SME
General Chemistry II
Nuclear chemistry
Use this for “I-131 has a half-life of 8.02 days. A sample reads 2.00 × 10⁶ Bq. What will it read after 20 days?”
Diagram
After 20 days, 2.49 half-lives have passed: 2 × 10⁶ × (1/2)²·⁴⁹ = 355,000 Bq left.
In the grid 18 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.
k (day⁻¹)
=
ln
2
÷
t½ (days)
ln
A₀ (Bq)
A (Bq)
=
k (day⁻¹)
×
t (days)
n
Half-lives passed
worked out (not typed)
Formulas
k
=
ln(2)
t
1
2
0.08643
=
ln(2)
8.02
ln(
A₀
A
)
=
k
×
t
ln(
2.00 ×
10
6
3.55 ×
10
5
)
=
0.08643
×
20
n
=
t
t
1
2
2.49
=
20
8.02
Assumptions
•
Decay is first order: the activity A = kN is the number of nuclei times the decay constant, so A falls as N does.
•
One becquerel (Bq) is one decay a second; the half-life and the time use the same unit, days here.
•
Each half-life halves the activity, whatever it was at the start.
Step-by-step
WE KNOW
t½ = 8.02 days, A₀ = 2.00 × 10⁶ Bq, t = 20 days
FIND
decay constant (k), activity at time t (A), half-lives passed (n)
Step 1 · Find decay constant (k)
Use k = ln(2) ÷ t½
kt
1
2
=
ln
2,
so
divide
ln
2
by
the
half-life.
k
=
ln(2)
8.02
k
=
0.6931
8.02
k
=
0.08643
day⁻¹
Step 2 · Find activity at time t (A)
Use ln(A₀ ÷ A) = k × t
Undo
the
ln:
A
is
A₀
times
e
raised
to
−kt.
A
=
A₀
×
e
−
k
×
t
A
=
2.00 ×
10
6
×
e
−0.08643 × 20
A
=
(2.00
×
10
6
)
×
e
−1.7286
A
=
(2.00
×
10
6
)
×
0.1775
A
=
3.55 ×
10
5
Bq
Step 3 · Find half-lives passed (n)
Use n = t ÷ t½
Count how many half-lives fit in the time.
n
=
20
8.02
n
=
2.49
Check
Put the numbers back in. Both sides should match.
0.08643
=
ln(2)
8.02
✓
ln(
2.00 ×
10
6
3.55 ×
10
5
)
=
0.08643
×
20
✓
2.49
=
20
8.02
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.