Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate x millimeters of lead shielding is given by I(x) = 100e^−1.5x. What percentage of radiation, to the nearest tenth of a percent, will penetrate a lead shield that is 1 millimeter thick?
? %
(b) How many millimeters of lead shielding are required so that less than 0.02% of the radiation penetrates the shielding? Round to the nearest millimeter.
? mm
L(x) = 100*e^-1.5 = 100*0.22313=22.313 %
To find the percentage of radiation that will penetrate a 1 millimeter thick lead shield, we can substitute x = 1 into the given function I(x) = 100e^(-1.5x).
First, let's plug in x = 1 into the equation:
I(x) = 100e^(-1.5x)
I(1) = 100e^(-1.5(1))
I(1) = 100e^(-1.5)
I(1) ≈ 100(0.22313016014)
I(1) ≈ 22.313016014
So, approximately 22.3% (to the nearest tenth of a percent) of the radiation will penetrate a 1 millimeter thick lead shield.
Now, let's find the number of millimeters of lead shielding required for less than 0.02% of the radiation to penetrate the shielding. We need to solve the equation:
I(x) = 100e^(-1.5x) < 0.02
Since we need less than 0.02%, we can rewrite the equation as:
I(x) < 0.02, where I(x) is the percentage of radiation that penetrates the shielding.
Let's solve this equation:
100e^(-1.5x) < 0.02
Divide both sides by 100:
e^(-1.5x) < 0.0002
To solve for x, take the natural logarithm (ln) of both sides:
ln(e^(-1.5x)) < ln(0.0002)
Using the property of logarithms, ln(e^(-1.5x)) simplifies to -1.5x:
-1.5x < ln(0.0002)
Divide both sides by -1.5:
x > ln(0.0002) / -1.5
Using a calculator, we can find ln(0.0002) ≈ -8.51719319:
x > -8.51719319 / -1.5
x > 5.678128793
Since we are looking for the number of millimeters, we will round up to the nearest whole number. Therefore, less than 0.02% of the radiation will penetrate a lead shield that is 6 millimeters thick.