One-half of 𝛾-rays from a 99m technetium sample are absorbed by a 0.170-millimeter-thick lead shielding layer. Half of the 𝛾-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 𝛾-rays?
In this exercise statement, we’re told that we have two lead shielding layers that shield 𝛾 radiation. And the thickness of its layer is 0.170 millimetres; let’s call that thickness 𝑡. We want to solve for the total thickness of lead that will absorb all but one in 1000 𝛾-rays; we can call that thickness 𝑑. To begin our solution, let’s draw a diagram of this situation.
We have a source which emits 𝛾 radiation in the direction of two lead shields, each of thickness 𝑡, where 𝑡 is given to us. If the number of 𝛾-rays emitted is 𝑁, we’re told that the first shield cuts that in half. So that 𝑁 over two rays make it pass. The second shield has the same effect as the first, cutting in the number of 𝛾-rays in half. After passing through both shields, one-quarter of the original number of rays remain.
This is a case of radiation attenuation — 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 𝐼 sub 𝑖 times 𝑒 raised to the negative 𝜇 times 𝑥 power, where 𝜇 is an attenuation factor specific for each element and 𝑥 is the shield thickness.
We can apply this radiation attenuation equation to our scenario. If 𝑁 over four is the final intensity of the 𝛾 radiation from our source, that’s equal to the initial intensity 𝑁 times 𝑒 to the negative 𝜇 times two 𝑡, where 𝑡 is the thickness of each of our two lead shields. We don’t know 𝑁, but that’s okay because it cancels out of this equation.
We’re given the thickness 𝑡, but we don’t know 𝜇 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 𝜇 times two 𝑡. Rearranging this to solve for 𝜇, it equals the natural logarithm of one-quarter divided by negative two times 𝑡. We can now plug in for the value of 𝑡, the thickness of each one of our lead shields. When we do this and calculate this fraction, we find that 𝜇 equals roughly 407.7 inverse millimetres.
Now that we knew 𝜇, we can start to solve for 𝑑. Recall that 𝑑 is the total lead thickness such that only one in 1000 𝛾-rays makes it through the lead shield. We can again use our radiation attenuation relationship. The final amount of radiation 𝑁 divided by 1000 equals the initial amount 𝑁 times 𝑒 to the negative 𝜇 times 𝑑. Once again, 𝑁 cancels out.
And if we take the natural logarithm of both sides, the right side simplifies to negative 𝜇 times 𝑑. Therefore, 𝑑 equals the natural log of one one thousandths divided by negative 𝜇. Plugging in for 𝜇, when we calculate this value, we find that 𝑑 is 1.69 millimetres. This is the thickness of lead shielding that will cut down 99.9 percent of the incoming 𝛾-rays.