4

Basically, I'm trying to use amsmath to align the 2 possible outcomes of a reaction, which naturally does not include an equals sign.

\begin{equation}
\ce{^{2}_{1}H} + \ce{^{2}_{1}H} &\rightarrow \ce{^{1}_{1}H} + \ce{^{3}_{1}H} + 4.03 \text{MeV}\\
&\rightarrow \ce{^{1}_{0}n} + \ce{^{3}_{2}He} + 3.27 \text{MeV}\\
\end{equation}

The reaction has a branch, 2 possible outcomes occur and I'm trying to align the arrows. Any advice?

  • 3
    replace equation by align*. – Sigur Feb 12 '19 at 22:47
  • 3
    ... and consider using siunitx for typesetting MeV. – user121799 Feb 12 '19 at 22:50
  • @P227 I will substitute \mathrm{MeV} instead of \text{MeV} if you not will use siunitx. – Sebastiano Feb 12 '19 at 23:06
4

You could use split or a fancier typesetting.

\documentclass{article}
\usepackage{amsmath,siunitx}
\usepackage[version=4]{mhchem}

\DeclareSIUnit{\eV}{eV}

\begin{document}

\begin{equation}
\begin{split}
\ce{^{2}_{1}H} + \ce{^{2}_{1}H}
  &\rightarrow \ce{^{1}_{1}H} + \ce{^{3}_{1}H} + \SI{4.03}{\mega\eV} \\
  &\rightarrow \ce{^{1}_{0}n} + \ce{^{3}_{2}He} + \SI{3.27}{\mega\eV}
\end{split}
\end{equation}

\begin{equation}
\ce{^{2}_{1}H} + \ce{^{2}_{1}H}
\mathrel{\begin{array}{@{}c@{}}\nearrow\\\searrow\end{array}}
\begin{array}{@{}l@{}}
  \ce{^{1}_{1}H} + \ce{^{3}_{1}H} + \SI{4.03}{\mega\eV} \\[2ex]
  \ce{^{1}_{0}n} + \ce{^{3}_{2}He} + \SI{3.27}{\mega\eV}
\end{array}
\end{equation}

\end{document}

enter image description here

Even fancier, with tikz-cd:

\documentclass{article}
\usepackage{amsmath,siunitx}
\usepackage[version=4]{mhchem}
\usepackage{tikz-cd}

\DeclareSIUnit{\eV}{eV}

\begin{document}

\begin{equation}
\begin{tikzcd}[row sep=-1em,column sep=3em]
& \ce{^{1}_{1}H} + \ce{^{3}_{1}H} + \SI{4.03}{\mega\eV} \\
\ce{^{2}_{1}H} + \ce{^{2}_{1}H}
  \arrow[ur,start anchor=real east,end anchor=west]
  \arrow[dr,start anchor=real east,end anchor=real west] \\
& \ce{^{1}_{0}n} + \ce{^{3}_{2}He} + \SI{3.27}{\mega\eV}
\end{tikzcd}
\end{equation}

\end{document}

enter image description here

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.