Help Calculating


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Sara Richmond 2K
Posts: 94
Joined: Fri Aug 30, 2019 12:16 am

Help Calculating

Postby Sara Richmond 2K » Sat Nov 02, 2019 1:59 pm

I am kind confused when it comes to the Heisenberg uncertainty principle. Can you help me calculate this problem? It is from a worksheet made by the UA Karen Leung
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Diana_Diep2I
Posts: 101
Joined: Sat Aug 17, 2019 12:17 am

Re: Help Calculating

Postby Diana_Diep2I » Sat Nov 02, 2019 2:45 pm

1) manipulate your equation so that you are solving for ΔX.
2) ΔP is equal to mΔv because mass is constant and velocity is the only thing that changes. Set up the equation Δp=mΔv on the side to solve for Δp.
3) multiply your given velocity by 2 so that you get the entire error and not just one part of it. 2(0.55)=1.1 m/s
4) Convert your given marble mass to kg because Planck's constant involves kg and not g.
5) Now plug your mass and velocity into the Δp equation.
6) plug your Δp into the original Heisenberg's equation in which you manipulated your variables to solve for Δx

kevinchang_4I
Posts: 55
Joined: Tue Nov 13, 2018 12:17 am

Re: Help Calculating

Postby kevinchang_4I » Sat Nov 02, 2019 2:53 pm

In this problem we have to use 2 equations.
p*x = 1/2*hbar and p = m*v

m*v*x = 1/2*hbar

x(what we're looking for) = hbar / 2*m*v

We have all three components, but not in the right units. 0.55 m/s must be multiplied by 2 because we are given it as a range. 1.5 g must be 1.5 x 10^3 kg, and we have to remember hbar which is h/2pi, or 6.626 x 10^-34, or 1.054 x 10^-34.

(1.054 x 10^-34) / (1.5 x 10^-3) x (1.1) = 6.39134424 × 10^-32 m

This tiny value makes sense because there isn't a lot of uncertainty in the position of a ball.

Sara Richmond 2K
Posts: 94
Joined: Fri Aug 30, 2019 12:16 am

Re: Help Calculating

Postby Sara Richmond 2K » Wed Nov 06, 2019 1:22 pm

Thank you, the process makes more sense now. I was missing the equation for delta P. "ΔP is equal to mΔv because mass is constant and velocity is the only thing that changes. Set up the equation Δp=mΔv on the side to solve for Δp". Now that I know this, it makes more sense.


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