Law of radioactive decay, half-life and uses of radioactivity
The decay rate is proportional to the atoms present, so N = N₀ exp(−λt). In each half-life, T½ = 0.693/λ, half the remaining atoms decay.
The law of radioactive decay
The rate of decay does not depend on temperature, pressure or chemical state. It depends only on chance and on how many atoms are left.
- 1Law: the number of atoms decaying per second is proportional to the number N present: dN/dt ∝ N.
- 2So dN/dt = −λN. The constant λ is the decay constant; the minus sign shows that N falls.
- 3Rearrange: dN/N = −λ dt. Integrate: ln N = −λt + k.
- 4At t = 0, N = N₀, so k = ln N₀. Then ln(N/N₀) = −λt.
- 5★So, N = N₀ exp(−λt). N becomes zero only after an infinite time, so a sample never disappears completely.
★ key step
Half-life and decay constant
- 1The half-life T½ is the time in which half the radioactive atoms decay: at t = T½, N = N₀/2.
- 2Put this in N = N₀ exp(−λt): ½ = exp(−λT½), so λT½ = ln 2.
- 3ln 2 = 2.303 × log₁₀ 2 = 2.303 × 0.3010 = 0.693.
- 4★So, T½ = 0.693/λ: a larger decay constant means a shorter half-life.
- 5After n half-lives, the fraction left is (½)ⁿ: N₀/2, N₀/4, N₀/8, and so on.
- 6The average (mean) life is τ = 1/λ = 1.44 T½.
★ key step
Worked example: using half-lives
A radioactive sample has a half-life of 5 days. (a) What fraction of it is left after 15 days? (b) How much of 16 g is left after 20 days? (c) Find its decay constant.
- (a) 15 days = 3 half-lives, so the fraction left = (½)³ = 1/8.
- (b) 20 days = 4 half-lives: 16 g → 8 g → 4 g → 2 g → 1 g.
- (c) λ = 0.693/T½ = 0.693 ÷ 5 days = 0.139 per day.
- In seconds: 5 days = 5 × 86400 s = 4.32 × 10⁵ s, so λ = 0.693 ÷ 4.32 × 10⁵ s = 1.6 × 10⁻⁶ s⁻¹.
(a) 1/8; (b) 1 g; (c) λ ≈ 0.139 per day ≈ 1.6 × 10⁻⁶ s⁻¹
Worked example: carbon dating
A piece of old wood shows 5 decays per minute per gram of carbon, while living wood shows 15. The half-life of carbon-14 is 5730 years. Estimate the age of the wood. (log₁₀ 3 = 0.4771)
- From N = N₀ exp(−λt): t = (1/λ) ln(N₀/N).
- λ = 0.693/5730 per year, so 1/λ = 5730 ÷ 0.693 = 8268 years.
- N₀/N = 15/5 = 3, and ln 3 = 2.303 × 0.4771 = 1.099.
- t = 8268 × 1.099 ≈ 9090 years.
About 9090 years old, a little over one and a half half-lives
Uses of radioactivity
| Field | Use | How it works |
|---|---|---|
| Medicine | Treating cancer (radiotherapy) | Gamma rays from a cobalt-60 source destroy cancer cells (your NIOS lesson calls them X-rays, but cobalt-60 gives gamma rays) |
| Medicine | Tracers to find blockages | A little radioactive sodium-24 is injected; a detector follows it and finds where the flow stops |
| Agriculture | Better seeds and longer storage | Controlled gamma rays improve seed varieties and stop stored potatoes and onions from sprouting |
| Geology and archaeology | Finding age (dating) | Carbon-14 dates fossils and old wood; uranium-238 turning into lead-206 dates rocks, giving the Earth an age of about 4 to 4.5 billion years |
| Industry | Finding hidden flaws | Gamma rays pass more easily through a crack or an air bubble inside a metal casting, which then shows up on film |
How to draw the decay curve
- Draw the x-axis for time t, marked T½, 2T½, 3T½ and 4T½.
- Draw the y-axis for number of atoms N, marked N₀, N₀/2, N₀/4 and N₀/8.
- Draw a smooth curve starting at N₀ and falling to N₀/2 at T½, N₀/4 at 2T½ and N₀/8 at 3T½.
- Let it get closer and closer to the time axis without touching it.
- Title it radioactive decay curve: N = N₀ exp(−λt).
Carbon dating tells archaeologists the age of bones and charcoal from ancient sites such as those of the Indus Valley. In Indian hospitals, cobalt-60 gamma therapy and radioactive tracers help find and treat cancer.
3 marks: “State the law of radioactive decay and show that the half-life is T½ = 0.693/λ.” Rate of decay ∝ N: dN/dt = −λN, so N = N₀ exp(−λt). At t = T½, N = N₀/2, so exp(−λT½) = ½ and λT½ = ln 2 = 0.693. Hence T½ = 0.693/λ.
3 marks: “The activity of a radioactive sample falls to 1/8 of its first value in 24 hours. Find its half-life and decay constant.” 1/8 = (½)³, so 3 half-lives = 24 h and T½ = 8 h. λ = 0.693 ÷ 8 h ≈ 0.0866 per hour (about 2.4 × 10⁻⁵ s⁻¹).
2 marks: “What is carbon dating? Why is it useful?” Living things keep a fixed share of carbon-14; after death it decays with T½ = 5730 years. Comparing a sample's activity with that of living matter gives its age. It dates fossils and ancient objects.
Half, half, half again: after n half-lives (½)ⁿ is left, and T½ = 0.693/λ.