Uncertainty momentum position velocity

[jackxxny]

Homework Statement

I have a dilemma. My problem states that an electron that is in position 2.34 nm along the x axis, travels along the x-axis with a certain speed (A) and a with the uncertainty in the velocity (B). I am asked to calculate the minimum uncertainty in the position.

Homework Equations

I know

delta(x)*delta(P)>= hbar/2
delta(P)= delta(v)*(mass electron)

for delta(v) I'm going to use the B

The Attempt at a Solution

I have done this

delta (x) = (hbar)/2*(delta(v))*(mass electron)

my question is, i don't need the position and the velocity then

 

[Doc Al]

I have done this

delta (x) = (hbar)/2*(delta(v))*(mass electron)

Careful with parentheses--it's hard to tell if you are multiplying or dividing by delta(P).

my question is, i don't need the position and the velocity then

That's right--you don't need that information.

 

[nlightner]

why do I need the uncertainty in both of them?



As a scientist, it is important to remember that uncertainty is a fundamental concept in quantum mechanics. The uncertainty principle states that the more precisely one quantity is known, the less precisely the other can be known. In this case, the uncertainty in position and velocity are related through the uncertainty principle. The equation you have used, delta(x)*delta(P)>= hbar/2, is a direct consequence of the uncertainty principle. Therefore, it is necessary to consider the uncertainty in both position and velocity in order to calculate the minimum uncertainty in position. This allows us to understand the limitations of our knowledge and make accurate predictions about the behavior of particles at the quantum level.