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.