Question

(a) Assume an electron in the ground state of the hydrogen atom moves at an average speed of 5.00 × 106 m/s. If the speed is known to an uncertainty of 1 percent, what is the minimum uncertainty in its position

Answer 1

Answer:

The minimum uncertainty in its position is 1.1587 nm

Explanation:

Given;

average speed of electron, v = 5.00 × 10⁶ m/s

percentage of speed uncertainty = 1%

Δv = 0.01( 5.00 × 10⁶ m/s) = 5.00 × 10⁴ m/s

Applying Heisenberg's uncertainty principle, to determine the uncertainty in its position.

ΔxΔP ≥ h/4π

Δx(mΔv)  ≥ h/4π

Δx = h/4πmΔv

where;

Δx is uncertainty in its position

h is Planck's constant

m is mass of electron

Δx ≥ $\frac{6.626 *10^{-34}}{4\pi *9.1*10^{-31}*5*10^4} = 1.1587*10^{-9} \ m$

Δx ≥ 1.1587 nm

Therefore, the minimum uncertainty in its position is 1.1587 nm