The uncertainty in momentum and uncertainty in the position of a particle are equal. So, $∆\mathrm{v}=∆\mathrm{x}$

According to the uncertainty principle,

The uncertainty principle states that we cannot know both the position and speed of a particle, such as a photon or electron, with perfect accuracy.

$$\begin{gathered} ∆\mathrm{p}.∆\mathrm{x}=\frac{\mathrm{h}}{4\mathrm{\pi }} \\ ∆\mathrm{p}=\mathrm{m}∆\mathrm{v} \\ \mathrm{If} ∆\mathrm{v}=∆\mathrm{x} \\ \mathrm{m}{\left(∆\mathrm{v}\right)}^2=\frac{\mathrm{h}}{4\mathrm{\pi }} \\ {\left(∆\mathrm{v}\right)}^2=\frac{\mathrm{h}}{4\mathrm{\pi m}} \\ ∆\mathrm{v}=\frac{1}{2}\sqrt{\frac{\mathrm{h}}{\mathrm{\pi m}}} \end{gathered}$$

If the uncertainty in velocity and uncertainty in the position of a particle are equal, then the uncertainty in its momentum would be given by $$\begin{gathered} \mathrm{m\Delta v}=\frac{\mathrm{m}}{2}\sqrt{\frac{\mathrm{h}}{\mathrm{\pi m}}} \\ \Rightarrow ∆\mathrm{p}=\frac{1}{2}\sqrt{\frac{\mathrm{mh}}{\mathrm{\pi }}} \end{gathered}$$.