SME
Grade 12 Science · Problem type
Bracketing a fossil layer between ash beds
Use this for “Which ash beds date a fossil layer, and how old can the fossils be?”
Diagram
90% of the uranium-235 is left, drawn as 90 of 100 atoms.
The shale is older than 100 and younger than 107 million years.
This is an example. Type your own values to start a new problem.
P
%
=
100%
×
(0.5)
÷
704
n
Half-lives passed
worked out (not typed)
u
Age of the upper ash
example · million years
w
Width of the age bracket
worked out (not typed) · million years
Formulas
P
=
100
×
0.5
n
90
=
100
×
0.5
0.152
t
=
n
×
704
107 = 0.152 × 704
w
=
t
−
u
7.01 = 107 − 100
Assumptions
•
Ash layers can be dated; sandstone and shale cannot.
•
A layer between two dated ash beds is younger than the one below and older than the one above.
•
The same index fossil marks rock of the same age anywhere.
•
Uranium-235 decays to lead-207 with a half-life of 704 million years.
Step-by-step
WE KNOW
P = 90%, u = 100 million years
FIND
half-lives passed (n), age of the lower ash (t), width of the age bracket (w)
Step 1 · Find half-lives passed (n)
Use P = 100 × 0.5ⁿ
Count
the
halvings:
n
is
the
power
of
2
that
equals
100
P
.
n
=
ln(
100
P
)/ln(2)
n
=
ln(
100
90
)/ln(2)
n
=
0.10536/0.69315
n
=
0.152
Step 2 · Find age of the lower ash (t)
Use t = n × 704
The age is the half-lives passed times the half-life.
t
=
0.152
×
704
t
=
107.01
million
years
Step 3 · Find width of the age bracket (w)
Use w = t − u
The
shale
lies
between
the
ash
beds,
so
it
is
from
u
to
t
million
years
old;
the
bracket
is
their
difference.
w
=
107.01
−
100
w
=
7.01
million
years
Check
Put the numbers back in. Both sides should match.
90
≈
100
×
0.5
0.152
✓
107.01 ≈ 0.152 × 704 ✓
7.01 = 107.01 − 100 ✓
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Any isotope: age from the daughter-to-parent ratio
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