If the amount of a radioactive element decreases from 3.2 g to 0.40 g in 18. days, what is its half-life? Be sure your answer has the correct number of
significant figures.
day(s) in one half-life
The formula to calculate half-life is:
Nf = Ni * (1/2)^(t/T)
Where:
Nf = final amount of radioactive element
Ni = initial amount of radioactive element
t = time elapsed
T = half-life
Plugging in the values given:
0.40 g = 3.2 g * (1/2)^(18/T)
0.40 = 0.13 * (1/2)^(18/T)
Dividing both sides by 0.13:
3.08 = (1/2)^(18/T)
Taking the natural logarithm of both sides:
ln(3.08) = ln((1/2)^(18/T))
ln(3.08) = (18/T) * ln(1/2)
ln(3.08) = -0.693 * (18/T)
ln(3.08) = -12.474/T
T = -12.474 / ln(3.08)
T ≈ 23.5 days
Therefore, the half-life of the radioactive element is approximately 23.5 days. The answer is rounded to two significant figures to match the given values.