The intensity of gamma rays passing through a material as a function of thickness,x is given by where n is the number of atoms per cm³ and $\sigma$ is the absorption cross section.It is observed that the intensity of gamma rays of a particular energy is reduced by half when they passed through 1 cm of lead.What is the thickness of lead needed to reduce the intensity to a quarter?

Question

Quizplus · Accepted Answer

We can use the fact that the intensity is reduced by a factor of 2 when passing through 1 cm of lead, which means that:

I(x) = I0/2

where I(x) is the intensity after passing through x cm of lead and I0 is the initial intensity. Substituting the given equation for I(x), we have:

11nF1exp(-F1x) = 11nF1exp(-F1(1)) / 2

Simplifying and solving for x, we get:

x = ln(2) / F1

Using the same equation again for reducing the intensity to a quarter, we have:

I(x) = I0/4

which means that:

11nF1exp(-F1x) = 11nF1exp(-F1(2))

Simplifying and solving for x, we get:

x = 2 ln(2) / F1

Taking the ratio of the two expressions for x, we get:

x2 / x1 = (2 ln(2) / F1) / (ln(2) / F1) = 2

So the thickness of lead needed to reduce the intensity to a quarter is twice the thickness needed to reduce it by half, which is 2.0 cm. Therefore, the answer is C.

The Intensity of Gamma Rays Passing Through a Material as a Function