Question

How much time is needed to measure kinetic energy of an electron whose speed is 10 m/s with an uncertainty of not more than 0.1% How much the electron will travel in this period of time?

Answer 1

1) From the uncertainty principle:

$$∆E∆t ≥ \frac{\hslash}{2}$$

$$\frac{\Delta E}{E}=\frac{\hslash}{2E\Delta t}=0.001$$

$$\Delta t=\frac{\hslash}{0.001m_e v^2}=1.16\ ms$$

2)

$$x=v\Delta t=(10)(0.00106)=0.0106\ m=1.06\ cm$$