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I'm trying to calculate the de Broglie wavelength of a particle with a known momentum $p = 980.93 \,\mathrm{GeV}/\mathrm s$. The de Broglie relation is: $$ \lambda = \frac{h}{p} $$

but I notice that if I use the value of $4.136\times 10^{-15}\, \mathrm{eV}\cdot\mathrm s$ for the Planck's constant, the answer that I get from the equation is in units of $\mathrm s^2$, not m. Specifically: $$ \lambda = \frac{4.136 \times 10^{-15} \mathrm{eV}\cdot \mathrm s}{980.93 \, \mathrm{GeV}/\mathrm s} = 4.25\times 10^{-27}\, \mathrm s^2 $$ which is not at all equivalent to the value that one gets when using $h = 6.626\times 10^{-34} \,\mathrm J\cdot\mathrm s$ instead (since $\mathrm J\cdot\mathrm s$ can be converted into $\frac{\mathrm{kg}\cdot \mathrm m^2}{\mathrm s^2}\cdot \mathrm s$). $$ \lambda = \frac{6.626\times10^{-34}\,\frac{\mathrm{kg}\cdot \mathrm m^2}{\mathrm s^2}\cdot \mathrm s}{5.237\times10^{-16}\frac{\mathrm{kg}\cdot \mathrm m}{\mathrm s}} = 1.27 \times 10^{-18} \,\mathrm m $$

What's going on here?

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You're running into trouble because in order to give momentum units of energy, you're setting the speed of light equal to 1, $c=1$. If you keep the units of $c$ the momentum should be given in units of $\text{eV}/c$. By dimensional analysis you can check for yourself that eV/s does not have units of momentum (kg$\cdot$m/s).

Therefore, in your case the momentum is actually given by $p = 980.93~ \text{GeV}/c$ which yields

$$\lambda = \frac{4.136 \times 10^{-15} \text{eV}\cdot \text{s}}{980.93 \, \text{GeV}/c} = \frac{4.136 \times 10^{-15} \text{eV}\cdot \text{s} \cdot 2.99 \times 10^8 \text{m/s}}{980.93 \, \text{GeV}} = 1.27\times 10^{-18}\, \text{m}. $$

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    $\begingroup$ Ah, I can't believe the source of my confusion for one hour straight was forgetting the units for momentum!! I feel positively ridiculous, but thank you :) $\endgroup$ – bream Apr 24 '18 at 9:22
  • $\begingroup$ In references (like the particle data booklet) for fields that make heavy use of these units you will find tabulated values for $hc$ and/or $\hbar c$ to easy the transition from natural units to SI and back again. $\endgroup$ – dmckee --- ex-moderator kitten Apr 24 '18 at 17:00
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The momentum $P$ is not homogeneous to eV/s but to eV.s/m which one can remember using the equation :

$$P^0 = \frac{E}{c}$$ Where $P^0$ is the time component of the quadri-momentum, $E$ the energy of the particle and $c$ the speed of light.

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