Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

LaTex help for nuclear reactions?

  1. Feb 21, 2004 #1
    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.

    [tex]

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

    [/tex]


    [tex]
    2.4048\times 10^{-25} kg + 1.5364\times 10^{-25} kg + 8.3812\times 10^{-29} kg [/tex]

    [tex] 3.941\times 10^{-25} kg

    [/tex]


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

    [tex]

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

    [/tex]

    [tex]

    3(m^{}_n) = 3(1.67\times 10^{-27}) = 5.01\times 10^{-27} kg
    [/tex]
    ---
    [tex]
    \sum {m^{}_{{}^3H}} = m^{}_p + m^{}_n = 4.033271\textrm{amu}[/tex]
    [/tex]

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

    [tex] 5.011265\textrm{amu} [/tex]
    ---

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

    [tex] \frac{E^{}_{\textrm{difference}}}_{E^{}_{\textrm{potential}}}} [/tex]

    [tex] \frac{E^{}_{\textrm{difference}}}_{E^{}_{{}^2H}+E^{}_{{}^3H}}} [/tex]

    mproton + mneutron = 1.007276 + 1.008665 = 2.015941 amu
     
    Last edited: Feb 23, 2004
  2. jcsd
  3. Feb 21, 2004 #2
    [tex]
    {}^4_2\textrm{He} + {}^{27}_{13}\textrm{Al} \to {}^{31}_{15}\textrm{P} \to {}^{30}_{15}\textrm{P} + {}^1_0\textrm{n}
    [/tex]

    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
     
    Last edited: Feb 21, 2004
  4. Feb 22, 2004 #3
    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.
     
  5. Feb 23, 2004 #4

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    Nice! :smile:

    - Warren
     
  6. Feb 23, 2004 #5
    yes, that is a very nice feature :)

    and another test...sorry...

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

    {}_{{}^3H}}} [/tex]

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

    [tex] 22168.125MeV [/tex]
     
    Last edited: Feb 23, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: LaTex help for nuclear reactions?
  1. Nuclear Reactions (Replies: 12)

  2. Nuclear reaction heat (Replies: 4)

Loading...