A small dust particle of mass 7.90 * 10^-6 g is being observed under a magnifying lens. Its position is determined to within 0.0050 mm. (1 y = 3.156 * 10^7 s, h = 6.626 * 10^-34 J * s) A) Find the minimum uncertainty in its velocity implied by the uncertainty in its position. correct ans = 4.6*10^-21 m/s. Show work. I'm getting 1.3 * 10^-21 m/s. B) Assuming the dust particle is moving at the speed you just found, how many years would it take for the particle to move 1.0 mm? correct ans= 6.9*10^9 yrs. Show work.

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Hello!
A. The main physical law to use here is the uncertainty principle with respect to momentum and position. The formula is
Delta p*Delta x gt= ħ/2,
where Delta p is an uncertainty in a momentum, Delta x is an uncertainty in a position of the same particle, and ħ is the reduced Planck's constant, h/(2 pi).  It is obvious that Delta p = m*Delta v, where m is a mass of a particle and Delta v is an uncertainty in its speed.
Therefore the minimum uncertainty in a velocity is  h/(4 pi)*1/(m*Delta x).
All values are given, but we have to make m from mm and kg from g. The numerical answer is
((6.626*10^(-34))/(4 pi))/(7.9*10^(-6)*10^(-3)*0.0050*10^(-3)) approx 6.626/(7.9*5*4*3.14) * 10^(-34)/10^(-15) approx
approx 0.0133*10^(-19) = 1.333*10^(-21) (m/s).
Thus I agree with your answer. I don't even have an idea where 4.6 may come from.
 
B. A time is a distance divided by a speed, the only problem is to use correct units. A time in seconds is  10^(-3)/(1.33*10^(-21)) approx 0.75*10^(18) (s). To get this time in years we have to divide it by the given number of seconds in a year, i.e. (0.75*10^(18))/(3.156*10^7) approx 0.238*10^(11) = 2.38*10^(10) (years).
If we would use the speed that stated as the "correct answer" for A, the answer would be that you want as "correct" for B.