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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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Borys Shumyatskiy eNotes educator | Certified Educator

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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.