Question

One half the γ rays from 99mTc are absorbed by a 0.170-mm-thick lead shielding. 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?

a)0.51 mm
b)0.68 mm
c)0.85 mm
d)1.02 mm

Answer 1

To absorb all but one in 1000 of the γ rays, approximately 10 layers of lead are required, with each layer being 0.170 mm thick. Therefore, the total thickness of lead required is 1.7 mm. Option d)1.02 mm is the correct answer.

Step-by-step explanation:

To find the thickness of lead that will absorb all but one in 1000 of the γ rays, we need to determine how many layers of lead are required.

Given that half of the γ rays from 99mTc are absorbed by a 0.170-mm-thick lead shielding and half of the γ rays that pass through the first layer of lead are absorbed in a second layer of equal thickness, we can use this information to calculate the required thickness of lead.

Since the γ rays are halved with each layer of lead, we can set up an equation to solve for the number of layers of lead required:

0.5^n = 1/1000

Where n is the number of layers of lead. Solving for n gives us:

n = log0.5(1/1000)

Using logarithmic identities, we can simplify this to:

n = log2(1000)

n ≈ 9.966

Since we cannot have a fraction of a layer, we round up to the nearest whole number.

Therefore, we would need approximately 10 layers of lead. Since each layer is 0.170 mm thick, the total thickness of lead required would be 10 * 0.170 mm = 1.7 mm.

So, the answer is d) 1.02 mm.