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\begin{document}
 \begin{center}
  {\bf {\underline{INDIAN INSTITUTE OF INFORMATION TECHNOLOGY, ALLAHABAD}}}\\
  \vspace{0.2cm}
  {\small{\bf End-Semester Examination, November-December 2017}}
  \end{center}

{\bf\leftline{Max Marks: 75 \hspace{10cm} Duration: 3 hours}}
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{\bf Note: Use of non-programmable calculator is allowed. Section A is compulsory, and attempt any seven questions from Section B.$ ~n_i^{Si}= 10^{10} cm^{-3}; m_e=9.1 \times 10^{-31} kg.$}
\vspace{0.5cm}

{\bf \hspace{7cm} Section A}
\begin{enumerate}

\item The position and momentum of a 1.0 KeV electron are simultaneously measured. If the position is located within $1 \mathring{A}$, what is the percentage of uncertainty in momentum? Given $h=6.62 \times 10^{-34}J.s; 1 eV=1.6 \times 10^{-19} J.$ \hfill[6]\\

\noindent\textbf{Solution}: According to Heisenberg's uncertainty principle: $\Delta x \Delta p\sim\frac{\hbar}{2}$.\\
Momentum of 1 KeV electron $p=\sqrt{2 m_e E}=\sqrt{2 \times 9.1 \times 10^{-31} \times 1000 \times 1.6 \times 10^{-19}}=1.71$ Kg m/s. \rh{[2]} 
The uncertainty in momentum $\Delta p=\frac{\hbar}{2\Delta x}=\frac{6.62 \times 10^{-34}}{4 \times \pi \times 1.0 \times 10^{-10}}=5.27 \times 10^{-25}.$ \rh{[3]} 

So the percentage of uncertainty in momentum is $\frac{\Delta p}{p}\times 100= \frac{5.27 \times 10^{-25}}{1.71\times 10^{-23}}\times 100=3.1\%.$ \rh{[1]} 

\item If no electron-hole pairs were produced in germanium (Ge) until the temperature
reached the value corresponding to the energy gap, at what temperature would Ge
become conductive? $E_g=0.72 \times 1.6 \times 10^{-19} J$; $K= 1.38 \times 10^{-23} J/K.$
\hfill[6]

\noindent\textbf{Solution}: Average kinetic energy of an electron (or hole) $E_{th}=\frac{3KT}{2}.$ \rh{[2]} 

Now equating $E_{th}=E_g,$ \rh{[1]}  we get,

\[T=\frac{2 \times 0.72 \times 1.6 \times 10^{-19} }{3 \times 1.38 \times 10^{-23}}=5565 K.\] \rh{[3]} 

\end{enumerate}


{\bf \hspace{7cm} Section B}

\begin{enumerate}
\item	A particle is acted upon a force $F = F(r){\bf \hat{r}}$ which is directed towards a fixed point. Show the motion is planar and angular momentum about the fixed point (origin) is conserved. Suppose $F(r) = k r^2$ , then write down the Lagrangian of the particle and deduce equation of motion. Write down the Hamiltonian of the system. [12]\hfill[12]

\noindent\textbf{Solution}: {\bf Answer may vary} (a) Let the length of the pendulum string is $l$ and the angle it makes with vertical at an instant t is $\phi$ (Fig-1). Then $x=a Cos \omega t+l Sin \phi,~ y=-a Sin \omega t +l Cos \phi.$ The Lagrangian is given by,

\[L= \frac{1}{2}\left(m a^2\omega^2 + m l^2 \dot{\phi}^2 \right)-m a \omega l \dot{\phi} Sin (\omega t -\phi) + mgl Cos \phi -mga Sin \omega t.\]
The last term can be omitted as it is a total derivative. The third term can be written as $ma\omega^2l Sin(\phi-\omega t)+$ total derivative.  \rh{[4]} 

