SME
Physical Chemistry II: Quantum & Spectroscopy · Problem type
The chance of finding it in part of the box
Use this for “An electron is in the ground state of a 1.00 nm box. What is the probability of finding it between 0.25 nm and 0.75 nm?”
Diagram
Eₙ
=
n
2
h
2
8mL²
:
the
levels
grow
as
n
2
,
and
ψ
=
0
at
both
walls.
The shaded area of |ψ₁|² is P = 0.8183; a classical particle would give 0.5.
This is an example. Type your own values to start a new problem.
n
Level
example
L
Box length
example
x₁
From
example
x₂
To
example
P
Probability
calculated
Pclassical
Classical probability
worked out (not typed)
Formulas
P
=
x₂
−
x₁
L
−
(sin(2
×
n
×
π
×
x₂
÷
L
)
−
sin(2
×
n
×
π
×
x₁
÷
L
))
÷
(2
×
n
×
π)
0.8183
=
0.75 − 0.25
1
−
(sin(2
×
1
×
π
×
0.75
÷
1)
−
sin(2
×
1
×
π
×
0.25
÷
1))
÷
(2
×
1
×
π)
Pclassical
=
x₂
−
x₁
L
0.5000
=
0.75 − 0.25
1
Assumptions
•
ψₙ = √(2/L) sin(nπx/L) is normalized, so the chance between x₁ and x₂ is the area under |ψ|² there.
•
The sines are in radians; the walls are at x = 0 and x = L.
•
A classical particle bouncing at one speed is equally likely anywhere: its chance is just (x₂ − x₁) ÷ L.
Step-by-step
WE KNOW
n = 1, L = 1 nm, x₁ = 0.25 nm, x₂ = 0.75 nm
FIND
probability (P), classical probability (Pclassical)
Step 1 · Find probability (P)
Use P = (x₂ − x₁) ÷ L − (sin(2 × n × π × x₂ ÷ L) − sin(2 × n × π × x₁ ÷ L)) ÷ (2 × n × π)
Integrate
(
2
L
)sin²(nπx/
L
)
from
x₁
to
x₂
:
the
stretch’s
share
of
L
,
less
the
sine
terms.
P
=
0.75 − 0.25
1
−
(sin(2
×
1
×
π
×
0.75
÷
1)
−
sin(2
×
1
×
π
×
0.25
÷
1))
÷
(2
×
1
×
π)
P
=
0.75
−
0.25
−
(sin(6.283
×
0.75)
−
sin(6.283
×
0.25))
÷
(2
×
π)
P
=
0.75
−
0.25
−
sin(4.712) − sin(1.571)
2 × π
P
=
0.75
−
0.25
−
(−1) − 1
6.283
P
=
0.75
−
0.25
−
−2
6.283
P
=
0.75
−
0.25
−
(−0.3183)
P
=
0.8183
Step 2 · Find classical probability (Pclassical)
Use Pclassical = (x₂ − x₁) ÷ L
A classical particle spends equal time everywhere, so its chance is the stretch’s share of the box.
Pclassical
=
0.75 − 0.25
1
Pclassical = 0.5000
Check
Put the numbers back in. Both sides should match.
0.8183
=
0.75 − 0.25
1
−
(sin(2
×
1
×
π
×
0.75
÷
1)
−
sin(2
×
1
×
π
×
0.25
÷
1))
÷
(2
×
1
×
π)
✓
0.5000
=
0.75 − 0.25
1
✓
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Particle in a box and harmonic oscillator
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A bond’s vibration and its zero-point energy
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