Question

When X-rays of a fixed wavelength strike a material \(x\) centimeters thick, the intensity \(I(x)\) of the X-rays transmitted through the material is given by \(I(x)=I_0 e^{-\mu x}\), where \(I_0\) is the initial intensity and \(\mu\) is a value that depends on the type of material and the wavelength of the X-rays. The table shows the values of \(\mu\) for various materials and X-rays of medium wavelength. $$ \begin{array}{|l|c|c|c|} \hline \text { Material } & \text { Aluminum } & \text { Copper } & \text { Lead } \\ \hline \text { Value of } \mu & 0.43 & 3.2 & 43 \\ \hline \end{array} $$a. Find the thickness of aluminum shielding that reduces the intensity of \(\mathrm{X}\)-rays to \(30 \%\) of their initial intensity. (Hint: Find the value of \(x\) for which \(I(x)=0.3 I_0\). b. Repeat part (a) for the copper shielding. c. Repeat part (a) for the lead shielding. d. Your dentist puts a lead apron on you before taking X-rays of your teeth to protect you from harmful radiation. Based on your results from parts (a)-(c), explain why lead is a better material to use than aluminum or copper.

Short answer

a) Thickness for aluminum is approximately \(x_{Al} = -\frac{\ln{0.3}}{0.43}\) cm. b) Thickness for copper is approximately \(x_{Cu} = -\frac{\ln{0.3}}{3.2}\) cm. c) Thickness for lead is approximately \(x_{Pb} = -\frac{\ln{0.3}}{43}\) cm. d) Lead is a better material for radiation shielding as it requires the least thickness to reduce the intensity of X-rays to 30% of their initial intensity.

Step-by-step solution

Derive the thickness equation

Starting with the equation \(I(x) = I_0 e^{-\mu x}\), and setting \(I(x) = 0.3I_0\), which indicates that the intensity of the X-rays is 30% of their initial intensity, we get \(0.3I_0 = I_0 e^{-\mu x}\). Cancelling \(I_0\) on both sides simplifies the equation to \(0.3 = e^{-\mu x}\). Taking natural logarithm on both sides gives us \(\ln{0.3} = -\mu x\), hence \(x = -\frac{\ln{0.3}}{\mu}\). This is the general formula for the thickness of a material that reduces the intensity of X-rays to 30% of their initial intensity.

Calculate the thickness for aluminum

Substitute \(\mu = 0.43\) (for aluminum) into the equation, we get \(x_{Al} = -\frac{\ln{0.3}}{0.43}\) cm.

Calculate the thickness for copper

Substitute \(\mu = 3.2\) (for copper) into the equation, we get \(x_{Cu} = -\frac{\ln{0.3}}{3.2}\) cm.

Calculate the thickness for lead

Substitute \(\mu = 43\) (for lead) into the equation, we get \(x_{Pb} = -\frac{\ln{0.3}}{43}\) cm.

Compare the thicknesses

After calculating the thicknesses in the previous steps, it will be observed that the thickness of lead needed to reduce the intensity of X-rays to 30% is considerably less than that of aluminum or copper, implying that less lead is needed to achieve the same level of protection. This explains why lead is a better material to use for radiation shielding.

Key concepts

Logarithmic Functions

Logarithmic functions play a crucial role in modeling exponential processes like decay, growth, and attenuation. An important aspect of logarithms is their ability to take inputs in the form of exponential results and convert them into a linear form, making complex data more manageable. In our exercise, we see that the intensity of X-rays decreases exponentially as it passes through a material. When the function is expressed as \(I(x)=I_0 e^{-\mu x}\), the goal is often to find a value for \(x\) that satisfies a condition, such as reducing intensity to 30% of its original value. By setting \(0.3 = e^{-\mu x}\) and applying the natural logarithm (ln) to both sides, we obtain \(\ln{0.3} = -\mu x\). This shows that ln can be used to linearize exponential equations, transforming them into forms that are easier to solve for specific variables, such as thickness \(x\). This demonstrates how logarithms serve as powerful tools in solving real-world problems like calculating protective material thickness against X-ray exposure.

X-ray Intensity

X-ray intensity describes the strength or magnitude of the X-ray's beam as it penetrates different materials. When X-rays pass through a substance, they gradually lose intensity due to absorption and scattering. This loss is described by the formula \(I(x) = I_0 e^{-\mu x}\), where \(I_0\) is the initial intensity, and \(\mu\) is the material-specific absorption coefficient.

• It is critical to understand that X-ray intensity diminishes exponentially rather than linearly.
• The degree of reduction depends on the thickness of the material \(x\) and also on the material's absorption properties.

The ability of a material to reduce X-ray intensity effectively is quantified by its absorption coefficient, \(\mu\). Higher values of \(\mu\) indicate a material that more effectively reduces X-ray intensity, therefore requiring less thickness to achieve the same reduction as a material with a lower \(\mu\) value. This is central to applications in medical and industrial settings where precise control over doses of X-ray exposure is required.

Material Absorption Coefficient

The material absorption coefficient, denoted by \(\mu\), is a key figure that determines how effectively a material attenuates or absorbs X-rays. It varies depending on the material and the wavelength of the X-rays. In our given example, the table shows distinct values of \(\mu\) for aluminum, copper, and lead:

• Aluminum: \(\mu = 0.43\)
• Copper: \(\mu = 3.2\)
• Lead: \(\mu = 43\)

You can see that lead has a significantly higher absorption coefficient compared to aluminum and copper, meaning it absorbs X-rays more effectively.
To understand why lead aprons are preferred for protection, consider that higher \(\mu\) values mean that materials can be thinner while still providing substantial attenuation. The formula \(x = -\frac{\ln{0.3}}{\mu}\) illustrates this well. As lead has the largest \(\mu\), the \(x\) required to achieve 30% intensity is the smallest among the three materials. This shows how crucial the material absorption coefficient is in practical applications like designing safety gear and shielding in radiology.