The equation of motion is 

\[ml^2\ddot{\phi}-ma\omega^2lCos (\phi -\omega t) +mgl Sin \phi=0.\] \rh{[1]}

(b) In this case $x=a Cos \omega t+l Sin \phi,~ y=l Cos \phi.$ The Lagrangian is  \[L= \frac{1}{2}\left(m a^2\omega^2 Sin^2 \omega t + m l^2 \dot{\phi}^2 \right)-m a \omega l \dot{\phi} Sin (\omega t) Cos \phi + mgl Cos \phi.\] \rh{[3]}

EOM \[ ml^2\ddot{\phi}=-ma\omega^2l Cos\omega t Cos \phi + mgl Sin \phi=0.\] \rh{[1]}

\begin{figure}[h]
 \centering
    \includegraphics[width=.5\textwidth]{fig-1.jpeg}
    
      %\caption{Penrose diagram of the spacetime formed by matching two charged Vaidya spacetimes along the thin null shell $S$ (thick green line). The thick red lines are the inner and the outer apparent horizons. The dotted lines are the event horizons of the background geometry (mass $m_f$, cf.~\eqref{mass_function}).}
   % \label{figure1}
\end{figure}

\item Find the path on which an object, in the absence of friction, will slide from one given point to another
in the shortest possible time under the effect of gravity. \hfill [9]

\noindent\textbf{Solution}: {\bf Answer may vary} $v=\frac{ds}{dt}=\sqrt{2 g y}.$

\[\int_0^{x_0} \frac{ds}{v}=\frac{1}{\sqrt{2g}}\int_0^{x_0} \sqrt{\frac{1+y'^2}{y}}dx.\]\rh{[5]}

After rearrangement one gets, $\frac{dy}{dx}=\sqrt{\frac{a-y}{y}}.$ \rh{[2]}

Now set, $y=a sin^2 \theta $
 and get \[x=\frac{a}{2}(2\theta -sin 2\theta)+c ~;\, y=\frac{a}{2}(1-cos 2\theta)\] \rh{[2]}
 
 These is equation of a cycloid. 
\item From the Schr̈\"{o}dinger equation derive the continuity equation:
\[\frac{\partial \rho}{\partial t} + \frac{\partial S}{\partial x}=0,\]
where $\rho=|\Psi(x, t)|^2$ is the ‘probability density’ and $S(x,t)$ 
is the 'probability current density'. \hfill[9]

\noindent\textbf{Solution}: By the definition of the probability density $\rho$, 
\[\frac{\partial \rho}{\partial t}=\frac{\partial}{\partial t}[\Psi^* \Psi]=\left[\Psi\frac{\partial \Psi^*}{\partial t}+\Psi^*\frac{\partial \Psi}{\partial t}\right].\] \rh{[2]}

Recall, $S=\frac{i\hbar}{2m}\left( \Psi \frac{\partial \Psi^*}{\partial x}-\Psi^*\frac{\partial \Psi}{\partial x}\right).$ \rh{[1]}

Now, \[\frac{\partial \Psi}{\partial t}=(1/i\hbar)H\Psi; \frac{\partial \Psi^*}{\partial t}=-(1/i\hbar)H\Psi^*\] gives

\[\frac{\partial \rho}{\partial t}=\frac{-1}{i\hbar}(H\Psi^*)\Psi + \frac{1}{i\hbar}\Psi^*H\Psi=\frac{1}{i\hbar}\left[\frac{\hbar^2}{2m}\Psi \partial^2_x \Psi^*-V\Psi^*\Psi-\frac{\hbar^2}{2m}\Psi^* \partial^2_x \Psi + V\Psi\Psi^*.\right]\] \rh{[4]}
Therefore, \[\frac{\partial \rho}{\partial t}=\frac{\partial}{\partial x}\frac{-i\hbar}{2m}\left( \Psi \frac{\partial \Psi^*}{\partial x}-\Psi^*\frac{\partial \Psi}{\partial x}\right)=-\frac{\partial S}{\partial x}.\] \rh{[2]}
\item A particle trapped in a box of length a, has as its initial wave function
an even mixture of the first two stationary states:
\[\Psi(x,0)=A[\psi_1(x)+\psi_2(x)]\]
(a) Normalize $\Psi(x, 0)$. (b) Find $\Psi(x,t)$ and $|\Psi(x,t)|^2$. \hfill[3+6]

