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diff --git a/ediffract.tex b/ediffract.tex index 96cbd58..e1da926 100644 --- a/ediffract.tex +++ b/ediffract.tex @@ -12,15 +12,15 @@ Broglie suggested in his dissertation that since light has both particle-like {\bf and} wave-like properties, perhaps all matter might also have wave-like properties. He postulated - that the wavelength of objects was given by $\lambda = h/p$, where $h$ is Planck's constant, and $p = + that the wavelength of objects was given by $\lambda = h/p$, where $h$ is Planck's constant and $p = mv$ is the momentum. {\it This was quite a revolutionary idea}, since there was no evidence at the time that matter behaved like waves. In 1927, however, Clinton Davisson and Lester Germer discovered experimental proof of the wave-like - properties of matter- particularly electrons (This discovery - was quite by mistake!). They were studying electron reflection - from a nickel target. They inadvertently crystallized their - target while heating it, and discovered that the scattered + properties of matter --- particularly electrons. This discovery + was quite by mistake: while studying electron reflection + from a nickel target, they inadvertently crystallized their + target, while heating it, and discovered that the scattered electron intensity as a function of scattering angle showed maxima and minima. That is, electrons were ``diffracting'' from the crystal planes much like light diffracts from a grating, @@ -39,13 +39,13 @@ Diffraction from atomic layers in a crystal.} \end{figure} \section*{Theory} - Consider planes of atoms in a {\bf crystal} as shown in Fig, \ref{ed1} + Consider planes of atoms in a {\bf crystal} as shown in Fig.~\ref{ed1} separated by distance $d$. Electron ``waves'' reflect from each of these planes. Since the electron is wave-like, the combination of the reflections from each -interface will lead to an interference pattern. This is completely analogous to +interface produces to an interference pattern. This is completely analogous to light interference, arising, for example, from different path lengths in the Fabry-Perot or Michelson interferometers. The de Broglie wavelength for the -electron is given by: $\lambda=h/p$, where $p$ can be calculated by knowing the +electron is given by $\lambda=h/p$, where $p$ can be calculated by knowing the energy of the electrons when they leave the ``electron gun'': \begin{equation}\label{Va} \frac{p^2}{2m}=eV_a, @@ -94,7 +94,7 @@ cannot be touched by accident while taking measurements. The diagram of the apparatus is given in Fig.\ref{ed2}. An electron gun (consisting of a heated filament to boil electrons off a cathode and an anode -to accelerate them to, similar to the e/m experiment) ``shoots'' electrons at a +to accelerate them, similar to the e/m experiment) ``shoots'' electrons at a carbon (graphite) target. The electrons diffract from the carbon target and the resulting interference @@ -104,8 +104,8 @@ pattern is viewed on a phosphorus screen. sheets in random orientations. Therefore, the constructive interference patterns will be seen as bright circular rings. For the carbon target, two rings (an outer and inner, corresponding to different crystal planes) will be -seen, corresponding to two spacings between atoms in the graphite arrangement, -see Fig. \ref{ed3} +seen, corresponding to two spacings between atoms in the graphite arrangement +(see Fig.~\ref{ed3}). \begin{figure} \centering \includegraphics[width=4in]{./pdf_figs/ed3} \caption{\label{ed3}Spacing of @@ -137,7 +137,11 @@ voltage $V_a$. Calculate the uncertainties for each $\theta$. To determine the crystalline structure of the target, one needs to carefully measure the diffraction angle $\theta$. It is easy to see (for example, from -Fig.~\ref{ed1} ) that the diffraction angle $\theta$ is 1/2 of the angle between the beam incident on the target and the diffracted beam to a ring, hence the $2\theta$ appearing in Fig.~\ref{ed4}. You are going to determine the diffraction angle $\theta$ for a given accelerated voltage from the simple geometrical ratio +Fig.~\ref{ed1} ) that the diffraction angle $\theta$ is 1/2 of the angle +between the beam incident on the target and the diffracted beam to a ring, +hence the $2\theta$ appearing in Fig.~\ref{ed4}. You are going to determine +the diffraction angle $\theta$ for a given accelerated voltage from the +approximate geometrical ratio \begin{equation} L\sin{2\theta} = R\sin\phi, \end{equation} @@ -145,7 +149,11 @@ where the distance between the target and the screen $L = 0.130$~m is controlled The ratio between the arc length and the distance between the target and the radius of the curvature of the screen $R = 0.066$~m gives the angle $\phi$ in -radian: $\phi = s/2R$. To measure $\phi$ carefully place a piece of masking tape on the tube so that it crosses the ring along the diameter. Mark the position of the ring for each accelerating voltage, and then remove the masking tape and measure the arc length $s$ corresponding to each ring. You can also make these markings by using the thin paper which cash register receipts are printed on. +radian: $\phi = s/2R$. To measure $\phi$ carefully place a piece of +masking tape on the tube so that it crosses the ring along the diameter. +Mark the position of the ring for each accelerating voltage, and then +remove the masking tape and measure the arc length $s$ corresponding to +each ring. You can also make these markings by using the thin paper on which cash register receipts are printed. \begin{figure} @@ -167,16 +175,15 @@ Eq.(\ref{bragg}): Make a plot of $1/\sqrt{V_a}$ versus $\sin\theta$ for the inner and outer rings (both curves can be on the same graph). Fit the linear dependence and measure -the slope for both lines. From the values of the slope find the distance +the slope for both lines. From the values of the slopes find the distances between atomic layers $d_{inner}$ and $d_{outer}$. -Compare your measurements to the accepted values : $d_{inner}=d_{10} = .213$~nm +Compare your measurements to the accepted values: $d_{inner}=d_{10} = .213$~nm and $d_{outer}=d_{11}=0.123$~nm. - \fbox{Looking with Electrons} - -\noindent -\begin{boxedminipage}{\linewidth} The resolution of ordinary optical microscopes +\section*{Looking with Electrons} + +The resolution of ordinary optical microscopes is limited (the diffraction limit) by the wavelength of light ($\approx$ 400 nm). This means that we cannot resolve anything smaller than this by looking at it with light (even if we had no limitation on our optical instruments). Since @@ -186,7 +193,6 @@ scale. This is why ``scanning electron microscopes'' (SEMs) are used to look at very small features. The SEM is very similar to an optical microscope, except that ``light'' in SEMs is electrons and the lenses used are made of magnetic fields, not glass. -\end{boxedminipage} %\begin{tabular}{lll} %Filament Voltage& $V_F$&6.3 V ac/dc (8.0 V max.)\\ |