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\author{Konrad\_Z (StudentKZ)}
\pdfinfo{
  /Title (aqa-a-level-physics-discovery-of-the-electron.pdf)
  /Creator (Cheatography)
  /Author (Konrad\_Z (StudentKZ))
  /Subject (AQA A-Level Physics: Discovery of The Electron Cheat Sheet)
}

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    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{AQA A-Level Physics: Discovery of The Electron Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Konrad\_Z (StudentKZ)} via \textcolor{DarkBackground}{\uline{cheatography.com/188038/cs/42165/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Konrad\_Z (StudentKZ) \\
  \uline{cheatography.com/studentkz} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 24th January, 2024.\\
   Updated 2nd February, 2024.\\
   Page {\thepage} of \pageref{LastPage}.
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  Measure your website readability!\\
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Cathode Ray Tube}}  \tn
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\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/studentkz_1706032576_download (1).png}}} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Production of Cathode Rays}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{When a high potential difference (p.d) is applied across a discharge tube with a low pressure gas inside, the tube will begin to glow. This glow was brightest at the cathode and was called a cathode ray.} \tn 
% Row Count 5 (+ 5)
% Row 1
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\mymulticolumn{1}{x{5.377cm}}{1. The high p.d pulls electrons off the gas atoms, forming positive ion and electron pairs.} \tn 
% Row Count 7 (+ 2)
% Row 2
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\mymulticolumn{1}{x{5.377cm}}{2. The positive gas ions are accelerated towards the cathode and upon collision, release more electrons.} \tn 
% Row Count 10 (+ 3)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{3. These electrons are accelerated across the tube due to the low pressure of the gas. They collide with gas atoms causing them to become excited. These atoms will de-excite and release photons of light.} \tn 
% Row Count 15 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{The Electron Gun}}  \tn
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\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/studentkz_1706081963_download (2).png}}} \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Thermionic Emission}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{Thermionic emission (TE) is when a metal is heated until the free electrons on the surface gain enough energy and are emitted.} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Electron guns use a p.d in order to accelerate electrons. Once the electrons leave the surface via TE, they are accelerated towards an anode with a small gap.} \tn 
% Row Count 7 (+ 4)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Once the electron reaches the anode. Its kinetic energy is equal to the work done on the electron by the electric field.} \tn 
% Row Count 10 (+ 3)
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\mymulticolumn{1}{x{5.377cm}}{1/2mv\textasciicircum{}2\textasciicircum{} = eV \newline mv\textasciicircum{}2\textasciicircum{} = 2eV \newline v\textasciicircum{}2\textasciicircum{} = 2eV / m \newline v = sqrt(2eV / m) \newline sqrt = Square root}  \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Fine Beam Tube}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Deflection in a magnetic field}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{The fine beam tube contains a low pressure gas and has a uniform magnetic field passing through.} \tn 
% Row Count 2 (+ 2)
% Row 1
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\mymulticolumn{1}{x{5.377cm}}{1. Electrons are accelerated using an electron gun and enter the fine beam tube perpendicular to the direction of the field.} \tn 
% Row Count 5 (+ 3)
% Row 2
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\mymulticolumn{1}{x{5.377cm}}{2. The magnetic force on the electrons acts perpendicular to their motion and therefore the electrons move in a circular path.} \tn 
% Row Count 8 (+ 3)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{3. As the electrons move through the fine beam tube, they collide with gas atoms causing them to become excited. The gas atoms then de-excite releasing photons of light meaning that the path of the electrons is visible and the radius of the path can be measured.} \tn 
% Row Count 14 (+ 6)
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\mymulticolumn{1}{x{5.377cm}}{mv\textasciicircum{}2\textasciicircum{}/r = Bev \newline mv/r = Be \newline v = Ber / m \newline v = sqrt (2eV/m) \newline sqrt (2eV/m) = Ber / m \newline 2eV / m = B\textasciicircum{}2\textasciicircum{}e\textasciicircum{}2\textasciicircum{}r\textasciicircum{}2\textasciicircum{} / m\textasciicircum{}2\textasciicircum{} \newline 2V = B\textasciicircum{}2\textasciicircum{}er\textasciicircum{}2\textasciicircum{} / m \newline e/m = 2V / B\textasciicircum{}2\textasciicircum{}r\textasciicircum{}2\textasciicircum{} \newline  \newline sqrt = Square root}  \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Thomson's Crossed Fields}}  \tn
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Balancing Electric and Magnetic Fields}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{This apparatus involves magnetic and electric fields which are perpendicular to each other. The electric and magnetic fields deflect the electrons in opposite directions.} \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{1. Electrons are accelerated using an electron gun and enter the apparatus perpendicular to the direction of both fields. The electrons will be deflected upwards due to electric field and downwards due to the magnetic field.} \tn 
