SME
University Physics III: Waves, Optics & Modern · Problem type
The uncertainty principle: the least spread in momentum
Use this for “An electron is confined to a region 0.1 nm wide. What is the least uncertainty in its momentum and in its speed?”
This is an example. Type your own values to start a new problem.
Δx (nm)
·
Δp (kg·m/s)
≥
ħ
÷
2
m
Particle mass
example · kg
Δv
Least uncertainty in speed
calculated · m/s
Formulas
Δp
=
1.0546 ×
10
-34
2 × Δx ×
10
-9
5.273 ×
10
-25
=
1.0546 ×
10
-34
2 × 0.1 ×
10
-9
Δv
=
Δp
m
5.7888 ×
10
5
=
5.273 ×
10
-25
9.109 ×
10
-31
Assumptions
•
Heisenberg: Δx·Δp ≥ ħ ÷ 2, with ħ = h ÷ 2π = 1.0546 × 10⁻³⁴ J·s; Δx and Δp are standard deviations.
•
The answers are least values: the spreads can be larger, never smaller.
•
Slow next to light, so Δv = Δp ÷ m. Electron 9.109 × 10⁻³¹ kg, proton 1.673 × 10⁻²⁷ kg.
Step-by-step
WE KNOW
m = 9.109 × 10⁻³¹ kg, Δx = 0.1 nm
FIND
least uncertainty in momentum (Δp), least uncertainty in speed (Δv)
Step 1 · Find least uncertainty in momentum (Δp)
Use Δp = 1.0546 × 10⁻³⁴ ÷ (2 × Δx × 10⁻⁹)
Δx·Δp
is
at
least
ħ
2,
so
the
least
Δp
is
ħ
2Δx
,
with
Δx
in
meters
(nm
×
10
-9
).
Δp
=
1.0546 ×
10
-34
2 × 0.1 ×
10
-9
Δp
=
1.0546 ×
10
-34
2 ×
10
-10
Δp
≥
5.273 ×
10
-25
kg·
m
/s
Step 2 · Find least uncertainty in speed (Δv)
Use Δv = Δp ÷ m
Slow
next
to
light
p
=
mv,
so
a
spread
in
p
is
a
spread
in
v
times
m
.
Δv
=
5.273 ×
10
-25
9.109 ×
10
-31
Δv
≥
5.7888 ×
10
5
m
/s
Check
Put the numbers back in. Both sides should match.
5.273 ×
10
-25
=
1.0546 ×
10
-34
2 × 0.1 ×
10
-9
✓
5.7888 ×
10
5
=
5.273 ×
10
-25
9.109 ×
10
-31
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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