- #1
Vodka
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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 going to 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
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 going to 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: