An electron has a speed of 400 m/sec with uncertainty of 0.01%. Find the accuracy in its position.

Given: \(v=400\) m/s, so \(\Delta v=0.01\%\) of 400 \(=\dfrac{0.01}{100}\times400=0.04\) m/s. \(m=9.1\times10^{-31}\) kg.

Uncertainty principle: \(\Delta x\,\Delta p\ge\dfrac{h}{4\pi}\)

\[ \Delta x=\frac{h}{4\pi m\Delta v}=\frac{6.63\times10^{-34}}{4\pi\times9.1\times10^{-31}\times0.04}=\frac{6.63\times10^{-34}}{4.574\times10^{-31}}=1.45\times10^{-3}\ \text{m} \]

The position can be fixed only to within \(\Delta x\approx1.45\times10^{-3}\) m ≈ 1.45 mm.

If the form \(\Delta x\Delta p\ge\hbar\) is used, \(\Delta x\approx2.9\times10^{-3}\) m.