Industrial radiography · Gamma sources
Half‑life
What happens inside a gamma source as it ages, and how to work out what's left.
An Ir-192 pellet, magnified. Each flash is one atom decaying.
Zoom in
A source is a pellet of unstable atoms
Each atom decays once.
When it does, it releases gamma rays. Watch the dots.
Ir-192, unstable
Decayed, now stable
Gamma ray
Decayed: 0 of 420
Activity
Activity = disintegrations per second
Activity = λ × N. For one isotope, λ is fixed, so activity depends only on N, the number of radioactive atoms.
100 Ci
Pellet ≈ 3.0 mm
0 flashes / s
50 Ci
Pellet ≈ 3.0 mm
0 flashes / s
Ir-192
Decayed: stable Pt-192 / Os-192
Same pellet, same size, same focal spot. Half its Ir-192 atoms have decayed into stable atoms that stay inside.
Ir-192 · each dot = billions of atoms · pellet sizes for natural Ir-192
Units
Curies and becquerels
1 Bq = 1 disintegration / second
1 Ci = 3.7 × 1010 Bq
100 Ci
Equals
3.7 × 1012 Bq
About 3.7 trillion atoms decaying every second.
The experiment
Watch a source decay
A pellet of 512 radioactive atoms. Advance one half-life at a time.
Ir-192 (radioactive): 512
Stable Pt-192 / Os-192: 0
Total in pellet: 512
Elapsed0 days
Half-lives0.00
Ir-192 left512
Activity100 Ci
Half-life 75 daysGamma energy 0.31–0.61 MeV, unchanged
What actually changes
Fewer gamma rays. Same energy.
The decay curve
Half-life: the time for activity to drop by half
Ir-192, starting at 100 Ci.
Activity (Ci) vs. days
Spot the pattern
Divide by 2, again and again
After 1 HLA0 ÷ 2
After 2 HLA0 ÷ (2 × 2)
After 3 HLA0 ÷ (2 × 2 × 2)
A0 = Original Activity
New Activity = Original Activity (A0) ÷ 2nn = No. of days ÷ Half-life
Three steps every time: find n, work out 2n, then divide.
Your turn · 1
150 days later
Ir-192Half-life = 75 daysOriginal Activity (A0) = 100 Ci
Step 1 · Number of half-livesn = No. of days ÷ Half-life of Ir-192n = 150 ÷ 75 = ?
Step 2 · Work out 2n2n = 2n = ?
Step 3 · New activityNew Activity = Original Activity (A0) ÷ 2nNew Activity = 100 ÷ 2n = ? Ci
Your turn · 2
What about 112 days?
Ir-192Half-life = 75 daysOriginal Activity (A0) = 100 Ci
n doesn't have to be a whole number.
Step 1 · Number of half-livesn = No. of days ÷ Half-life of Ir-192n = 112 ÷ 75 = ?
Step 2 · Work out 2n2n = 2n = ?
Step 3 · New activityNew Activity = Original Activity (A0) ÷ 2nNew Activity = 100 ÷ 2n = ? Ci
Your turn · 3
Cobalt-60 works the same way
Co-60Half-life = 5.27 yearsOriginal Activity (A0) = 40 CiNew Activity = 5 Ci
This time, work backwards: how long until it reads 5 Ci?
Step 1 · Work out 2n2n = Original Activity (A0) ÷ New Activity2n = 40 ÷ 5 = ?
Step 2 · Number of half-lives2n = ?, so n = ?
Step 3 · TimeTime = n × Half-life of Co-60Time = n × 5.27 = ? years
Practice set
Five more, same three steps
Problem
Ir-192 vs. Co-60
One year on the shelf
Ir-192 has a short half-life, so shops replace it every few months. Co-60 stays useful for years.
Activity (Ci) vs. days, both starting at 100 Ci
Wrap-up
Try any source
50 Ci
n = 75 ÷ 75 = 1 · A = 100 ÷ 2¹
Your opening guess: none. After 75 days an Ir-192 source reads 50 Ci.
- 1Activity = disintegrations per second, in Ci or Bq.
- 2Each half-life halves the activity. Energy and pellet size stay the same.
- 3New Activity = Original Activity (A0) ÷ 2n, with n = No. of days ÷ Half-life.
Ir-19275 daysCo-605.27 years1 Ci3.7 × 1010 Bq
Ir-192's measured half-life is 73.8 days. Training and exam problems round it to 75.