Nuclear Physics. Help Me Solve This Question.?

A radiologist finds a radioactive with no date of acquisition on it. He places the sample in front of a canter and records an activity of 16000Bq. He then places a similar sample that has just arrived in front of the canter and records the activity of 128000Bq. How long the unlabelled sample has been in the lab, if the 1/2 life of the material is 6 days?

In 6 days, ½ * 128,000Bq = 64,000

In 6 more days, ½ * 64,000 = 32,000

In 6 more days, ½ * 32,000 = 16,000

In 3 half lives, 18 days, the activity decreased from 128,000 to16,000 Bq

16,000 = (½)^x * 128,000

(½)^x = 16,000/128,000

(½)^x = 1/8

x = 3 half lives = 3 * 6 = 18 days

Half Life Equation:

N = N0 * e^(-λ * t)

λ = 0.693/ half life = 0.693/6

16000 = 128000 * e^(-0.693/6 * t)

1/8 = e^(-0.693/6 * t)

take natural log of both sides

ln 1/8 = -0.693/6 * t

t = ln 1/8 ÷ -0.693/6

t = 18 days

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