SME
Quantum Mechanics · Problem type
Hydrogen-like systems with a moving nucleus
Use this for “Find the ground-state energy of muonic hydrogen (a muon, 206.8 mₑ, round a proton).”
Chart
A log scale: each grid line up is 10 times the one below.
Ground-state energy E₁ = −2,529 eV
Ground-state radius r₁ = 0.2846 pm
This is an example. Type your own values to start a new problem.
μ
=
×
+
=
μ (mₑ)
E₁
=
−13.606
eV
×
=
E₁ (
)
r₁
Ground-state radius
calculated · pm
Formulas
μ
=
m₁
×
m₂
m₁
+
m₂
185.9
=
206.8
×
1,836.15
206.8 + 1,836.15
E₁
=
−13.606
×
μ
−2,529 = −13.606 × 185.9
r₁
=
52.9
μ
0.2846
=
52.9
185.9
Assumptions
•
Two bodies of charge −1 and +1 circle their center of mass; masses are in electron masses mₑ.
•
Their relative motion is one body of reduced mass μ = m₁m₂ ÷ (m₁ + m₂), so hydrogen’s results hold with μ for mₑ.
•
13.606 eV and a₀ = 52.9 pm are for an electron round an infinitely heavy nucleus. A proton is 1836.15 mₑ, a muon 206.8 mₑ.
Step-by-step
WE KNOW
m₁ = 206.8 mₑ, m₂ = 1,836.15 mₑ
FIND
reduced mass (μ), ground-state energy (E₁), ground-state radius (r₁)
Step 1 · Find reduced mass (μ)
Use μ = m₁ × m₂ ÷ (m₁ + m₂)
The product over the sum: a little less than the lighter mass, and half of it when the two are equal.
μ
=
206.8
×
1,836.15
206.8 + 1,836.15
μ
=
206.8
×
1,836.15
2,042.95
μ
=
379,715.82
2,042.95
μ
=
185.9
mₑ
Step 2 · Find ground-state energy (E₁)
Use E₁ = −13.606 × μ
The
energy
grows
in
step
with
the
mass:
13.606
eV
for
each
electron
mass
of
μ
.
E₁
=
−13.606
×
185.9
E₁
=
−2,529
eV
Step 3 · Find ground-state radius (r₁)
Use r₁ = 52.9 ÷ μ
The
orbit
shrinks
as
the
mass
grows:
a₀
=
52.9
pm
over
μ
in
electron
masses.
r₁
=
52.9
185.9
r₁
=
0.2846
pm
Check
Put the numbers back in. Both sides should match.
185.9
=
206.8
×
1,836.15
206.8 + 1,836.15
✓
−2,529 = −13.606 × 185.9 ✓
0.2846
=
52.9
185.9
✓
Topic check
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