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vlabodo [156]
3 years ago
10

Why can’t we overcome the uncertainty predicted by Heisenberg’s principle by building more precise devices to reduce the error i

n measurements below the h/4π limit?
Chemistry
1 answer:
inna [77]3 years ago
4 0

Answer:

Explanation:

This limit is a consequence of Heisenberg´s uncertainty principle:

Δp x Δx > =  h

This state that the product of the uncertainty in momentum ( or  velocity since p = mv ) times the uncertainty in position, Δx , must be greater or equal to Planck´s constant ( 6.626 x 10⁻³⁴ J·s ).

Later models refined this equation to:

Δp x Δx > =   h/4π

This is the consequence of duality wave matter of the electron and Schrodinger´s equation, in which we can talk of probabilities of finding an electron and not confined to specific distances from the nucleus as in the Bohr atom.

Now think of think of this relation in terms of the uncertainty it describes. If we know the position of the electron with great exactitude, the velocity of the particle will be very high since the mass of hte electron is very small.

This a principle in nature and has nothing to do with the precision of our instruments for particles at the subatomic level.

The reason we do not observe this effect  with everyday objects is that the obbects have masses so large compare to  subatomic particles that the term mΔv becomes large enough, allowing us to know the position and velocity of macroscopic objects with small uncertainties:

Δp  x  Δx > = h/4π,  Δp  very large  ( because the mass is very big ) then  Δx is very small

The same does not have with small masses of the subatomic levels.

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