(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 kno wn to an uncertainty of 1 percent, what is the minimum uncertainty in its position
1 answer:
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 ≥
Δx ≥ 1.1587 nm
Therefore, the minimum uncertainty in its position is 1.1587 nm
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