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## Main Question or Discussion Point

Hello all

I have a question concerning The Heisenberg Uncertainty Principle. The principle mathematically looks like this-

[tex]

\Delta x\Delta p \geq \hbar/2

[/tex]

The principle states that you can not measure more than two quantities simultaneously. If you know a particle's position very precisely, then you won't know its momentum very precisely and vise versa. I want to know how it is used, so lets set up an example.

I have a caliper that can measure to an accuracy of ±.05mm. I confine an electron which has a mass of 9.11*10^-31 kilograms in a space of 10mm. Predict the error of velocity. So lets start with.

[tex]

\Delta x\Delta p \geq \hbar/2

[/tex]

Lets solve for

[tex]

\Delta v

[/tex]

[tex]

\Delta x\Delta p \geq \hbar/2

[/tex]

[tex]

\Delta x\Delta m\Delta v \geq \hbar/2

[/tex]

[tex]

\Delta v \geq \hbar/(2 \Delta x\Delta m)

[/tex]

Ok now I just plug in values. If I do something wrong or stupid, don't be afraid to say something. I want to see how this works.

[tex]

\Delta v \geq \hbar/(2*.05*9.11*10^-31)

[/tex]

For an answer I get 1.15759767 m / s. Now I have a question about this answer. Does it mean that the electron will have a velocity of v±1.15759767? I don't know what the final answer is suppose to represent.

Thanks for your help!!

I have a question concerning The Heisenberg Uncertainty Principle. The principle mathematically looks like this-

[tex]

\Delta x\Delta p \geq \hbar/2

[/tex]

The principle states that you can not measure more than two quantities simultaneously. If you know a particle's position very precisely, then you won't know its momentum very precisely and vise versa. I want to know how it is used, so lets set up an example.

I have a caliper that can measure to an accuracy of ±.05mm. I confine an electron which has a mass of 9.11*10^-31 kilograms in a space of 10mm. Predict the error of velocity. So lets start with.

[tex]

\Delta x\Delta p \geq \hbar/2

[/tex]

Lets solve for

[tex]

\Delta v

[/tex]

[tex]

\Delta x\Delta p \geq \hbar/2

[/tex]

[tex]

\Delta x\Delta m\Delta v \geq \hbar/2

[/tex]

[tex]

\Delta v \geq \hbar/(2 \Delta x\Delta m)

[/tex]

Ok now I just plug in values. If I do something wrong or stupid, don't be afraid to say something. I want to see how this works.

[tex]

\Delta v \geq \hbar/(2*.05*9.11*10^-31)

[/tex]

For an answer I get 1.15759767 m / s. Now I have a question about this answer. Does it mean that the electron will have a velocity of v±1.15759767? I don't know what the final answer is suppose to represent.

Thanks for your help!!