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What is the uncertainty in the position, [tex]\Delta x[/tex], of an electron moving near the nucleus at [tex]6.00 \times 10^6 \, \text{m/s}[/tex]? Assume the relative uncertainty in the speed of the electron, [tex]\Delta v[/tex], is 1%; that is, [tex]6.00 \times 10^4 \, \text{m/s}[/tex]. Report your answer in meters.

Answer :

The uncertainty in the position, Δx, of an electron moving near the nucleus at 6.00×106 m/s is 7.24 x 10⁻¹⁰ m

What is relative uncertainty?

Relative uncertainty is the measuring uncertainty fraction by the absolute value of the measured quantity value.

By Heisenberg uncertainty principle,

Δx x Δv = h/(4 xpi x m)

Putting the values in the formula

Δv = 8 x 10⁴ m/s

mass, m = 9.11 x 10⁻³¹ Kg

Putting values,

Δx x (8 x 10⁴) = (6.626 x 10⁻³⁴)/(4 x 3.14 x 9.11 x 10⁻³¹)

Δx x (8 x 10⁴) = 5.791 x 10⁻⁵

Δx = 7.239 x 10⁻¹⁰ m

Therefore, the uncertainty in the position is 7.24 x 10⁻¹⁰ m

To learn more about relative uncertainty, refer to the link:

https://brainly.com/question/24121107

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Rewritten by : Barada

To determine the uncertainty in the position of an electron with a given speed and relative uncertainty, the Heisenberg uncertainty principle is applied, including calculations with the electron's mass and reduced Planck's constant.

To find the uncertainty in the position, \\Delta x\\, of an electron moving near the nucleus at \(6.00\times10^6 m/s\) with a relative uncertainty in the speed, \\Delta v\\, of \(1\%\) or \(6.00\times10^4 m/s\), we apply the Heisenberg uncertainty principle. This principle states that the uncertainty in position (\\Delta x\\) and uncertainty in momentum (\\Delta p\\), and hence velocity (\\Delta v\\), of a particle are inversely related.

We can express the uncertainty in momentum as \(\Delta p = m \times \Delta v\), where \(m\) is the mass of the electron (\(9.11 \times 10⁻³¹\) kg). Then, we employ the formula \(\Delta x \times \Delta p \geq \frac{\hbar}{2}\) where \(\hbar\) is the reduced Planck's constant, to solve for \(\Delta x\).

By substituting the mass of the electron and the given \(\Delta v\), we can estimate the minimum \(\Delta x\) as follows:

\(\Delta p = (9.11 \times 10⁻³¹} \text{ kg}) \times (6.00 \times 10^4 \text{ m/s})\)

\(\Delta x \geq \frac{\hbar}{2\Delta p}\)

Using \(\hbar = 1.055 \times 10^{-34} \text{ Js}\), we find:

\(\Delta x \geq \frac{1.055 \times 10⁻³⁴}}{2 \times 9.11 \times 10³¹times 6.00 \times 10⁴) meters.

After calculating this equation, we obtain a numerical value for \(\Delta x\) that represents the minimum uncertainty in the position of the electron in meters, which can be compared with the approximate size of an atom.