Technetium-99m is a radioactive element and is used to diagnose brain, thyroid, liver, and kidney diseases. This element has a half-life of 6 hours. If there are 200 mg of this technetium present, how much will be left in 36 hours? Ans.(3.12 mg)



Data Given:


Half-life of Technetium-99m = T₁/₂ = 6 hours 

Time =36 hours

Original quantity = Nâ‚’ = 200 mg


To Find:


The remaining amount of original radioactive isotope = N = ?


Solution:

From the given data it is clear that during 36 hours, Six Half-lives (6 T₁/₂ ) elapsed. hence using the formula

Remaining fraction = Original Fraction ï½˜ `\frac{1}{2^t}`

N = Nâ‚’ ï½˜ `\frac{1}{2^t}`

by putting values 

N = 200 mg ï½˜ `\frac{1}{2^6}`

N = `\frac{200 mg}{64}` 

N = 3.125 mg ---- Ans