Chapter 6.6, Problem 74E

(a)

To determine

To find: The thickness of aluminum shielding that reduces the intensity of $X$ -rays to $30\%$ .

Answer to Problem 74E

The thickness of aluminum shielding is $2.8$ .

Explanation of Solution

Given information: The given equation is $f(x)=I_0{e}^{-\mu x}$ and $I(x)=0.3I_0$

Calculation: Since $I(x)=0.3I_0$ ,

Hence, $x$ will be

  $\begin{array}{l}{I}_0{e}^{-\mu x}=0.3I_0\\ e^{-\mu x}=0.3\\ \ln e^{-\mu x}=\ln 0.3\\ -\mu x=\ln 0.3\\ x=-\frac{\ln 0.3}{\mu }\end{array}$

For aluminum, $\mu =0.43$

  $\begin{array}{l}x=-\frac{\ln 0.3}{0.43}\\ x\approx \frac{-1.2039728043}{0.43}\\ x\approx 2.8\end{array}$

(b)

To determine

To find: The thickness of copper shielding that reduces the intensity of $X$ -rays to $30\%$ .

Answer to Problem 74E

The thickness of copper shielding is $0.376$ .

Explanation of Solution

Given information: The given equation is $f(x)=I_0{e}^{-\mu x}$ and $I(x)=0.3I_0$

Calculation: Since $I(x)=0.3I_0$ ,

Hence, $x$ will be

  $\begin{array}{l}{I}_0{e}^{-\mu x}=0.3I_0\\ e^{-\mu x}=0.3\\ \ln e^{-\mu x}=\ln 0.3\\ -\mu x=\ln 0.3\\ x=-\frac{\ln 0.3}{\mu }\end{array}$

For copper, $\mu =3.2$

  $\begin{array}{l}x=-\frac{\ln 0.3}{3.2}\\ x\approx \frac{-1.2039728043}{3.2}\\ x\approx 0.376\end{array}$

(c)

To determine

To find: The thickness of lead shielding that reduces the intensity of $X$ -rays to $30\%$ .

Answer to Problem 74E

The thickness of lead shielding is $0.028$ .

Explanation of Solution

Given information: The given equation is $f(x)=I_0{e}^{-\mu x}$ and $I(x)=0.3I_0$

Calculation: Since $I(x)=0.3I_0$ ,

Hence, $x$ will be

  $\begin{array}{l}{I}_0{e}^{-\mu x}=0.3I_0\\ e^{-\mu x}=0.3\\ \ln e^{-\mu x}=\ln 0.3\\ -\mu x=\ln 0.3\\ x=-\frac{\ln 0.3}{\mu }\end{array}$

For lead, $\mu =43$

  $\begin{array}{l}x=-\frac{\ln 0.3}{43}\\ x\approx \frac{-1.2039728043}{43}\\ x\approx 0.028\end{array}$

(d)

To determine

To explain: The reason why lead is better material than aluminum or copper.

Explanation of Solution

Given information: Dentist use a lead apron before taking X-rays of your teeth.

Lead is the best material to protect the harmful X-rays because its shielding capacity is the best of the three materials. It reduces the intensity of the X-rays when it is transmitted through it. It requires less thickness than other materials, so doctors use it as a protector.