\noindent\textbf{Solution}: (a) $|\Psi(x,0)|^2=|A|^2(\psi_1^*\psi_1+\psi_1^*\psi_2+\psi_1\psi_2^*+\psi_2^*\psi_2)$. \rh{[1]}\\
Now, $1=\int_0^a |\Psi(x,0)|^2 dx=|A|^2\int_0^a (\psi_1^*\psi_1+\psi_2^*\psi_2) dx =2|A|^2,$ Since $\psi_1, \psi_2$ constitute an orthonormal set of eigenfunctions. [Or, $\int_0^a Sin(\pi x/a)Sin(2\pi x/a)=0$]
Hence, $A={1\over {\sqrt{2}}}.$ \rh{[2]}

(b) The complete solution of this problem is, $\Psi(x,t)=\Sigma_{1,2}A \sqrt{2 \over a}\left[Sin\left(\frac{n \pi x}{a}\right)e^{-i E_n t/\hbar}\right].$ Which gives,

$\Psi(x,t)=\frac{1}{\sqrt{a}}\left[ Sin\left(\frac{\pi x}{a}\right)e^{\frac{-i E_1 t}{\hbar}}+Sin\left(\frac{2\pi x}{a}\right)e^{\frac{-i 4E_1 t}{\hbar}}\right],$ where $E_n=\frac{n^2\pi^2\hbar^2}{2 m a^2}.$ \rh{[4]}

Finally after little bit of algebra, 

$|\Psi(x,t)|^2=\frac{1}{a}\left[Sin^2\left(\frac{\pi x}{a}\right)+ Sin^2\left(\frac{2\pi x}{a}\right)+2Sin\left(\frac{\pi x}{a}\right)Sin\left(\frac{2\pi x}{a}\right)Cos(3\omega t) \right].$ Where $\omega=\frac{\pi^2\hbar}{2 m a^2}.$ \rh{[2]}
%\newpage \rh{[4]}

\item Draw the Energy band diagram of an n-type extrinsic semiconductor. Find the position of Fermi level $E_F$ at 300 K for Germanium crystal having $5 \times 10^{22}$ atoms/$m^3$. \hfill[2+7]

\noindent\textbf{Solution}: $n=2\left(\frac{2 \pi m K T}{h^2}\right)^{3/2}e^{-(E_C-E_F)/KT}=5 \times 10^{22}.$
Then, $E_F-E_C=KT ~ln \left[\frac{n}{2\left(\frac{2 \pi m K T}{h^2}\right)^{3/2}}\right].$ \rh{[4]}

Putting all the supplied values one gets, $E_C-E_F={\bf 0.161 eV}.$ Also, $2\left(\frac{2 \pi m K T}{h^2}\right)^{3/2}=2.51\times 10^{25},$ and $E_C-E_F=KT \times 6.2186.$ \rh{[3]}

\item What is the concentration of holes in valence bad when energy is low ($E-E_F<<KT)$. 
Band gap of Si depends on the temperature as 
\[E_g= 1.17 eV -4.73 \times 10^{-4}\frac{T^2}{T+636}\]
Find a concentration of electrons in the conduction band of intrinsic
(undoped) Si at $T= 77 $ K if at $300$ K, $n_i=1.05 \times 10^{10} cm^{-3}$. \hfill[2+7]

\noindent\textbf{Solution}:
The hole concentration should be $e^{(E-E_F)/kT}.$ \rh{[2]}

