# LaTex help for nuclear reactions?

1. Feb 21, 2004

### Vodka

LaTex help for nuclear reactions? - got it, thanks

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]

$$5.011265\textrm{amu}$$
---

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

$$\frac{E^{}_{\textrm{difference}}}_{E^{}_{\textrm{potential}}}}$$

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

mproton + mneutron = 1.007276 + 1.008665 = 2.015941 amu

Last edited: Feb 23, 2004
2. Feb 21, 2004

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

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.

Last edited: Feb 21, 2004
3. Feb 22, 2004

### 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.

4. Feb 23, 2004

### chroot

Staff Emeritus
$$\newcommand{\nucl}[3]{ \ensuremath{ \phantom{\ensuremath{^{#1}_{#2}}} \llap{\ensuremath{^{#1}}} \llap{\ensuremath{_{\rule{0pt}{.75em}#2}}} \mbox{#3} } } \nucl{235}{92}{U}$$

Nice!

- Warren

5. Feb 23, 2004

### Vodka

yes, that is a very nice feature :)

and another test...sorry...

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

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

$$22168.125MeV$$

Last edited: Feb 23, 2004