# 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.

Borys Shumyatskiy | Certified Educator

calendarEducator since 2015

starTop subjects are Math and Science

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.

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user2962394 | Student

Thank you for sharing.