% Row Count 9 (+ 5)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{2. The strengths of these fields are adjusted until the electron beam passes through the crossed fields undeflected. Therefore, the electric and magnetic forces are equal and opposite.} \tn 
% Row Count 13 (+ 4)
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\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Magnetic Force: F=Bev \newline Electric Force: F = Ee where E = V/d so F = Ve / d \newline Bev = Ve / d \newline v = V / Bd \newline v = sqrt (2eV / m) \newline sqrt (2eV / m) = V / Bd \newline 2eV / m = V\textasciicircum{}2\textasciicircum{} / B\textasciicircum{}2\textasciicircum{}d\textasciicircum{}2\textasciicircum{} \newline e / m = V\textasciicircum{}2\textasciicircum{} / 2B\textasciicircum{}2\textasciicircum{}d\textasciicircum{}2\textasciicircum{}V \newline  \newline sqrt = Square root}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Electric Field Only}}  \tn
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\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/studentkz_1706117189_download (3).png}}} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Deflection in an electric field}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{Electrons of a known speed are fired into a uniform electric field of known length.} \tn 
% Row Count 2 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Time Taken = Plate length / Electron speed (t = d/v) \newline Upwards Acceleration (a) = F / m = eV / dm \newline Vertical Distance (s) = 1/2at\textasciicircum{}2\textasciicircum{} because u = 0 hence ut = 0 \newline s = 1/2ad\textasciicircum{}2\textasciicircum{} / v\textasciicircum{}2\textasciicircum{} \newline a = eV / dm \newline e/m = ad / V}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Milikan's apparatus}}  \tn
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\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/studentkz_1706118095_download (4).png}}} \tn 
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\begin{tabularx}{5.377cm}{x{1.73926 cm} x{1.69349 cm} x{1.14425 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{5.377cm}}{\bf\textcolor{white}{Milikan's oil drop experiment}}  \tn
% Row 0
\SetRowColor{LightBackground}
This experiment was formed in order to calculate the charge of an electron. & {\bf{Electric Field Disabled:}} & {\bf{Electric Field Enabled:}} \tn 
% Row Count 5 (+ 5)
% Row 1
\SetRowColor{white}
An atomizer is used to spray tiny negatively charged droplets of oil into a uniform electric field. The droplets will experience an electric force upwards and a weight downwards. The p.d can be adjusted to where these two quantities become equal. & F = 6πηrv & F = EQ \tn 
% Row Count 22 (+ 17)
% Row 2
\SetRowColor{LightBackground}
However, we must find the mass on the droplet and we can't use a mass balance so we turn off the p.d across the plates to remove the electric force and let the oil drop fall freely. & F = mg & F = mg \tn 
% Row Count 35 (+ 13)
\end{tabularx}
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\vfill
\columnbreak
\begin{tabularx}{5.377cm}{x{1.73926 cm} x{1.69349 cm} x{1.14425 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{5.377cm}}{\bf\textcolor{white}{Milikan's oil drop experiment (cont)}}  \tn
% Row 3
\SetRowColor{LightBackground}
Now the oil drop will experience a downwards weight like before but it will now experience a viscous drag force upwards which can be calculated using Stokes' Law. At terminal velocity, this viscous force and the weight will be equal. & At terminal velocity: mg = 6πηrv & EQ = mg \tn 
% Row Count 16 (+ 16)
% Row 4
\SetRowColor{white}
Milikan observed that the charge of the oil droplets was an integer value of 1.6x10\textasciicircum{}-19\textasciicircum{}. This is significant because it shows the charge was quantised meaning it exists in discrete packets of 1.6x10\textasciicircum{}-19\textasciicircum{}. & m = vρ  (where ρ is density) & E = V / d hence QV / d = mg \tn 
% Row Count 30 (+ 14)
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\vfill
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\begin{tabularx}{5.377cm}{x{1.73926 cm} x{1.69349 cm} x{1.14425 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{5.377cm}}{\bf\textcolor{white}{Milikan's oil drop experiment (cont)}}  \tn
% Row 5
\SetRowColor{LightBackground}
 & v = 4/3πr\textasciicircum{}3\textasciicircum{} (treating the oil drop as a sphere) &  \tn 
% Row Count 4 (+ 4)
% Row 6
\SetRowColor{white}
 & m = 4/3πr\textasciicircum{}3\textasciicircum{} ρ hence W = 4/3πr\textasciicircum{}3\textasciicircum{} ρg &  \tn 
% Row Count 7 (+ 3)
% Row 7
\SetRowColor{LightBackground}
 & 6πηrv = 4/3πr\textasciicircum{}3\textasciicircum{} ρg &  \tn 
% Row Count 9 (+ 2)
% Row 8
\SetRowColor{white}
 & 6ηv = 4/3r\textasciicircum{}2\textasciicircum{} ρg &  \tn 
% Row Count 11 (+ 2)
% Row 9
\SetRowColor{LightBackground}
 & 9ηv / 2 = r\textasciicircum{}2\textasciicircum{} ρg &  \tn 
% Row Count 13 (+ 2)
% Row 10
\SetRowColor{white}
 & r\textasciicircum{}2\textasciicircum{} = 9ηv / 2ρg hence r = sqrt (9ηv / 2ρg) &  \tn 
% Row Count 17 (+ 4)
% Row 11
\SetRowColor{LightBackground}
 & mg = 6πηrv hence m = 6πηrv / g therefore m = 6πηv x sqrt (9ηv / 2ρg) &  \tn 
% Row Count 23 (+ 6)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\SetRowColor{LightBackground}
\mymulticolumn{3}{x{5.377cm}}{sqrt = Square Root}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}---}
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% That's all folks
\end{multicols*}

\end{document}