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\author{MostAncientDream}
\pdfinfo{
  /Title (alvl-p2-radioactive-decay-ch11.pdf)
  /Creator (Cheatography)
  /Author (MostAncientDream)
  /Subject (Alvl P2: radioactive decay (ch11) Cheat Sheet)
}

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    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Alvl P2: radioactive decay (ch11) Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{MostAncientDream} via \textcolor{DarkBackground}{\uline{cheatography.com/168994/cs/39402/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}MostAncientDream \\
  \uline{cheatography.com/mostancientdream} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 26th January, 2024.\\
   Updated 29th June, 2023.\\
   Page {\thepage} of \pageref{LastPage}.
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  Measure your website readability!\\
  www.readability-score.com
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Equations}}  \tn
% Row 0
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decay rate & ∆N/∆t = -λN \tn 
% Row Count 2 (+ 2)
% Row 1
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activity & A = λN \tn 
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half life & T = ln2/λ \tn 
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activity & ln2N/T \tn 
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% Row 4
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Number of atoms after decay & N = N`o`e\textasciicircum{}-λt \tn 
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activity after decay & A = A`o`e\textasciicircum{}-λt \tn 
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% Row 6
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mass & mol x RAM \tn 
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\seqsplit{spontaneous} & can't be influenced/independent \tn 
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% Row 1
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random & can't predict when it will happen \tn 
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% Row 2
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decay constant & probability of a nucleus decaying per unit time \tn 
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% Row 3
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activity & number of disintegrations (or emissions) per unit time \tn 
% Row Count 8 (+ 2)
% Row 4
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metastable state & when an atom/nucleus exists for an extended time in a state other than ground state \tn 
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% Row 5
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daughter nucleus & product of the decay of a radioactive ('parent') nucleus \tn 
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half life & time taken for half of a sample of radioactive nuclei to decay \tn 
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\mymulticolumn{4}{x{5.377cm}}{\bf\textcolor{white}{Decay models of unstable nuclei}}  \tn
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Alpha α & Beta minus β- & Beta plus β+ & Gamma γ \tn 
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% Row 1
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 & N decay to P, emitting e and /Ve & (K capture) & nucleu in excited state emits \seqsplit{photon-some} after β decay \tn 
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% Row 2
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\textasciicircum{}241\textasciicircum{}`95`Am -\textgreater{} \textasciicircum{}237\textasciicircum{}`93`Np + \textasciicircum{}4\textasciicircum{}`2`α & \textasciicircum{}63\textasciicircum{}`28`Ni -\textgreater{} \textasciicircum{}63\textasciicircum{}`29`Cu +`-1`e + /Ve & \textasciicircum{}124\textasciicircum{}`53`I `+1`e-\textgreater{} \textasciicircum{}124\textasciicircum{}`52`Te + Ve & \textasciicircum{}99\textasciicircum{}`42`Mo -\textgreater{} \textasciicircum{}99m\textasciicircum{}`43`Tc + `-1`e + /Ve \tn 
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 &  & m -\textgreater{} \seqsplit{metastable} state & \textasciicircum{}99m\textasciicircum{}`43`Tc -\textgreater{} \textasciicircum{}99\textasciicircum{}`43`Tc + γ \tn 
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 &  & nucleus in high excited state for extended time & time (grater than a billionth of a second) \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{radioactive decay}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{-{}-\textgreater{} the significance of the - sign in -λN is that the number of radioactive nuclei in a sample material decreases over time  \newline % Row Count 3 (+ 3)
{\bf{Example Question:}} \newline % Row Count 4 (+ 1)
. Lanthanum-139 is the more abundant \newline % Row Count 5 (+ 1)
isotope and makes up 99.911\% of naturally occurring lanthanum. The remaining 0.089\% is the radioisotope lanthanum-138 \newline % Row Count 8 (+ 3)
. Lanthanum-138 has a decay constant of 2.0 × 10−19 s−1; and 139g of lanthanum contain 6 × 10\textasciicircum{}23 atoms. \newline % Row Count 11 (+ 3)
{\emph{Calculate the activity of a 40g sample of Lanthanum}} \newline % Row Count 13 (+ 2)
The number of atoms in 40g of lanthanum is \newline % Row Count 14 (+ 1)
6 x10\textasciicircum{}23\textasciicircum{} x 40/139 = 1.73 x10\textasciicircum{}23\textasciicircum{}  \newline % Row Count 15 (+ 1)
However, only 0.089\% of these are lanthanum-138. \newline % Row Count 16 (+ 1)
So the number of lanthanum-138 nuclei is \newline % Row Count 17 (+ 1)
N =  1.73x10\textasciicircum{}23\textasciicircum{} x 0.089/100 = 1.54x10\textasciicircum{}20\textasciicircum{} \newline % Row Count 18 (+ 1)
	 so \newline % Row Count 19 (+ 1)
A=λN \newline % Row Count 20 (+ 1)
= 2.0x10\textasciicircum{}-19\textasciicircum{} s\textasciicircum{}-1\textasciicircum{} x 1.54x10\textasciicircum{}20\textasciicircum{} \newline % Row Count 21 (+ 1)
= 31Bq% Row Count 22 (+ 1)
} \tn 
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 & - uses \textasciicircum{}14\textasciicircum{}C (half life of 5700 years) \tn 
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 & - The ratio of Carbon-14 remaining indicates the times since the death of a living substance \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Nuclear Instability}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{- every element has many different isotopes \newline % Row Count 1 (+ 1)
- however most isotopes are unstable, and decay by emission of radiation to become more stable \newline % Row Count 3 (+ 2)
- this can be plotted on a no.Neutrons to no.Protons graph% Row Count 5 (+ 2)
} \tn 
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% That's all folks
\end{multicols*}

\end{document}