Lead shielding is used to contain radiation. The percentage...

Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate $x$ millimeters of lead shielding is given by $I(x)=100 e^{-1.5 x}$ a. 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 $0.05 \%$ of the radiation penetrates the shielding? Round to the nearest millimeter.

Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate $x$ millimeters of lead shielding is given by $I(x)=100 e^{-1.5 x}$
a. 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 $0.05 \%$ of the radiation penetrates the shielding? Round to the nearest millimeter.

Key Concepts

Exponential Decay Function

An exponential decay function describes a process where a quantity decreases at a rate proportional to its current value. In various scientific fields, especially physics, it is used to model phenomena like radioactive decay or radiation attenuation, where the intensity of a substance or signal diminishes rapidly over time or distance.

Logarithms

Logarithms are the inverse of exponential functions and are essential for solving equations where the unknown variable is in the exponent. In the context of exponential decay problems, logarithms allow us to isolate and compute the value of the variable representing time, distance, or thickness needed to reach a specified level of decay.

Radiation Attenuation and Shielding

Radiation attenuation refers to the reduction in the intensity of radiation as it passes through a material. Lead shielding is a common application of this concept, where lead's high density and atomic number effectively absorb or scatter radiation, reducing its impact. The mathematical modeling of this process often employs exponential decay functions to represent how the percentage of radiation decreases with the material's thickness.

Transcript

00:01 Hi there.

00:01 In this problem, we're going to be looking at an application of exponential functions.

00:06 In the upper left -hand corner, you will see the function r of x, below at a table that we'll be using in just a moment, and on the right, a graph of that same function.

00:15 In the lower left -hand corner, we're going to give a description of our input and output and kind of talk about what this problem is about.

00:24 So we're going to look at a problem here that involves radiation, and more specifically, how much radiation is able to penetrate a shield? if you've ever gone in for any kind of x -rays, they might have put a shield over your chest or the body part that they're using, right? they use a lead shield to try to eliminate the amount of radiation that can get through and get into your body.

00:51 So our function is going to be looking at the percent of radiation that can actually get through or penetrate that lead shield, depending on how many millimeters thick that shielding is.

01:04 So i'm going to start by looking at my y intercept, which is also going to show me the initial value here in my function, right? and it's 100, meaning that if you have zero shield, 100 % of that radiation is penetrating, right? we have full penetration there when there is no shielding whatsoever.

01:27 Ever.

01:29 As we start increasing the thickness of the shield, the lead shield, the amount of radiation that can get through decreases, decreases, and actually approaches zero, right? we probably can't eliminate it completely, but we are the thicker that we make the lead shield, the closer we can get to eliminating most of or nearly all of that radiation that gets through.

01:55 So we start with 100 % if you have no shield, and as we increase that thickness, that percent drops.

02:04 All right.

02:04 So there's a situation.

02:05 There's the context.

02:08 What we're going to look at here first is some different depths, i guess.

02:16 If we make the shield, the lead shield, one millimeter thick, right? just one millimeter of lead shielding.

02:24 We can see that we reduce the amount of radiation significantly.

02:28 I can show you that using the table.

02:32 I can click on the coordinate on the graph.

02:35 Algebraically, you could input one for x and evaluate this expression here to find this value.

02:42 But we come up with a little over 22%...