LaTex help for nuclear reactions?

[Vodka]

i need a guideline for to get images for nuclear equations, all my attempts thus far were failures. i tried

^4_2 He + ^27_13 Al becomes ^31_15 P becomes ^30_15 P + ^1_0 n ,
but it didn't work. i don't know why, nor do i have the time to learn it before this paper is due. i don't have any other way to do it [short of making my own in MSPaint heh], so if somebody could give me a guideline from which i could substitute letters and numbers as needed, it would be great and i can delete this thread. thanks :D

ediT: thanks a lot :) i'm gonna keep this here for a little bit longer to reference again if i need, but this thread should be gone in a day or so.



\frac{216MeV}{22 168.125MeV} = 0.00974 \times 100 = 0.974





2.4048\times 10^{-25} kg + 1.5364\times 10^{-25} kg + 8.3812\times 10^{-29} kg

3.941\times 10^{-25} kg




E = (3.941\times 10^{-25} kg)(3.00\times 8 \ ms^{-1})^2
[/tex]



F = \frac{k q^{}_1 q^{}_2}_{r^2}





3(m^{}_n) = 3(1.67\times 10^{-27}) = 5.01\times 10^{-27} kg

---

\sum {m^{}_{{}^3H}} = m^{}_p + m^{}_n = 4.033271\textrm{amu}
[/tex]


\sum {m^{}_{{}^2H}} = m^{}_p + m^{}_n = 3.024606\textrm{amu}
[/tex]

5.011265\textrm{amu}
---

E^{}_{{}^3H} = 6.012\times 10^{-10}J
E^{}_{{}^2H} = 4.505\times 10^{-10}J

\frac{E^{}_{\textrm{difference}}}_{E^{}_{\textrm{p otential}}}}

\frac{E^{}_{\textrm{difference}}}_{E^{}_{{}^2H}+E^ {}_{{}^3H}}}

mproton + mneutron = 1.007276 + 1.008665 = 2.015941 amu

[cookiemonster]

{}^4_2\textrm{He} + {}^{27}_{13}\textrm{Al} \to {}^{31}_{15}\textrm{P} \to {}^{30}_{15}\textrm{P} + {}^1_0\textrm{n}


You can make empty characters with {} (two brackets, no space), and you can apply sub and superscripts to empty characters. Click on the image to see what I typed in.

cookiemonster

[GRQC]

Here's a nice custom command which I use:

\newcommand{\nucl}[3]{
\ensuremath{
\phantom{\ensuremath{^{#1}_{#2}}}
\llap{\ensuremath{^{#1}}}
\llap{\ensuremath{_{\rule{0pt}{.75em}#2}}}
\mbox{#3}
}
}

It must be implemented in math mode. So, if you want the chemical symbol for U-235, you would type $\nucl{235}{92}{U}$.

Works great.

[chroot]

Originally posted by GRQC
So, if you want the chemical symbol for U-235, you would type $\nucl{235}{92}{U}$.
\newcommand{\nucl}[3]{
\ensuremath{
\phantom{\ensuremath{^{#1}_{#2}}}
\llap{\ensuremath{^{#1}}}
\llap{\ensuremath{_{\rule{0pt}{.75em}#2}}}
\mbox{#3}
}
}
\nucl{235}{92}{U}


Nice! [:)]

- Warren

[Vodka]

yes, that is a very nice feature :)

and another test...sorry...

\frac{E^{}_{\textrm{difference}}}_{E^{}_{{}^2H}+E^

{}_{{}^3H}}}

E^{}_{{}^2H} = 4.505\times 10^{-10}J = 281.56
E^{}_{{}^3H} = 6.012\times 10^{-10}J = 375.75

22168.125MeV