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\subsubsection*{1.2.II Photo electric effect}

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    \includegraphics[width=0.6\linewidth]
    {images/feb7aa7a495511ed.png}
\end{center}
As change $I$ and $\omega$ of incident light
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    \includegraphics[width=0.6\linewidth]
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\end{center}
classical expectation:
\begin{enumerate}[(i)]
    \item incident light carries $E \propto I$ as
        $I$ increases there is enough $E$ to break
        the bond of $e^-$ with atom $\forall
        \omega$.
    \item emission rate should be constant as $I$
        increases
\end{enumerate}
surprising facts:
\begin{enumerate}[(1)]
    \item Below $\omega_{\text{min}}$ no $e^-$
        emission
    \item $E_{\text{max}}$ depended on $\omega$
        not on $I$
    \item emission rate increases as $I$ increases
\end{enumerate}
1905 Einstein
\begin{itemize}
    \item light quantified in small quanta, called
        photon
    \item each photon carries
        \[ E = \hbar \omega \]
        \[ \bf{P} = \hbar \bf{K} \]
    \item phenomenon of $e^-$ emission comes from
        scattering of single photon off single
        $e^-$.
\end{itemize}
\[ E_{\text{min}} = 0 = \hbar \omega_{\text{min}}
- \phi \]
($\phi$ is the binding energy of $e^-$ with atom
of metal)
\[ E_{\text{max}} = \hbar \omega_{\text{max}} -
\phi \]
as $I$ increases, the number of protons increases,
so the amount of scattering increases, so there is
a higher $e^-$ emission rate.

\subsubsection*{!.2.III Compton scattering}

1923: X-rays scattering off free electron
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    \includegraphics[width=0.6\linewidth]
    {images/067cef16495811ed.png}
\end{center}
Recall Dynamics and Relativity example sheet 4
question 7:
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    \includegraphics[width=0.6\linewidth]
    {images/1f66f520495911ed.png}
\end{center}
\[ 2 \sin^2 \frac{\theta}{2} = \frac{mc}{|\bf{q}|}
- \frac{mc}{|\bf{p}|} \]
Why is this the peak?
\begin{align*}
    E
    &= \hbar \omega \\
    \bf{P}
    &= \hbar \bf{K}
    \implies |\bf{P}| = \hbar |\bf{K}| = \hbar
    \frac{\omega}{c} \\
    \bf{q}
    &= \hbar \bf{K}' \implies |\bf{q}| = \hbar
    |\bf{K}'| = \hbar \frac{\omega'}{c}
\end{align*}
Take (2) and plug in (1)
\[ \frac{1}{\omega'} = \frac{1}{\omega} +
\frac{\hbar}{mc}(1 - \cos\theta) \]
\begin{note*}
    $\hbar \to 0$, $\omega' \to \omega$.
\end{note*}

\subsection{Atomic spectra}

1897: Thomson, plum-pudding model of atoms.
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    \includegraphics[width=0.6\linewidth]
    {images/41bcb100495911ed.png}
\end{center}
1909: Rutherford
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\end{center}
scattering pattern $\to$ Rutherford model
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    \includegraphics[width=0.6\linewidth]
    {images/8f3601f2495911ed.png}
\end{center}
The Rutherford model did not work because
\begin{enumerate}[(i)]
    \item $e^-$ moves on circular orbits would
        radiate
    \item $e^-$ would collapse on nucleus due to
        Coulomb force
    \item model did not explain spectra measured.
\end{enumerate}
\[ \omega_{\text{min}} = 2\pi c R_0 \left(
\frac{1}{n^2} - \frac{1}{m^2} \right) \]
($c$ is the speed of light, $R_0$ is the Rydberg
constant, $\omega_{\text{min}}$ is the light
emitted by atoms when hit by light and $n, m \in
\NN$) \myskip
1913 (Bohr): $e^-$ orbits around nucleus are
quantised so that $L$ ($=$ orbital angular
momentum) takes discrete values
\[ L_n = n\hbar \]

\begin{proposition*}
    Quantisation of $L$ $\implies$ quantisation of
    $r$, $v$, $E$.
\end{proposition*}

\begin{proof}
    \[ L \equiv m_e vr \implies v = \frac{L}{m_e
    r} \implies v_n = n \frac{\hbar m_e r}{} \]
    Coulomb force:
    \[ \bf{F}^{\text{Coul}} = \frac{e^2}{4 \pi
    \eps_0} \frac{1}{r^2} \bf{e}_r \]
    Newton's second law:
    \[ \bf{F}^{\text{Coul}} = m_e a_r \bf{e}_r \]
    \[ \implies \frac{e^2}{4 \pi \eps_0}
    \frac{1}{r^2} = m_e \frac{v^2}{r} \implies r
    \equiv r_n = \frac{4 \pi \eps_0 \hbar^2}{m_e e^2}
    n^2 \]
    \[ \implies r_0 = \frac{4 \pi \eps_0
    \hbar^2}{m_e e^2} \]
    (min radius / Bohr radius)
    \begin{align*}
        E_n = \half m_e v_n^2 - \frac{e^2}{4\pi
        \eps_0} \frac{1}{r_n} \\
        &= - \frac{e^2}{8\pi \eps_0 r_0}
        \frac{1}{n^2}
    \end{align*}
    \begin{hiddenflashcard}[bohr-radius]
        \prompt{Bohr radius?}
        \[ r_0 = \cloze{\frac{4\pi \eps_0
        \hbar^2}{m_e e^2}} \]
    \end{hiddenflashcard}
    \begin{hiddenflashcard}[coulomb-force]
        \prompt{Coulomb force?}
        \[ \bf{F}^{\text{Coul}} =
        \cloze{\frac{e^2}{4\pi \eps_0}
        \frac{1}{r^2} e_r} \]
    \end{hiddenflashcard}
    \begin{hiddenflashcard}[bohr-energy]
        \prompt{Bohr energy?}
        \[ E_n = \cloze{-\frac{e^2}{8\pi \eps_0 r_0}
        \frac{1}{n^2}} \]
    \end{hiddenflashcard}
\end{proof}

\noindent
$n = 1$, $E_1 = -13.6 eV$ GROUND LEVEL.
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    \includegraphics[width=0.6\linewidth]
    {images/3537a2ee495b11ed.png}
\end{center}
\[ \omega_{\text{min}} = \frac{\Delta
E_{\text{min}}}{\hbar} = 2\pi c \left(
\frac{e^2}{4\pi \eps_0 \hbar c} \right)^2 \left(
\frac{1}{n^2} - \frac{1}{m^2} \right) \]

\subsection{The wave-like behaviour of particles}

1923: De Broglie hypothesis: $\forall$ particle of
$\forall$ mass associated with $Q$ wave having
\[ \omega = \frac{E}{\hbar} \]
\[ \bf{K} = \frac{\bf{p}}{\hbar} \]
1927: Davison and Geemer $e^-$ off crystals
interference pattern was consistent with De
Broglie.