SME
Historical Geology
Radiometric dating
Use this for “A zircon holds 1 lead-206 atom for every 2 uranium-238 atoms, and U-238’s half-life is 4,468 Myr. How old is it?”
Diagram
After 2,614 Ma, 0.585 half-lives have passed: 100 × (1/2)⁰·⁵⁸⁵ = 66.7% left.
In the grid 67 of 100 Parent atoms are left; each half-life, half of those still there decay.
This is an example. Type your own values to start a new problem.
t (Ma)
=
ln(1
+
D/P
)
÷
λ
t½
Half-life
example · Myr
p
Parent left
calculated · %
Formulas
λ
=
ln(2)
t
1
2
1.5514 ×
10
-4
=
ln(2)
4,468
t
=
ln(1 + D/P)
λ
2,614
=
ln(1 + 0.5)
1.5514 ×
10
-4
p
=
100
1 + D/P
66.7
=
100
1 + 0.5
Assumptions
•
The mineral held no daughter atoms when it formed and lost none since: a closed system.
•
The half-life is fixed (U-238 to Pb-206: 4,468 Myr), so λ = ln 2 ÷ t½ is the share decaying each million years.
•
Each daughter atom was once a parent atom, so N₀ = P + D and t = ln(1 + D ÷ P) ÷ λ.
•
Earth is about 4,600 million years old, so no rock here is older.
Step-by-step
WE KNOW
t½ = 4,468 Myr, D/P = 0.5
FIND
decay constant, per million years (λ), age (t), parent left (p)
Step 1 · Find decay constant, per million years (λ)
Use λ = ln(2) ÷ t½
After
one
half-life
e
−
λt
1
2
=
1
2
,
so
λt
1
2
=
ln
2:
divide
ln
2
by
the
half-life.
λ
=
ln(2)
4,468
λ
=
0.69315
4,468
λ
=
1.5514 ×
10
-4
Step 2 · Find age (t)
Use t = ln(1 + D/P) ÷ λ
The
parent
fell
from
P
+
D
to
P,
so
e
λt
=
1
+
D
P
:
take
ln,
then
divide
by
λ
.
t
=
ln(1 + 0.5)
1.5514 ×
10
-4
ln(1 + 0.5) = 0.4055
t
=
2,614
Ma
Step 3 · Find parent left (p)
Use p = 100 ÷ (1 + D/P)
For
each
parent
atom
left
there
are
D
P
daughters,
so
the
parent
is
1
of
every
1
+
D
P
atoms.
p
=
100
1 + 0.5
p
=
100
1.5
p
=
66.7%
Check
Put the numbers back in. Both sides should match.
1.5514 ×
10
-4
=
ln(2)
4,468
✓
2,614
=
ln(1 + 0.5)
1.5514 ×
10
-4
✓
66.7
=
100
1 + 0.5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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