How would you calculate Δx for an electron with Δv = 0.340 m/s?

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Answer 1 · 2015-12-02T16:41:20

$=>Deltax>=1.70xx10^(-4)m$

The Heisenberg's uncertainty principle is given by:

$Deltax*Delta(mv)>=h/(4pi)$

$=>Deltax>=h/(4pi*Delta(mv)$

$=>Deltax>=h/(4pi*mDeltav)$

$=>Deltax>=(6.626xx10^(-34)cancel(kg)*m^(cancel(2))*cancel(s^(-2))*cancel(s))/(4pixx9.11xx10^(-31)cancel(kg)xx0.340cancel(m)*cancel(s^(-1)))$

$=>Deltax>=1.70xx10^(-4)m$

Note that $Delta(mv)=m*Deltav$ because $m$ is the mass of the electron which will remain constant at all time.