Uncertainty in position of an electron ( mass of an electron is $$ = 9.1 \times {10^{ - 28}}g$$   ) moving with a velocity of $$3 \times {10^4}cm/s$$    accurate upto $$0.001\% $$  will be ( use $$\frac{h}{{4\pi }}$$  in uncertainty expression where $$h = 6.626 \times {10^{ - 27}}erg\,s$$     )

Solution :
According to Heisenberg’s uncertainty principle
$$\eqalign{ & \Delta x \times \Delta v = \frac{h}{{4\pi m}} \cr & {\text{Here, }}\Delta x = {\text{uncertainty in position}} \cr & \Delta v = {\text{uncertainty in velocity}} \cr & h = {\text{Planck's constant}}\,\,\,\left( {6.626 \times {{10}^{ - 27}}Js} \right) \cr & m = {\text{mass of electron }}\left( {9.1 \times {{10}^{ - 28}}kg} \right) \cr & {\text{Here,}}\,\,\Delta v = 0.001\% \,\,{\text{of}}\,\,3 \times {10^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.001}}{{100}} \times 3 \times {10^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.3\,cm/s \cr & \therefore \,\,\Delta x = \frac{h}{{4\pi m\,\Delta v}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 27}}}}{{4 \times 3.14 \times 9.1 \times {{10}^{ - 28}} \times 0.3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.93\,cm \cr} $$