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If the electron is to be located within 5 x 10⁻⁵ A°, what will be the uncertainty in its velocity?

(Mass of the electron = 9.1 x 10⁻³¹ kg)

User Chirag B
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We can see that the uncertainty in the velocity of the electron would be at least 1.053 x 10¹⁹ m/s.

To calculate the uncertainty in the velocity of an electron within a given location, we can use Heisenberg's uncertainty principle.

According to the principle, the uncertainty in position (Δx) and the uncertainty in velocity (Δv) of a particle are related by the equation:

Δx × Δv ≥ h / (4π)

Where h is Planck's constant (approximately 6.626 x 10⁻³⁴ J·s).

Given that the electron is to be located within 5 x 10⁻⁵ A° (angstroms), we need to convert this value to meters. One angstrom is equal to 10⁻¹⁰ meters.

So, the uncertainty in position (Δx) would be:

Δx = 5 x 10⁻⁵ A° * 10⁻¹⁰ m/A° = 5 x 10⁻¹⁵ m

Now, we can substitute the values into the uncertainty principle equation and solve for the uncertainty in velocity (Δv):

(5 x 10⁻¹⁵ m) * Δv ≥ (6.626 x 10⁻³⁴ J·s) / (4π)

To solve for Δv, we need to rearrange the equation:

Δv ≥ (6.626 x 10⁻³⁴ J·s) / (4π * 5 x 10⁻¹⁵ m)

Calculating this expression gives us:

Δv ≥ 1.053 x 10¹⁹ m/s

Therefore, the uncertainty in the velocity of the electron would be at least 1.053 x 10¹⁹ m/s.

User Armin Sadeghi
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