# Why doesn't calculating the de Broglie wavelength work with h in eV·s?

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

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

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.