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Physics 7 Online
OpenStudy (anonymous):

For a particle In a one dimension box, if uncertainty in position of the particle is 5% then what will be the percentage uncertainty in the energy of the particle?

OpenStudy (anonymous):

@Vincent-Lyon.Fr

OpenStudy (anonymous):

@Abhisar @perl

OpenStudy (anonymous):

Relative uncertainty cannot be used in the uncertainty relation. you must know the size of the box. 5% of what? a meter, a micron, a femto-meter each will have a different absolute uncertainty in position and therefore in momentum and energy

OpenStudy (anonymous):

@gleem is right, but if I had to hazard a guess I'd imagine that the uncertainty is 5% of the length of the box, in which case the uncertainty in momentum can be calculated. Knowledge of the uncertainty in energy, however, is a different matter.

OpenStudy (anonymous):

take the length of the box to be L. Now how to do it?

OpenStudy (anonymous):

You can calculate the uncertainty in momentum to be \[ \Delta p = \frac{\hbar}{2\Delta x} = \frac{\hbar}{2 \cdot 0.05 \cdot L}\] But a knowledge of the uncertainty in energy requires a knowledge of the state that the particle occupies.

OpenStudy (anonymous):

And of course, this is the minimum uncertainty - to be more correct, replace the first \(=\) with \(\geq\).

OpenStudy (anonymous):

there is some math processing error, i cant see it

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