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
22.3% for the Part A.
I too am struggling with Part B.
To find the percentage of radiation that can penetrate a 1 millimeter thick lead shield, we need to evaluate the function I(x) for x = 1.
I(x) = 100e^(-1.5x)
I(1) = 100e^(-1.5*1) [substituting x = 1]
I(1) = 100e^(-1.5)
To calculate this value, we can use a calculator or mathematical software. Evaluating e^(-1.5) gives us approximately 0.22313.
Therefore, I(1) ≈ 100 * 0.22313 = 22.313%
So, approximately 22.3% of the radiation will penetrate a lead shield that is 1 millimeter thick.
Now let's move on to the next part of the question.
To find the number of millimeters of lead shielding required so that less than 0.02% of the radiation penetrates, we need to solve the equation I(x) ≤ 0.02.
0.02 = 100e^(-1.5x) [substituting I(x) with 0.02]
e^(-1.5x) = 0.02/100 [dividing both sides by 100]
e^(-1.5x) = 0.0002
Now, to solve for x, we need to take the natural logarithm (ln) of both sides:
ln(e^(-1.5x)) = ln(0.0002)
-1.5x = ln(0.0002) [using the property ln(e^a) = a]
To find x, we need to divide both sides by -1.5:
x = ln(0.0002) / -1.5
Again, this calculation can be done using a calculator or mathematical software to get the value of x. Evaluating ln(0.0002) / -1.5 gives us approximately 7.936.
Therefore, around 7.936 millimeters of lead shielding are required so that less than 0.02% of the radiation penetrates the shielding. Rounding to the nearest millimeter, we get 8 millimeters.