MLE5104 Assignment 1

原来的模板的代码段好丑,用了SyntaxHighlighter才好点。
这又是用LaTeX写的作业。

[code lang=”latex”]\documentclass[12pt]{article}
\usepackage[margin=1.25in]{geometry}
\usepackage{fancyhdr}
\usepackage{tocloft}
\usepackage{setspace}
\pagestyle{fancy}
\usepackage[fleqn]{amsmath} %[fleqn]
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{cases}

\title{Assignment 1}
\lhead{A0146525E}
\chead{Zhou Mingyue}
\rhead{Assignment 1}

\begin{document}
\doublespacing
\bf
\begin{enumerate}
\item[Q.1]
Calculate the wave number of the following particles:
\begin{enumerate}
{ \item A photon with an energy of 100 eV.
\item An electron after an acceleration of 1000 kV.
\item A proton with a velocity of 1000 m/s}
\end{enumerate}
\end{enumerate}

\paragraph{ Solution:}
$\because k=2\pi/\lambda$\\
(a)
\begin{eqnarray*}
amp;&\because E = h\nu, c=\nu \lambda \\
&&\therefore k=2\pi\nu/c=2\pi E /(hc)= \frac{2*\pi*100*1.602*10^{-19}}{6.626*10^{-34}*3*10^8}=5.064 * 10^8\quad m^{-1}
\end{eqnarray*}
(b)
\begin{eqnarray*}
&&\because P = h/\lambda, P=\sqrt{2m_eE} \\
&&\therefore \lambda=h/P=h/\sqrt{2m_eE} \\
&&\therefore k=2\pi \sqrt{2m_eE}/h= \frac{2*\pi* \sqrt{2*9.109*10^{-31}*1.6*10^{-19}*1000*10^3}}{6.626*10^{-34}}=5.120*10^{12}\quad m^{-1}
\end{eqnarray*}
(c)
\begin{eqnarray*}
&&\because P = h/\lambda, P=m_p v \\
&&\therefore \lambda=h/P=h/m_pv \\
&&\therefore k=2\pi m_pv/h= 2*\pi* \frac{1.672*10^{-27}*1000}{6.626*10^{-34}}=1.585 * 10^{10}\quad m^{-1}
\end{eqnarray*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{enumerate}
\item[Q.2]
Calculate:
\begin{enumerate}
{ \item The energy and momentum of a photon with a wavelength of 450 nm.
\item The kinetic energy of a proton is the same of the energy of the photon with a wavelength of
450 nm. What is the expected velocity?}
\end{enumerate}
\end{enumerate}

\paragraph{ Solution:}
$\because E = h\nu, P=h/\lambda$\\
(a)
\begin{eqnarray*}
&&\because c=\nu \lambda \\
&&\therefore E=hc/\lambda=\frac{6.626*10^{-34}*3*10^8}{450*10^{-9}}=4.417*10^{-19} J, \\
&&P=h/\lambda=\frac{6.626*10^{-34}}{450*10^{-9}}=1.472*10^{-27}\quad kg ms^{-1}
\end{eqnarray*}
(b)
\begin{eqnarray*}
&&\because KE=m_{p}v^2/2 \\
&&\therefore v=\sqrt{2KE/m_{p}}= \sqrt{\frac{2*4.417*10^{-19}}{1.672*10^{-27}}}=22985.8\quad m s^{-1}
\end{eqnarray*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{enumerate}
\item[Q.3]
A proton is confined in a one-dimensional well with a length of 10 nm.
\begin{enumerate}
{ \item Calculate the ground level energy.
\item What is the longest wavelength of a photo, if it can excite the proton from the ground level to
third energy level?}
\end{enumerate}
\end{enumerate}

