A measurement of an electron's speed is v = 2.0 x 10^6 m/s and has an uncertainty of 10%. What is the minimum uncertainty in its position? (h = 6.626 x 10^-34 J*s, m_el = 9.11 x 10^-31 kg). Show steps. I got .91 nm, is it correct?

Hello!
The uncertainty principle (one of them) states that it is impossible to exactly measure the speed and the position of a body. This inexactness is called uncertainty. Of course it is significant only for very small particles of matter, for example for electrons.
The main formula for the speed-position uncertainty is
Delta p*Delta x gt= bar h/2,
where Delta p is the uncertainty of the momentum, Delta x is the uncertainty of the position and bar h is the reduced Plank's constant  h/(2pi).
The momentum is the mass multiplied by the speed and 10% is 0.1. Thus we obtain the inequality
0.1*m_(el)*v*Delta x gt= h/(4pi),
and therefore
Delta x gt= h/(4pi)*10/(m_(el)*v) =(6.626*10^(-34))/(4pi)*10/(9.11*10^(-31)*2*10^6)=
=(6.626)/(4pi*9.11*2)*10^(-8) approx0.0289*10^(-8) = 2.89*10^(-10) (m).
In nanometers it is 0.289 nm.
The answer: the minimum uncertainty in position is about 0.289 nm.

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