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Minchanka [31]
4 years ago
15

The mass of an electron is 9.11× 10–31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 106 m/s with

an uncertainty of Δv = 0.01 × 106 m/s?
Chemistry
1 answer:
Kruka [31]4 years ago
6 0
Just use the Heisenberg Uncertainty principle: 

<span>ΔpΔx = h/2*pi </span>

<span>Δp = the uncertainty in momentum </span>
<span>Δx = the uncertainty in position </span>
<span>h = 6.626e-34 J s (plank's constant) </span>

<span>Hint: </span>

<span>to calculate Δp use the fact that the uncertainty in the momentum is 1% (0.01) so that </span>

<span>Δp = mv*(0.01) </span>

<span>m = mass of electron </span>
<span>v = velocity of electron </span>

<span>Solve for Δx </span>

<span>Δx = h/(2*pi*Δp) </span>

<span>And that is the uncertainty in position. </span>
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