The half-life of Radon-222 is 3.83 days. How many days will it take for a sample to decay to 69% of its original mass? Round your answer to two decimal places.
2^⁻t/3.83 = .69
ln2^-t/3.83 =ln(.69)
-t/3.83In2=In(.69)
t=3.83(In.69)/In2
t=-2.05032
t=2.05 days
check my work please ..thanks
you are correct
To solve this problem, we can use the formula for exponential decay:
A(t) = A0 * (1/2)^(t/h)
where A(t) is the mass at time t, A0 is the initial mass, t is the time elapsed, and h is the half-life.
In this case, we are given that the half-life of Radon-222 is 3.83 days. We want to find the time it takes for a sample to decay to 69% of its original mass.
Let's assume the initial mass is A0, and after t days, the mass is 69% of A0. Mathematically, we can write:
A(t) = 0.69 * A0
Substituting this into the exponential decay formula, we get:
0.69 * A0 = A0 * (1/2)^(t/3.83)
Dividing both sides by A0:
0.69 = (1/2)^(t/3.83)
Now, we can take the natural logarithm of both sides to solve for t:
ln(0.69) = ln((1/2)^(t/3.83))
Using the property of logarithms, we can bring down the exponent:
ln(0.69) = (t/3.83) * ln(1/2)
Dividing both sides by ln(1/2):
t/3.83 = ln(0.69) / ln(1/2)
Now, we can solve for t:
t = (3.83 * ln(0.69)) / ln(1/2)
Using a calculator to evaluate this expression, we get:
t ≈ 2.05 days
Therefore, it will take approximately 2.05 days for the sample to decay to 69% of its original mass.