum gate length () of MOSFET from the answer (a). (Hint: use tunneling effect.)
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核心觀念
本題考查電子的德布羅意物質波波長,以及量子穿隧效應對 MOSFET 閘極長度的限制。電子的物質波波長為
其中 是電子的動量。當粒子遇到能量低於位能障壁的區域時,量子力學中的波函數仍可穿入障壁並延伸至另一側,因此電子有非零機率穿越障壁,這就是穿隧效應。
解題方法
圖中 MOSFET 有源極、汲極與閘極;源極至汲極的電子運輸路徑依序標為 I、II、III,II 是由閘極建立的位能障壁,圖中障壁範圍由 延伸至 。先用電子的質量與速度求出物質波波長,再以波長尺度判斷閘極長度縮短時穿隧效應的影響。
(a) 電子物質波波長
代入 、 與 :
所以電子的物質波波長約為 。代回檢查,,符合德布羅意關係式。