- #1
hover
- 343
- 0
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 let's 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 let's 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 let's 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 let's 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!