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
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To find the percentage of radiation that will penetrate a lead shield that is 1 millimeter thick, we can use the given equation: I(x) = 100e^(-1.5x).
(a) Substituting x = 1 into the equation, we have:
I(1) = 100e^(-1.5 * 1)
= 100e^(-1.5)
≈ 100(0.223)
≈ 22.3
Therefore, approximately 22.3% of the radiation will penetrate a lead shield that is 1 millimeter thick.
(b) To find the number of millimeters of lead shielding required so that less than 0.02% of the radiation penetrates, we need to find the value of x for which I(x) ≤ 0.02.
We can set up the equation as follows:
0.02 = 100e^(-1.5x)
To solve for x, we need to isolate the exponential term on one side. Divide both sides by 100:
0.0002 = e^(-1.5x)
Next, take the natural logarithm (ln) of both sides:
ln(0.0002) = -1.5x
Now, divide by -1.5 to solve for x:
x = ln(0.0002) / -1.5
≈ -8.5174 / -1.5
≈ 5.678
Therefore, approximately 5.678 millimeters of lead shielding are required for less than 0.02% of the radiation to penetrate. Rounded to the nearest millimeter, this would be 6 millimeters.