A smoke detector holds 0.52 mg of americium-241, with a half-life of 432 years…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A smoke detector holds of americium-241, with a half-life of . Find the mass left after . Find also the time, in years, after which only 10 per cent of the americium is left.
Need a reminder? The key ideaShow
The actinoids fill 5f. All are radioactive. In each half-life, half of the sample decays: . In a nuclear equation the mass numbers add up on both sides, and so do the atomic numbers. A beta emission raises the atomic number by one and leaves the mass number unchanged.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Half-lives passed. .
Step 2. Mass left. .
Step 3. The 10 per cent point. . Take logarithms: , so .
Step 4. The time. .
Answer: is left after , and pass before only 10 per cent remains.
Check. After 3 half-lives, one eighth ( per cent) is left. After 4, one sixteenth ( per cent) is left. So 10 per cent comes in between, at a little more than 3 half-lives.
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