SME
University Physics III: Waves, Optics & Modern · Problem type
The de Broglie wavelength of an accelerated particle
Use this for “Electrons in a microscope are accelerated from rest through 2500 V. What is their de Broglie wavelength?”
Chart
Period 122.7 ≈ 122.7 pm
This is an example. Type your own values to start a new problem.
m
Particle mass
example · kg
V
Accelerating voltage
example · V
K
Kinetic energy
calculated · eV
p
Momentum
calculated · kg·m/s
λ
de Broglie wavelength
calculated · pm
Formulas
K
=
V
100 = 100
p
=
√
2 ×
m
×
K
× 1.602 ×
10
-19
5.4023 ×
10
-24
=
√(2
×
(9.109
×
10
-31
)
×
100
×
1.602 ×
10
-19
)
λ
=
6.626 ×
10
-34
p
×
10
12
122.7
=
6.626 ×
10
-34
5.4023 ×
10
-24
×
10
12
Assumptions
•
A particle of charge ±e accelerated from rest through V gains K = eV, so K in eV is V in volts; for a neutron, type K.
•
It stays slow next to light (an electron below about 10 kV), so K = p² ÷ (2m) and p = √(2mK), with K in joules (eV × 1.602 × 10⁻¹⁹).
•
Every moving particle has a wavelength λ = h ÷ p, with h = 6.626 × 10⁻³⁴ J·s.
•
Masses: electron 9.109 × 10⁻³¹ kg, proton 1.673 × 10⁻²⁷ kg, neutron 1.675 × 10⁻²⁷ kg.
Step-by-step
WE KNOW
m = 9.109 × 10⁻³¹ kg, V = 100 V
FIND
kinetic energy (K), momentum (p), de Broglie wavelength (λ)
Step 1 · Find kinetic energy (K)
Use K = V
Each
volt
gives
a
charge
e
one
electronvolt,
so
K
in
eV
is
the
voltage
in
volts.
K
=
100
eV
Step 2 · Find momentum (p)
Use p = √(2 × m × K × 1.602 × 10⁻¹⁹)
Slow
next
to
light,
K
=
p
2
2
m
:
so
p
=
√
2
mK
,
with
K
turned
into
joules.
p
=
√(2
×
(9.109
×
10
-31
)
×
100
×
1.602 ×
10
-19
)
p
=
√((1.8218
×
10
-30
)
×
100
×
(1.602
×
10
-19
))
p
=
√((1.8218
×
10
-28
)
×
(1.602
×
10
-19
))
p
=
√
2.9185 ×
10
-47
p
=
5.4023 ×
10
-24
kg·
m
/s
Step 3 · Find de Broglie 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
5.4023 ×
10
-24
×
10
12
λ
=
6.626 ×
10
-34
5.4023 ×
10
-24
×
10
12
λ
=
(1.2265
×
10
-10
)
×
10
12
λ
=
122.7
pm
Check
Put the numbers back in. Both sides should match.
100 = 100 ✓
5.4023 ×
10
-24
=
√(2
×
(9.109
×
10
-31
)
×
100
×
1.602 ×
10
-19
)
✓
122.7
=
6.626 ×
10
-34
5.4023 ×
10
-24
×
10
12
✓
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The photoelectric effect and the stopping potential
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Hydrogen’s levels and spectral lines (Bohr)
Topic check
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