SME
University Physics III: Waves, Optics & Modern · Problem type
The decay law: nuclei left and activity
Use this for “A sample holds 4 × 10¹² nuclei of phosphorus-32, whose half-life is 14.3 days. How many are left after 30 days, and what is the activity then?”
Diagram
After 24.06 days, 3 half-lives have passed: 1 × (1/2)³ = 0.125 left.
In the grid 13 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.
N
=
N₀
×
1
2
days
/
days
With λ = ln 2 ÷ T½, N₀e−λt is N₀ × (1/2)t ÷ T½: one halving per half-life. The chart shows the share left, N ÷ N₀, which starts at 1.
λ
Decay constant
calculated · s⁻¹
t/T½
Half-lives passed
worked out (not typed)
N/N₀
Share left
worked out (not typed)
A
Activity
calculated
Formulas
λ
=
ln(2)
T
1
2
× 86,400
1.0003 ×
10
-6
=
ln(2)
8.02 × 86,400
t
/T
1
2
=
t
T
1
2
3
=
24.06
8.02
N
=
N₀
×
e
−
λ
×
t
× 86,400
1.25 ×
10
14
=
1 ×
10
15
×
e
−(1.0003 ×
10
-6
) × 24.06 × 86,400
N
N₀
=
N
N₀
0.125
=
1.25 ×
10
14
1 ×
10
15
A
=
λ
×
N
1.2504 ×
10
8
=
1.0003 ×
10
-6
×
(1.25
×
10
14
)
Assumptions
•
Each nucleus decays at random, with the same chance λ each second however old it is: λ = ln 2 ÷ T½.
•
So the number left falls exponentially, N = N₀e−λt, halving once every half-life.
•
The activity A = λN counts decays each second, in becquerels (Bq); it falls with N.
•
Times are in days and λ is per second (s⁻¹), so A comes out in Bq: each day is 86,400 s.
Step-by-step
WE KNOW
T½ = 8.02 days, N₀ = 1 × 10¹⁵, t = 24.06 days
FIND
decay constant (λ), half-lives passed (t/T½), nuclei left (N), share left (N/N₀), activity (A)
Step 1 · Find decay constant (λ)
Use λ = ln(2) ÷ (T½ × 86,400)
After
one
half-life
e
−λT
1
2
=
1
2
,
so
λT
1
2
=
ln
2:
divide
ln
2
by
the
half-life
in
seconds.
λ
=
ln(2)
8.02 × 86,400
λ
=
ln(2)
692,928
λ
=
0.69315
692,928
λ
=
1.0003 ×
10
-6
s
-1
Step 2 · Find half-lives passed (t/T½)
Use t/T½ = t ÷ T½
How
many
half-lives
fit
in
the
time:
N₀
is
halved
that
many
times.
t
/T
1
2
=
24.06
8.02
t
/T
1
2
=
3
Step 3 · Find nuclei left (N)
Use N = N₀ × e−λ × t × 86,400
Solving
dN/dt
=
−λN
gives
N
=
N₀
e
−
λt
:
the
share
e
−
λt
is
still
undecayed.
N
=
10
15
×
e
−(1.0003 ×
10
-6
) × 24.06 × 86,400
N
=
10
15
×
e
(−2.4067 ×
10
-5
) × 86,400
N
=
10
15
×
e
−2.0794
N
=
10
15
×
0.12501
N
=
1.25 ×
10
14
Step 4 · Find share left (N/N₀)
Use N/N₀ = N ÷ N₀
What
is
left
as
a
share
of
the
start:
(
1
2
)
to
the
power
of
the
half-lives
passed.
N
N₀
=
1.25 ×
10
14
10
15
N
N₀
=
0.125
Step 5 · Find activity (A)
Use A = λ × N
Each
second
a
share
λ
of
the
nuclei
left
decays.
A
=
1.0003 ×
10
-6
×
(1.25
×
10
14
)
A
=
1.2504 ×
10
8
Bq
Check
Put the numbers back in. Both sides should match.
1.0003 ×
10
-6
=
ln(2)
8.02 × 86,400
✓
3
=
24.06
8.02
✓
1.25 ×
10
14
=
10
15
×
e
−(1.0003 ×
10
-6
) × 24.06 × 86,400
✓
0.125
=
1.25 ×
10
14
10
15
✓
1.2504 ×
10
8
=
1.0003 ×
10
-6
×
(1.25
×
10
14
)
✓
Topic check
See what you know: 5 questions on this topic
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