Protons and the Heisenberg principle
You are using the Heisenberg uncertainty principle to relate the uncertainty in position $x$ to the uncertainty in velocity $v$.

However, the quantitative version of this principle actually is

$$\Delta x\cdot\Delta p\geqslant\tfrac12\hbar $$

where $\Delta x$ is the uncertainty in position $x$ and $\Delta p$ is the uncertainty in momentum $p$.

Certainly, since
$$p=m\cdot v$$
or
$$v=\frac pm$$
where $m$ is mass and $v$ is velocity, you may also use the uncertainty principle to compare the uncertainty in position $x$ to the uncertainty in velocity $v$.

However, the mass of a proton is $m_\mathrm p=1.672\,621\,923\,69(51)\times10^{-27}\ \mathrm{kg}$[CODATA2018], whereas the mass of an electron is only $m_\mathrm e=9.109\,383\,7015(28)\times10^{-31}\ \mathrm{kg}$[CODATA2018]; i.e. a proton is over a thousand times heavier than an electron.

Therefore, a given uncertainty in position $x$ and the corresponding uncertainty in momentum $p$ result in a smaller uncertainty in velocity $v$ for the proton.