Question

An electron has an uncertainty in its position of 471 pm .

Answer 1

`$\Delta x . \Delta p \geq (h)/(4 \pi)$`

∆x is the Uncertainty in the position of the particle

∆p is the Uncertainty in the momentum of the particle

`$\Delta p=(h)/(4 \pi * m * \Delta x)$`

`$=\frac{6.626 * 10^(-34) \mathrm{kgm}^(2) s^(-1)}{4 * 3.14 * 9.109 * 10^(-31) \mathrm{kg} * 471 * 10^(-12) \mathrm{m}}$`

`$=\frac{6.626 * 10^(-34) \mathrm{ms}^(-1)}{53887 * 10^(-43)}$`

`=1.23*10^5 ms^(-1)` is the answer