$n_i^2=N_cN_v e^{-E_g/KT}\sim T^3 e^{-E_g/KT}. $ Therefore
\[n_i(T_2)=n_i(T_1)\left(T_2/T_1\right)^{3/2}\exp{\left(-\frac{E_g(T_2)}{2 K T_2}+\frac{E_g(T_1)}{2 K T_1}\right)}.\] \rh{[5]}

Now, $E_g(300K)=1.06 eV,\, E_g(77K)=1.166 eV, \left(T_2/T_1\right)^{3/2}=0.13, exp(-67.02)=7.82 \times 10^{-30}. $ Then putting the values of $n_i$, we get $n_i(77K)=1.069 \times 10^{20}.$ \rh{[2]}

\item What is mean free time and how it is related with mobility of carriers? What is the hole diffusion constant in a piece of silicon doped with $3 \times 10^{15} cm^{-3}$ of donors and $7 \times 10^{15} cm^{-3}$ of acceptors at 300 K? at 400 K? 
Given, $\mu_p=410~cm^2/V.s$ at $300 K$ and $220~ cm^2/V.s$ at $400 K$. \hfill [2+7] 

\noindent\textbf{Solution}: The mean time taken by the carriers between two consecutive collisions. It is proportional to mobility. \rh{[2]}

$N_a + N_d = 3 \times 10^{15} + 7 \times 10^{15} = 1 \times 10^{16} cm^{-3}.$ \rh{[2]}

\[D_p = (kT/q) \mu_p = 26 mV \times 410cm^2 /V·s = 10.66cm^2 /s. 
\] \rh{[2]}

For the 400 K, \[D_p = \mu_p (kT/q) = 220cm^2 /V·s \times 26 mV \times (400 K/300 K) = 7.6 cm^2 /V·s.\] \rh{[3]}



\item What is built in potential in a p-n junction device? Find the built-in potential for a p-n Si junction at room temperature if the bulk
resistivity of Si is $1 \Omega$ cm. Electron mobility in Si at room temperature (300 K) is 1400 cm$^2$ V$^{-1}$ s$^{-1}$ ; $\mu_n/\mu_p = 3.1$;
$n_i = 1.05 \times 10^{10} cm^{-3}$ .\hfill[2+7]

\noindent\textbf{Solution}:  Presence of a non uniform electric field at the depletion layer of a p-n junction material indicates some potential difference has developed across the layer. \rh{[2]}

By definition, $φ_{bi} =\frac{KT}{q} ln\left(\frac{np}{n_i^2}\right)$. \rh{[2]}

Now $\sigma_n=n q \mu_n\Rightarrow n=\frac{1}{q \rho \mu_n }.$ Similarly, $p=\frac{1}{q \rho \mu_p }.$ \rh{[3]}

Now putting the values we get $\phi_{bi}=.026 ln \left(\frac{1}{q^2 \mu_n\mu_p n_i^2}\right)\approx .70 V.$ \rh{[2]}

\item The slope of the inverse square Capacitance versus Voltage graph is $2\times  10^{23} F^{-2} V^{-1}$ , the intercept is $0.84 V$,
and the area of the PN junction is $1 \mu m^2$. Find the lighter doping concentration, $N_l$ , and the
heavier doping concentration, $N_h$. $ \epsilon_{Si}=12 \times 8.85 \times 10^{-14}$. \hfill[9]

\noindent\textbf{Solution}: From the expression $\frac{1}{C_{dep}^2}=\frac{(\phi_{bi}+V_r)}{q N \epsilon_s A^2},$ \rh{[2]}

we get

\[N_1=\frac{2}{slope \times q \epsilon_s A^2}=5.9 \times 10^{23} cm^{-3}.\] \rh{[2]}

Then using, $\phi_{bi}=\frac{kT}{q} ln \frac{N_1N_2}{n_i^2}$ we get, $N_2=\frac{10^{20}}{5.9 \times 10^{23}}\times e^{.84/.026}=1.77 \times 10^{10} cm^{-3}.$ \rh{[3]}

Therefore $N_1=N_h$ and $N_2=N_l$.

\end{enumerate}

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