After 1.84 days, the activity of a sample of an unknown type radioactive material has decreased to 85.8% of the initial activity. What is the half-life of this material? (in days)
Let T be the half life, in days.
(1/2)^(1.84/T) = 0.858
Take logs of both sides.
(1.84/T)*(-0.30103) = -0.06651
1.84/T = 0.22095
T = 8.33 days
To find the half-life of the radioactive material, we can use the formula:
A = A0 * (1/2)^(t / t_1/2)
Where:
A0 is the initial activity
A is the activity after a certain time (in this case, after 1.84 days)
t_1/2 is the half-life of the material
t is the time elapsed
In this case, we know that the activity after 1.84 days is 85.8% of the initial activity, which means A/A0 = 0.858. We can substitute these values into the formula:
0.858 = (1/2)^(1.84 / t_1/2)
To solve for t_1/2, we can take the logarithm of both sides:
log(0.858) = log((1/2)^(1.84 / t_1/2))
Using the property of logarithms, we can bring the exponent down:
log(0.858) = (1.84 / t_1/2) * log(1/2)
To solve for t_1/2, divide both sides by log(1/2):
(1.84 / t_1/2) = log(0.858) / log(1/2)
Now, divide 1.84 by log(0.858) / log(1/2) to find t_1/2:
t_1/2 = 1.84 / (log(0.858) / log(1/2))
Using a calculator, we can find that t_1/2 is approximately 0.7757 days. Therefore, the half-life of the radioactive material is 0.7757 days.