SME
General Chemistry I · Problem type
The de Broglie wavelength of a moving mass
Use this for “Find the de Broglie wavelength of an electron moving at 2.00 × 10⁶ m/s.”
Chart
Period 364 ≈ 364 pm
This is an example. Type your own values to start a new problem.
λ (pm)
=
h
÷
m (kg)
÷
v (m/s)
p
Momentum
calculated · kg·m/s
Formulas
p
=
m
×
v
1.82 ×
10
-24
=
9.109 ×
10
-31
×
2,000,000
λ
=
6.626 ×
10
-34
p
×
10
12
364
=
6.626 ×
10
-34
1.82 ×
10
-24
×
10
12
Assumptions
•
The speed is below a tenth of light’s, so p = mv holds without relativity.
•
Any moving mass has a wavelength λ = h ÷ p, but a ball’s is far too small to see or measure.
•
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, v = 2,000,000 m/s
FIND
momentum (p), wavelength (λ)
Step 1 · Find momentum (p)
Use p = m × v
Momentum is mass times speed.
p
=
9.109 ×
10
-31
×
2,000,000
p
=
1.82 ×
10
-24
kg·
m
/s
Step 2 · Find wavelength (λ)
Use λ = 6.626 × 10⁻³⁴ ÷ p × 10¹²
de
Broglie:
Planck’s
constant
over
the
momentum
is
λ
in
meters;
×
10
12
turns
it
into
picometers.
λ
=
6.626 ×
10
-34
1.82 ×
10
-24
×
10
12
λ
=
6.626 ×
10
-34
1.82 ×
10
-24
×
10
12
λ
=
(3.641
×
10
-10
)
×
10
12
λ
=
364
pm
Check
Put the numbers back in. Both sides should match.
1.82 ×
10
-24
=
9.109 ×
10
-31
×
2,000,000
✓
364
=
6.626 ×
10
-34
1.82 ×
10
-24
×
10
12
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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