Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Part A You are using a Geiger counter to measure the activity of a radioactive s

ID: 988459 • Letter: P

Question

Part A

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 78.4 minutes , what is the half-life of this substance?

Express your answer with the appropriate units.

Part B

An unknown radioactive substance has a half-life of 3.20 hours . If 26.8 g of the substance is currently present, what mass A0 was present 8.00 hours ago?

Express your answer with the appropriate units.

Part C

Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 45.0 % of an Am-241 sample to decay?

Express your answer with the appropriate units.

Part D

A fossil was analyzed and determined to have a carbon-14 level that is 80 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?

Express your answer with the appropriate units.

Explanation / Answer

Part A

Since it is an radioactive reaction decay, it will be first order reaction

ln(Ao/At) = kt

ln(400/100) = k * 78.4

k = 2ln2/78.4 = ln2/39.2

Half life of the reaction is given by ln2/k = 39.2 minutes

Part B

Half life = ln2/k, so k = ln2/3.2

ln(Ao/At) = kt

ln(Ao/At) = ln2/3.2 * 80 = 2.5ln(2)

Ao/At = 2^(2.5)

Ao = 26.8 gms * 5.6568 = 151.6024 grams

Part C

ln(Ao/At) = kt

ln(100/55) = ln2/432 * t

t = 432 * ln(1.818181)/ln(2) = 372.59 years

Part D

ln(Ao/At) = kt

ln(100/80) = ln2/5730 * t

t = ln(1.25)/ln(2) * 5730 = 1844.64 years