# Video: Absorption of Ionizing Radiation

One-half of πΎ-rays from a βΉβΉα΅Tc sample are absorbed by a 0.170-mm-thick lead shielding layer. Half of the πΎ-rays that pass through the first layer of lead are absorbed in a second layer of equal thickness. What thickness of lead will absorb all but one in 1000 of these πΎ-rays?

04:24

### Video Transcript

One-half of πΎ-rays from a 99m technetium sample are absorbed by a 0.170-millimeter-thick lead shielding layer. Half of the πΎ-rays that pass through the first layer of lead are absorbed in a second layer of equal thickness. What thickness of lead will absorb all but one in 1000 of these πΎ-rays?

In this exercise statement, weβre told that we have two lead shielding layers that shield πΎ radiation. And the thickness of its layer is 0.170 millimetres; letβs call that thickness π‘. We want to solve for the total thickness of lead that will absorb all but one in 1000 πΎ-rays; we can call that thickness π. To begin our solution, letβs draw a diagram of this situation.

We have a source which emits πΎ radiation in the direction of two lead shields, each of thickness π‘, where π‘ is given to us. If the number of πΎ-rays emitted is π, weβre told that the first shield cuts that in half. So that π over two rays make it pass. The second shield has the same effect as the first, cutting in the number of πΎ-rays in half. After passing through both shields, one-quarter of the original number of rays remain.

This is a case of radiation attenuation β radiated energy being absorbed by shielding material. This process is described by a mathematical relationship, which says that the final intensity of the radiated rays equals the initial intensity πΌ sub π times π raised to the negative π times π₯ power, where π is an attenuation factor specific for each element and π₯ is the shield thickness.

We can apply this radiation attenuation equation to our scenario. If π over four is the final intensity of the πΎ radiation from our source, thatβs equal to the initial intensity π times π to the negative π times two π‘, where π‘ is the thickness of each of our two lead shields. We donβt know π, but thatβs okay because it cancels out of this equation.

Weβre given the thickness π‘, but we donβt know π the attenuation factor. To solve for it, we start by taking the natural logarithm of both sides of the equation, which simplifies the right side to negative π times two π‘. Rearranging this to solve for π, it equals the natural logarithm of one-quarter divided by negative two times π‘. We can now plug in for the value of π‘, the thickness of each one of our lead shields. When we do this and calculate this fraction, we find that π equals roughly 407.7 inverse millimetres.

Now that we knew π, we can start to solve for π. Recall that π is the total lead thickness such that only one in 1000 πΎ-rays makes it through the lead shield. We can again use our radiation attenuation relationship. The final amount of radiation π divided by 1000 equals the initial amount π times π to the negative π times π. Once again, π cancels out.

And if we take the natural logarithm of both sides, the right side simplifies to negative π times π. Therefore, π equals the natural log of one one thousandths divided by negative π. Plugging in for π, when we calculate this value, we find that π is 1.69 millimetres. This is the thickness of lead shielding that will cut down 99.9 percent of the incoming πΎ-rays.