Answer:
[tex]$\Delta x . \Delta p \geq \frac{h}{4 \pi}$[/tex]
∆x is the Uncertainty in the position of the particle
∆p is the Uncertainty in the momentum of the particle
[tex]$\Delta p=\frac{h}{4 \pi \times m \times \Delta x}$[/tex]
[tex]$=\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} s^{-1}}{4 \times 3.14 \times 9.109 \times 10^{-31} \mathrm{kg} \times 471 \times 10^{-12} \mathrm{m}}$[/tex]
[tex]$=\frac{6.626 \times 10^{-34} \mathrm{ms}^{-1}}{53887 \times 10^{-43}}$[/tex]
[tex]=1.23\times10^5 ms^{-1}[/tex] is the answer
