SME
Quantum Mechanics · Problem type
The chance the 1s electron is within R
Use this for “What is the chance of finding hydrogen’s ground-state electron within one Bohr radius of the nucleus?”
Chart
Shaded: 0 ≤ x ≤ 1
Chance within R P = 0.3233
This is an example. Type your own values to start a new problem.
x
=
R
÷
a₀
=
x
P
=
1
−
e
−2x
(1
+
2x
+
2x²)
=
P
R
Radius of the sphere
calculated · pm
Pₒᵤₜ
Chance beyond R
worked out (not typed)
Formulas
x
=
R
52.9
1
=
52.9
52.9
Pₒᵤₜ
=
e
−2 ×
x
×
(1
+
2
×
x
+
2
×
x
2
)
0.6767
=
e
−2 × 1
×
(1
+
2
×
1
+
2
×
1
2
)
P
=
1
−
Pₒᵤₜ
0.3233 = 1 − 0.6767
Assumptions
•
Hydrogen in its ground state (1s), ψ = (1 ÷ √(πa₀³))e−r/a₀, with a₀ = 52.9 pm and the nucleus held still.
•
The chance per unit r of finding the electron at r is P(r) = (4 ÷ a₀³)r²e−2r/a₀; its whole area is 1.
•
The chance inside R is that area from 0 to R, integrated by parts twice.
Step-by-step
WE KNOW
x = 1
FIND
radius of the sphere (R), chance beyond R (Pₒᵤₜ), chance within R (P)
Step 1 · Find radius of the sphere (R)
Use x = R ÷ 52.9
Each Bohr radius is 52.9 pm.
R
=
x
×
52.9
R
=
1
×
52.9
R
=
52.9
pm
Step 2 · Find chance beyond R (Pₒᵤₜ)
Use Pₒᵤₜ = e−2 × x × (1 + 2 × x + 2 × x²)
The
area
of
P
(r)
from
R
out
to
far
away:
the
exponential,
times
what
integrating
by
parts
leaves.
Pₒᵤₜ
=
e
−2 × 1
×
(1
+
2
×
1
+
2
×
1
2
)
Pₒᵤₜ
=
e
-2
×
(1
+
2
+
2)
Pₒᵤₜ
=
e
-2
×
5
Pₒᵤₜ = 0.13534 × 5
Pₒᵤₜ = 0.6767
Step 3 · Find chance within R (P)
Use P = 1 − Pₒᵤₜ
The electron is somewhere: inside and beyond add to 1.
P
=
1
−
0.6767
P
=
0.3233
Check
Put the numbers back in. Both sides should match.
1
=
52.9
52.9
✓
0.6767
=
e
−2 × 1
×
(1
+
2
×
1
+
2
×
1
2
)
✓
0.3233 = 1 − 0.6767 ✓
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