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

Given:

$I(x) = I_0e^{-n\sigma x}$

$\frac{I(x)}{I_0} = \frac{1}{2}$ when $x=1$cm

We need to find $x$ such that $\frac{I(x)}{I_0} = \frac{1}{4}$.

Using the equation,

$\frac{1}{2} = e^{-n\sigma (1)}$

$\implies n\sigma = \frac{\ln 2}{-1}$

$\implies I(x) = I_0 e^{-\frac{\ln 2}{-1}x}$

$\implies \frac{I(x)}{I_0} = \frac{1}{4}$

When $x = ?$

$\frac{1}{4} = e^{-\frac{\ln 2}{-1}x}$

$\implies x = \frac{\ln 4}{\ln 2} \approx 2.0 	ext{ cm}$

Therefore, the thickness of lead needed to reduce the intensity to a quarter is 2.0 cm. The correct answer is (C).

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