An electron is allowed to move freely in a closed cubic box of length of side 10 cm. The uncertainty in its velocity will be:

$$\begin{gathered} ∆x.∆p=\frac{h}{4\mathrm{\pi }} \\ m∆v.∆x=\frac{h}{4\mathrm{\pi }} \\ ∆v=\frac{h}{4\mathrm{\pi }∆\mathrm{xm}} \end{gathered}$$

Substitute the values:

$$\begin{gathered} ∆\mathrm{v}=\frac{6.63\times {10}^{-34}}{4\times 3.14\times 9.1\times {10}^{-31}\times \underset{\mathrm{diagonal} \mathrm{of} \mathrm{cube} \sqrt{3}\mathrm{a}}{\underbrace{\sqrt{3}\times 0.1}}} \\ ∆\mathrm{v}\cong 3.34\times {10}^{-4} \mathrm{m}/\sec \end{gathered}$$