A radioactive substance decays from 20 g to 15 g in 7 hours. Determine the half-life of the
substance.
The half-life of a radioactive substance is the time it takes for half of the substance to decay.
In this case, the substance decays from 20 g to 15 g in 7 hours.
To find the half-life, we can set up the equation:
20g * (1/2)^(t/h) = 15g
Where t is the time in hours and h is the half-life.
Simplifying the equation, we have:
(1/2)^(t/h) = 15/20
(1/2)^(t/h) = 3/4
Take the logarithm of both sides to solve for t/h:
log((1/2)^(t/h)) = log(3/4)
(t/h) * log(1/2) = log(3/4)
Using the logarithmic property log_b(x^y) = y * log_b(x):
(t/h) * log(1/2) = log(3) - log(4)
Since log(1/2) is negative, we can flip the equation to get a positive value for t/h:
(t/h) = (log(4) - log(3)) / log(1/2)
Calculating the right side of the equation:
(t/h) = (log(4/3)) / log(1/2)
(t/h) = -log(3/4) / log(1/2)
Using the logarithmic property log_b(x / y) = log_b(x) - log_b(y):
(t/h) = -(log(3) - log(4)) / log(1/2)
(t/h) = -(log(3) - log(4)) / -log(2)
(t/h) = (log(4) - log(3)) / log(2)
Calculating the right side of the equation:
(t/h) = (log(4/3)) / log(2)
(t/h) = log(4/3) / log(2)
Using the change of base formula log_b(x) = log_c(x) / log_c(b):
(t/h) = ln(4/3) / ln(2)
Using a calculator to evaluate the right side of the equation:
(t/h) ≈ 0.2877
Therefore, the half-life of the substance is approximately 0.2877 hours.