SME
General Chemistry I · Problem type
The least uncertainty in speed
Use this for “An electron is located to within 0.100 nm. What is the least uncertainty in its speed?”
This is an example. Type your own values to start a new problem.
Δx (nm)
×
m (kg)
×
Δv (m/s)
=
h
÷
(4π)
Δp
Least momentum uncertainty
calculated · kg·m/s
Formulas
Δp
=
6.626 ×
10
-34
4π × Δx ×
10
-9
5.27 ×
10
-25
=
6.626 ×
10
-34
4π × 0.1 ×
10
-9
Δv
=
Δp
m
5.79 ×
10
5
=
5.27 ×
10
-25
9.109 ×
10
-31
Assumptions
•
Heisenberg: Δx·Δp ≥ h ÷ (4π). The page works the equality, so Δp and Δv are the least uncertainties possible.
•
The speed stays well below light’s, so p = mv and the spread in momentum is Δp = mΔv.
•
An electron’s mass is 9.109 × 10⁻³¹ kg; h = 6.626 × 10⁻³⁴ J·s.
Step-by-step
WE KNOW
m = 9.109 × 10⁻³¹ kg, Δx = 0.1 nm
FIND
least momentum uncertainty (Δp), least speed uncertainty (Δv)
Step 1 · Find least momentum uncertainty (Δp)
Use Δp = (6.626 × 10⁻³⁴) ÷ (4π × Δx × 10⁻⁹)
Heisenberg
at
its
limit:
Δp
=
h
4πΔx
,
with
Δx
in
meters
(nm
×
10
-9
).
Δp
=
6.626 ×
10
-34
4π × 0.1 ×
10
-9
Δp
=
5.27 ×
10
-25
kg·
m
/s
Step 2 · Find least speed uncertainty (Δv)
Use Δv = Δp ÷ m
Δp = mΔv, so divide the momentum spread by the mass.
Δv
=
5.27 ×
10
-25
9.109 ×
10
-31
Δv
=
5.79 ×
10
5
m
/s
Check
Put the numbers back in. Both sides should match.
5.27 ×
10
-25
=
6.626 ×
10
-34
4π × 0.1 ×
10
-9
✓
5.79 ×
10
5
=
5.27 ×
10
-25
9.109 ×
10
-31
✓
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Effective nuclear charge (Slater’s rules)
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