Question

if error in measurement of momentum and position are equal .find error in momentum ​

Answer 1

∆x is the error in position measurement and ∆p is the error in the measurement of momentum, then

∆X × ∆p ≥ \frac{h}{4\pi }

4π

h

Since momentum, p = mv, Heisenberg’s uncertainty principle formula can be alternatively written as-

∆X × ∆mv ≥ \frac{h}{4\pi }

4π

h

or ∆X × ∆m × ∆v ≥ \frac{h}{4\pi }

4π

h

Where, ∆V is the error in the measurement of velocity and assuming mass remaining constant during the experiment,

∆X × ∆V ≥ \frac{h}{4\pi m}

4πm

h

.

Accurate measurement of position or momentum automatically indicates larger uncertainty (error) in the measurement of the other quantity.

Applying the Heisenberg principle to an electron in an orbit of an atom, with h = 6.626 ×10-34Js and m= 9.11 ×10-31Kg,

∆X × ∆V ≥ \frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 9.11\times {{10}^{-31}}}

4×3.14×9.11×10

−31

6.626×10

−34

= 10-4 m2 s-1.

If the position of the electron is measured accurately to its size (10-10m), then the error in the measurement of its velocity will be equal or larger than 106m or 1000Km.

Heisenberg principle applies to only dual-natured microscopic particles and not to a macroscopic particle whose wave nature is very small.

Answer 2

Answer: Error in ∆p = 1/2√h/π

Explanation:

Error in measurement of momentum (∆p) = Error in measurement of position (∆x)

By Heisenberg's uncertainty principle,

∆x.∆p ≥ h/4π

=> ∆x² ~ h/4π

=> ∆x ~ √(h/4π)

=> ∆x = ∆p ~ 1/2√(h/π)

More:

This principle is applicable to all moving particles, especially microscopic particles. However, this principle is not applicable for macroscopic objects, e.g.,cricket ball. We can clearly determine the momentum of the ball and position simultaneously by using observations and equations.

For an e-,

∆x.∆p ≥ h/4π

=> ∆x.m∆v ~ h/4π

=> ∆x.∆v ~ h/4πm

=> ∆x.∆v ≈ 10-⁴ m-² s-¹