typos fixed, thanks to Jordan

1 files changed, 8 insertions, 7 deletions

diff --git a/pe-effect.tex b/pe-effect.tex
index 760ef13..6a78e76 100644
--- a/pe-effect.tex
+++ b/pe-effect.tex
@@ -21,7 +21,7 @@
 \section*{Theory}
 
 \begin{figure}[h]
-\centering \includegraphics[height=3in]{./pdf_figs/pefig1} \caption{\label{pefig1} A simplified schematic of the photo-electric apparatus.}
+\centering \includegraphics[height=3in]{./pdf_figs/pefig1} \caption{\label{pefig1} A simplified schematic of the photoelectric apparatus.}
 \end{figure}
  Consider an apparatus as outlined in Figure \ref{pefig1} (it is the
 apparatus which Heinrich Hertz used to inadvertently discover the ``photoelectric effect''). Light of a
@@ -29,7 +29,7 @@ frequency $\nu$ strikes a cathode, causing electrons to be emitted with velocity
 between the anode and the cathode can accelerate emitted electrons towards the positive anode, producing an
 electrical current in the circuit. A reverse bias potential applied between the anode and cathode will slow down
 the electrons, and even stop them from reaching the anode by matching their kinetic energy. This is the way to
-carefully measure the kinetic energies of the photo electrons.
+carefully measure the kinetic energies of the photoelectrons.
 \begin{equation}
 KE_{max} = (\frac{1}{2}mv^2)_{max} = eV_0
 \end{equation}
@@ -38,7 +38,7 @@ where $KE_{max}$ is the maximum kinetic energy, $m$ and $e$ are the mass and the
 $V_0$ is a potential required to stop the electrons (known as the stopping potential).
 
 From the point of view of a wave theory of light, the energy carried by the light wave is proportional to its
-intensity, and independent of light frequency. Thus, it was logical to expect that stronger light should
+intensity and independent of light frequency. Thus, it was logical to expect that stronger light should
 increase the energy of photoelectrons. However, the experiments of Philipp Lenard in 1900 showed something
 different: although the maximum current increased with light intensity, the stopping potential $V_0$ was
 independent of light intensity. This meant that increasing the rate of energy falling onto the cathode does not
@@ -77,7 +77,7 @@ for the emission of electrons with particular spin. See: www.jlab.org
 
 \section*{Procedure}
 
-\textbf{Equipment needed}: Pasco photo-electric apparatus, Hg lamp, digital
+\textbf{Equipment needed}: Pasco photoelectric apparatus, Hg lamp, digital
 voltmeter. The Pasco apparatus contains a circuit which automatically determines the stopping potential, which you measure off of a voltmeter, so there is no need to adjust the stopping potential yourself or measure the current (lucky you!). Read the brief description of its operation in the appendix.
 
 \begin{figure}
@@ -92,7 +92,7 @@ lines in both the first and the second diffraction orders on both sides. Keep in
 Often the first/second order lines on one side are brighter than on the other - check your apparatus and
 determine what orders you will be using in your experiment.
 
-After that install the $h/e$ Apparatus and focus the light from the Mercury Vapor Light Source onto the slot in
+After that, install the $h/e$ Apparatus and focus the light from the Mercury Vapor Light Source onto the slot in
 the white reflective mask on the $h/e$ Apparatus. Tilt the Light shield of the Apparatus out of the way to
 reveal the white photodiode mask inside the Apparatus. Slide the Lens/Grating assembly forward and back on its
 support rods until you achieve the sharpest image of the aperture centered on the hole in the photodiode mask.
@@ -123,11 +123,12 @@ the $h/e$ Apparatus.
 	we don't know the work function. 
 	You'll measure it in the second part of the lab. 
 \begin{itemize}
-	\item For now switch to violet color and measure the stopping
+	\item For now, switch to violet color and measure the stopping
 		potential for this color.
 	
 	\item compute $h c$ using  $\Delta
-		V_0$ between the two measurements and  values of light wavelength for this
+		V_0$ between the two measurements and  values of light
+		wavelength for these
 		two cases.  Does your data agree with the accepted value?  For this
 		computation, it might be helpful to realize that $eV_0$ is the electron
 		charge times the voltage. If you measured $V_0=\unit[1]{V}$ then $eV_0 =1$