Question
The half-life of a radioactive isotope is 8 days. A sample initially contains 6.4×1020 atoms. How many atoms remain after 24 days, and what is the activity at that time? Take ln2=0.693.
Solution — Step by Step
Number of half-lives elapsed:
n=T1/2t=824=3
After n half-lives, the number of remaining atoms is:
N=2nN0=236.4×1020=86.4×1020=8×1019atoms
λ=T1/2ln2=8×864000.693s−1
λ=6912000.693≈1.003×10−6s−1
Activity A=λN:
A=1.003×10−6×8×1019≈8.02×1013Bq
Final: N=8×1019 atoms, activity ≈8×1013 Bq.
Why This Works
The 2n shortcut works only when t is an integer multiple of T1/2. For non-integer multiples (say, 10 days), use the general law:
N(t)=N0e−λt
Activity always falls in step with N — both decay with the same exponential time constant.
Alternative Method
Using the general formula for t=24 days =24×86400 s:
N=N0e−λt=6.4×1020×e−3ln2=6.4×1020×81=8×1019
Identical answer — confirms that e−nln2=2−n.
Common Mistake
Students forget to convert days to seconds when computing λ for activity in becquerels (Bq = decays per second). The formula A=λN requires λ in s−1. If you keep λ in day−1, your activity will be in decays per day — wrong unit.