SME
Quantum Mechanics · Problem type
Reflection at a potential step
Use this for “Electrons of 9 eV meet a potential step 5 eV high. What fraction is reflected, and what fraction goes on?”
Diagram
Past
the
step
the
wave
is
longer:
k₁
k₂
=
√(
E
E − U₀
)
=
2.
R
=
(
k₁ − k₂
k₁ + k₂
)²
=
0.1111;
a
classical
particle
would
never
reflect.
This is an example. Type your own values to start a new problem.
R
=
E (eV)
−
−
U₀ (eV)
E
+
−
2
=
k₁/k₂
Wavenumber ratio
worked out (not typed)
T
Transmission
worked out (not typed)
Formulas
k₁/k₂
=
√(
E
E
−
U₀
)
2
=
√(
4
4 − 3
)
R
=
(
k₁/k₂ − 1
k₁/k₂ + 1
)²
0.1111
=
(
2 − 1
2 + 1
)²
T
=
1
−
R
0.8889 = 1 − 0.1111
Assumptions
•
The step is sudden and lower than the energy, U₀ < E; k goes as √(kinetic energy), so k₁ ∝ √E before it and k₂ ∝ √(E − U₀) after.
•
ψ and its slope match at the step, which gives R = ((k₁ − k₂) ÷ (k₁ + k₂))²; dividing by k₂, only the ratio k₁/k₂ matters.
•
R + T = 1: what is not reflected goes on. A classical particle with E > U₀ would never reflect.
Step-by-step
WE KNOW
E = 4 eV, U₀ = 3 eV
FIND
wavenumber ratio (k₁/k₂), reflection (R), transmission (T)
Step 1 · Find wavenumber ratio (k₁/k₂)
Use k₁/k₂ = √(E ÷ (E − U₀))
k
goes
as
√
kinetic energy
:
E
before
the
step,
E
−
U₀
after
it.
k₁/k₂
=
√(
4
4 − 3
)
k₁/k₂
=
√
4
k₁/k₂ = 2
Step 2 · Find reflection (R)
Use R = ((k₁/k₂ − 1) ÷ (k₁/k₂ + 1))²
Divide
the
top
and
the
bottom
of
k₁ − k₂
k₁ + k₂
by
k₂,
then
square.
R
=
(
2 − 1
2 + 1
)²
R
=
(
1
3
)
2
R
=
0.3333
2
R
=
0.1111
Step 3 · Find transmission (T)
Use T = 1 − R
What is not reflected goes on past the step.
T
=
1
−
0.1111
T
=
0.8889
Check
Put the numbers back in. Both sides should match.
2
=
√(
4
4 − 3
)
✓
0.1111
=
(
2 − 1
2 + 1
)²
✓
0.8889 = 1 − 0.1111 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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