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Suppose an unknown radioactive substance produces 4800 counts per minute on a Geiger counter at a certain time, and only 600 counts per?
Suppose an unknown radioactive substance produces 4800 counts per minute on a Geiger counter at a certain time, and only 600 counts per minute 8 days later. Assuming that the amount of radioactive substance is proportional to the number of counts per minute, determine the half-life of the radioactive substance.
The radioactive substance has a half-life of
days.
1 Answer
- billrussell42Lv 74 years ago
Half life
N(t) = N₀(1/2)^(t/th)
N₀ is initial amount
N(t) is the amount remaining after time t
th is the half life time, ie, time for half the amount to decay
solving for t or th, t/th = –log₂(N/N₀)
N(t) = 4800(1/2)^(8/th) = 600
(1/2)^(8/th) = 1/8
log both sides
(8/th) log (1/2) = log(1/8)
8/th = -0.903 / -0.301 = 3
th = 8/3 = 2.67 days
