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\begin{center}
\large{\textbf{Indian Institute of Information Technology Allahabad}}
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\large{\textbf{Recapitulation Test on Quantum Mechanics}}
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\hfill \footnotesize{Date: 17/09/2018}
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\noindent
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\begin{center}
\footnotesize{B.Tech. IT, ECE - Semester I}
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\begin{center}
\footnotesize{Paper: Engineering Physics}
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%\footnotesize{Paper Setter: Abdullah Bin Abu Baker \& Sumit Kumar Upadhyay}
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\noindent
\footnotesize{Max Marks: 15} \hfill \small{Duration: 1 hour}
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\begin{enumerate}

 \item An electron is described by the wave function \\
 \begin{center}
 $\psi(x) = \left\{\begin{array}{ll} 
                    0 & \text{if }x < 0\\ 
                    C e^{-x}(1- e^{-x}) & \text{if }x \geq 0
            \end{array}\right. ,$ \end{center}
            where $C$ is a constant and $x$ is measured in nm.
    \begin{enumerate}
    \item Determine the value of $C$ that normalizes $\psi(x)$.\\
    
    \noindent {\bf Ans.}  $ \int_0^{\infty} \psi^{*}(x) \psi(x) dx =1\implies |C|^2\int_0^{\infty} e^{-2x}(1- e^{-x})^2=1$  \hfill[1]\\
    $\implies |C|^2 \times \frac{1}{12}=1\implies |C|=\sqrt{12}.$\hfill[1]
    
    \item Where is the electron most likely to be found? That is, for what value of $x$ is the probability of finding the electron the largest? Plot the wave function.\\
     \noindent {\bf Ans.}$ \frac{d}{dx} \psi(x)=0 \implies x=ln 2$. Also, $\frac{d^2}{dx^2} \psi(x)\Big{|}_{x=ln 2 }<0$ \hfill[2]\\
     
     Plot of $\psi/|C|$:
     \begin{figure}[h]
 \centering
    \includegraphics[width=.4\textwidth]{Plot_psi}
    \hfil[1]
 \end{figure}
     
         \item Calculate the average position $<x>$ for the electron. Compare this result with the most likely position, and comment on the difference. Calculate $<x^2>$ and then the dispersion $(\Delta x)^2=<x^2> - <x>^2$.\\
         
 \noindent {\bf Ans.} $<x>=\int x |\psi(x)|^2 dx =  \frac{13}{12}~ nm.$ \hfill[1]\\
 $<x^2>=\int x^2 |\psi(x)|^2 dx=\frac{115}{72} ~nm.$ \hfill[2]\\
    $(\Delta x)^2=<x^2> - <x>^2=\frac{115}{72}- \frac{169}{144}=\frac{61}{144}~ nm^2.$ \hfill[1]
    
    \item Calculate the dispersion in momentum and verify whether the given system satisfies the Heisenberg's uncertainty principle or not. \\
    
    \noindent {\bf Ans.} $<p>=\int \psi^{*}(x)\left(-i \hbar \frac{d}{d x}\right) \psi(x) dx=0$ \hfill[1]\\
    $<p^2>=-\int \psi^{*}(x)\left(\hbar^2 \frac{d^2}{d x^2}\right) \psi(x) dx=2\hbar^2.$ \hfill[2]\\
    $(\Delta p)^2=<p^2> - <p>^2=2\hbar^2.$ \hfill[1]\\
    $\Delta x \Delta p=\frac{\hbar \sqrt{122}}{12}>\frac{\hbar}{2}.$\hfill[2]
    
\end{enumerate}

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