Question

An electron is travelling at a speed of $1 \cdot 00 \times 10^{8} cms ^{-1}$. The uncertainty in the position of electron is $1.0 \times 10^{-11} cm$. Calculate the uncertainty in the momentum of the electron. $\left(m_{e}=9.1 \times 10^{-28} g \right)$. Comment on the result.

Answer 1

$p=m v=9 \cdot 10 \times 10^{-28} \times 1 \cdot 0 \times 10^{8} g \cdot cm s ^{-1}=9 \cdot 1 \times 10^{-20} g \cdot cm s ^{-1}$

$\\\because \quad \Delta p=\frac{h}{4 \pi \Delta x}=\frac{6 \cdot 626 \times 10^{-27}}{4 \times 3 \cdot 1416 \times 1 \cdot 0 \times 10^{-11}} g \cdot cm s ^{-1}=5 \cdot 27 \times 10^{-17} g \cdot cm s ^{-1}$

The error in the momentum measurement is approximately 600 times the value of the momentum itself, which means that we cannot measure the momentum at all.

This means that we cannot measure the speed either.

So our initial statement that an electron is travelling at a speed of $1.0 \times 10^{8} cm / s$ is not an experimental statement, only hypothetical.