The half-life for the radioactive decay of calcium-47 is 4.5 d. If a sample has an activity of 5.0 Ci after 27 d , what was the initial activity of the sample?

Answer 1

An easy way to do these types of problems is by using

Fraction remaining = 0.5ⁿ where n = # of half lives elapsed.

27 days x 1 half life4.5 days = 6 half lives = n

fraction remaining = 0.5⁶ = 0.015625

0.015625x = 5 Ci

x = 320 Ci

Answer 2

Let Ca =Ca₀ e^-kt for 1/2 life Ca/Ca₀ =.5 = e^-4.5k solve for k to get k=.154 go back to the original equation to find Ca₀ using the information in the problem 5 = Ca₀e^-.154•27 = Ca_0e^-4.159 solve for Ca₀ to get Ca₀ = 320 Ci