\paragraph{ Solution:}
$E_n=\frac{h^2n^2}{8ma^2}$\\
(a)
\begin{eqnarray*}
&&\therefore E_0=\frac{{(6.626*10^{-34})}^2}{8*1.672*10^{-27}*{(10*10^{-9})}^2}=3.282*10^{-25} \quad J
\end{eqnarray*}
(b)
\begin{eqnarray*}
&&\because \Delta E=hc/\lambda, \Delta E=\frac {h^2 (3^2-1)}{8ma^2} \\
&&\therefore \lambda=ma^2c/h=\frac{1.672*10^{-27}*{(10*10^{-9})}^2*3*10^8}{6.626*10^{-34}}=0.076\quad m
\end{eqnarray*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{enumerate}
\item[Q.4]
What is the tunneling probability when 1 eV electrons impinge onto a barrier 5 eV high and 0.5 nm wide?
\end{enumerate}
\textbf{Solution:}
\begin{eqnarray*}
&&\because T=T_0exp(-2a\alpha),T_0=16E(V_0-E)/V_0^2, \alpha=\sqrt{2m(V_0-E)/\hbar^2} \\
&&\therefore T_0=\frac{16*1.602*10^{-19}*(5-1)*1.602*10^{-19}}{(5*1.602*10^{-19})^2}\\
&&\therefore T=T_0*exp[-2*0.5*10^{-9}*\sqrt{\frac{2*9.109*10^{-31}*4*1.602*10^{-19}}{(6.626*10^{-34}/2*\pi)^2}}]\\
&&=9.091*10^{-5}
\end{eqnarray*}
%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{enumerate}
\item[Q.5]
The density of conduction electrons is 1 x $10^{23}$ electrons/cm3 for a metal.
\begin{enumerate}
{ \item Estimate the Fermi energy
\item Estimate the average velocity of conduction electrons}
\end{enumerate}
\end{enumerate}
Solution:
(a)
\begin{eqnarray*}
&&\because E_{F0}=(\frac{h^2}{8m_e})(\frac{3n}{\pi})^{2/3}\\
&&\therefore E_{F0}=(6.626*10^{-34})^2/(8*9.109*10^{-31})*(3*10^{23}*10^6/\pi)^{2/3}=7.86\quad eV
% &&\because n\approx g(E_f)f(E_f)dE, let dE=kT=0.026eV \\
% &&\because g(E_f)=8\pi\sqrt{2(m_e/h^2)^3E_f}, f(E_f)=1/2\\
% &&\therefore E_f\approx n^2h^6/(32\pi^2m_e^3)=(10^{23})^2*(6.626*10^{-34})^6/(32\pi^2(9.109*10^{-31})^3)\\
% &&=9.091*10^{-5}
\end{eqnarray*}
(b)
\begin{eqnarray*}
&&\because 1/2m_ev_e^2=3/5E_{F0}\\
&&\therefore v_e=\sqrt{6E_{F0}/(5m_e)}=\sqrt{6*7.86*1.602*10^{-19}/(5*9.109*10^{-31})}=1.288*10^6\quad ms^{-1}
\end{eqnarray*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{enumerate}
\item[Q.6]
Fermi energy of a metal is 5 eV. Estimate the occupation probability f (E) at 300K, if
\begin{enumerate}
{ \item E = 0.1 eV
\item E = 4 eV
\item E = 5 eV
\item E = 6 eV}
\end{enumerate}
\end{enumerate}
Solution: $\because f(E)=\frac{1}{1+exp(\frac{E-E_F}{kT})} $\\
(a)
\begin{eqnarray*}
&&\therefore f(0.1eV)=1/(1+exp(-4.9*1.602*10^{-19}/(1.381*10^{-23}*300)))\approx1
\end{eqnarray*}
(b)
\begin{eqnarray*}
&&\therefore f(4eV)=1/(1+exp(-1*1.602*10^{-19}/(1.381*10^{-23}*300)))\approx1
\end{eqnarray*}
(c)
\begin{eqnarray*}
&&\therefore f(5eV)=1/2
\end{eqnarray*}
(d)
\begin{eqnarray*}
&&\therefore f(6eV)=1/(1+exp(1*1.602*10^{-19}/(1.381*10^{-23}*300)))=1.610*10^{-17}
\end{eqnarray*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{enumerate}
\item[Q.7]
Fermi energy of a metal is 5 eV, and the width of the energy band is 10 eV. At a temperature of 0K,
\begin{enumerate}
{ \item What is the required photon energy, to eject an electron on the Fermi surface out of the metal surface?
\item What is the expected density of states at E = 4 eV?
\item What is the expected density of electrons ($n_E$) at E = 4eV?}
\end{enumerate}

Solution:\\
(a)
\begin{eqnarray*}
&& E=5\quad eV
\end{eqnarray*}
(b)
\begin{eqnarray*}
&&\because g(E)=8\pi\sqrt{2(m_e/h^2)^3E} \\
&&\therefore g(E)=8\pi\sqrt{2*(9.109*10^{-31}/(6.626*10^{-34})^2)^3*4*1.602*10^{-19}}\\
&&=1.362*10^{22}\quad cm^{-3}eV^{-1}
\end{eqnarray*}
(c)
\begin{eqnarray*}
&&\because f(E)=1 ,n_E=g(E)*f(E) \\
&&\therefore n_E=f(E)=1.362*10^{22}\quad cm^{-3}eV^{-1}
\end{eqnarray*}
\end{enumerate}
\end{document}[/code]

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