Chapter 4.6, Problem 60PS

Solution

a.

To find: the thickness of aluminum shielding.

  $x\approx 2.7997$ centimeter.

Given: The equation is $I(x)=I_0(x)\cdot e^{-\mu \cdot x}$ . Here $I_0(x)$ is the initial intensity which $\mu$ is a number that depends on the type of material and the wavelength of the $X-$ rays.

Concept used:

(1) A exponential model or function is defined as:

  $y=a\cdot {(b)}^{k\cdot x}$ ; where $a>0$ .

(2) By the product property of logarithm is as; ${\log }_a\left(mn\right)={\log }_{a}m+{\log }_{a}n$ , where $a$ is the base of the logarithm.

Calculation:

Now, simplify the above equation as follows: $I(x)=I_0(x)\cdot e^{-\mu \cdot x}$   ...... (1)

rewrite this equation ;

  $I=I_0\cdot e^{-\mu \cdot x}$

Take logarithms on both sides with a natural base.

  $\begin{array}{c}\ln I=\ln \left(I_0\cdot e^{-\mu \cdot x}\right)\\ \ln I=\ln I_0+\ln e^{-\mu \cdot x}\left[\ln ab=\ln a+\ln b\right]\\ \ln I-\ln I_0=\ln e^{-\mu \cdot x}\\ \ln \left(\frac{I}{I_0}\right)=\ln e^{-\mu \cdot x}\left[\ln a-\ln b=\ln \left(\frac{a}{b}\right)\right]\end{array}$

Thus,

  $\begin{array}{c}\ln \left(\frac{I}{I_0}\right)=-\mu \cdot x\ln e\left[\ln e=1\right]\\ \ln \left(\frac{I}{I_0}\right)=-\mu \cdot x\end{array}$

Therefore,

  $x=\frac{-1}{\mu }\cdot \ln \left(\frac{I}{I_0}\right)$  ...... (2)

Also given $I=0.3\cdot I_0$ and $\mu =0.43$

These values are substituted in the above equation (2);

  $\begin{array}{c}x=\frac{-1}{0.43}\cdot \ln \left(\frac{0.3\cdot I_0}{I_0}\right)\\ x=\frac{-1}{0.43}\cdot \ln \left(0.3\right)\\ x\approx \frac{-1}{0.43}\cdot \left(-1.2039\right)\left[\ln \left(0.3\right)\approx -1.2039\right]\end{array}$

Thus,

  $x\approx 2.7997$ centimeters.

Conclusion:

Hence, the thickness of aluminum shielding is $x\approx 2.7997$ centimeters.

b.

To find: the thickness of copper shielding.

  $x\approx 0.3762$ centimeters.

Given: The equation is $I(x)=I_0(x)\cdot e^{-\mu \cdot x}$ . Here $I_0(x)$ is the initial intensity which $\mu$ is a number that depends on the type of material and the wavelength of the $X-$ rays.

Concept used: (1) A exponential model or function is defined as:

  $y=a\cdot {(b)}^{k\cdot x}$ ; where $a>0$ .

(2) By the product property of logarithm is as; ${\log }_a\left(mn\right)={\log }_{a}m+{\log }_{a}n$ , where $a$ is the base of the logarithm.

Calculation:

Now, simplify the above equation as follows: $I(x)=I_0(x)\cdot e^{-\mu \cdot x}$   ...... (3)

rewrite this equation ;

  $I=I_0\cdot e^{-\mu \cdot x}$

Take logarithms on both sides with a natural base.

  $\begin{array}{c}\ln I=\ln \left(I_0\cdot e^{-\mu \cdot x}\right)\\ \ln I=\ln I_0+\ln e^{-\mu \cdot x}\left[\ln ab=\ln a+\ln b\right]\\ \ln I-\ln I_0=\ln e^{-\mu \cdot x}\\ \ln \left(\frac{I}{I_0}\right)=\ln e^{-\mu \cdot x}\left[\ln a-\ln b=\ln \left(\frac{a}{b}\right)\right]\end{array}$

Thus,

  $\begin{array}{c}\ln \left(\frac{I}{I_0}\right)=-\mu \cdot x\ln e\left[\ln e=1\right]\\ \ln \left(\frac{I}{I_0}\right)=-\mu \cdot x\end{array}$

Therefore,

  $x=\frac{-1}{\mu }\cdot \ln \left(\frac{I}{I_0}\right)$  ...... (4)

Also given $I=0.3\cdot I_0$ and $\mu =3.2$

These values are substituted in the above equation (2);

  $\begin{array}{c}x=\frac{-1}{3.2}\cdot \ln \left(\frac{0.3\cdot I_0}{I_0}\right)\\ x=\frac{-1}{3.2}\cdot \ln \left(0.3\right)\\ x\approx \frac{-1}{3.2}\cdot \left(-1.2039\right)\left[\ln \left(0.3\right)\approx -1.2039\right]\end{array}$

Thus,

  $x\approx 0.3762$ centimeters.

Conclusion:

Hence, the thickness of copper shielding is $x\approx 0.3762$ centimeters.

c.

To find: the thickness of lead shielding.

  $x\approx 0.028$ centimeters.

Given:

The equation is $I(x)=I_0(x)\cdot e^{-\mu \cdot x}$ . Here $I_0(x)$ is the initial intensity which $\mu$ is a number that depends on the type of material and the wavelength of the $X-$ rays.

Concept used:

(1) A exponential model or function is defined as:

  $y=a\cdot {(b)}^{k\cdot x}$ ; where $a>0$ .

(2) By the product property of logarithm is as; ${\log }_a\left(mn\right)={\log }_{a}m+{\log }_{a}n$ , where $a$ is the base of the logarithm.

Calculation:

Now, simplify the above equation as follows: $I(x)=I_0(x)\cdot e^{-\mu \cdot x}$   ...... (5)

rewrite this equation ;

  $I=I_0\cdot e^{-\mu \cdot x}$

Take logarithms on both sides with a natural base.

  $\begin{array}{c}\ln I=\ln \left(I_0\cdot e^{-\mu \cdot x}\right)\\ \ln I=\ln I_0+\ln e^{-\mu \cdot x}\left[\ln ab=\ln a+\ln b\right]\\ \ln I-\ln I_0=\ln e^{-\mu \cdot x}\\ \ln \left(\frac{I}{I_0}\right)=\ln e^{-\mu \cdot x}\left[\ln a-\ln b=\ln \left(\frac{a}{b}\right)\right]\end{array}$

Thus,

  $\begin{array}{c}\ln \left(\frac{I}{I_0}\right)=-\mu \cdot x\ln e\left[\ln e=1\right]\\ \ln \left(\frac{I}{I_0}\right)=-\mu \cdot x\end{array}$

Therefore,

  $x=\frac{-1}{\mu }\cdot \ln \left(\frac{I}{I_0}\right)$  ...... (6)

Also given $I=0.3\cdot I_0$ and $\mu =43$

These values are substituted in the above equation (2);

  $\begin{array}{c}x=\frac{-1}{43}\cdot \ln \left(\frac{0.3\cdot I_0}{I_0}\right)\\ x=\frac{-1}{43}\cdot \ln \left(0.3\right)\\ x\approx \frac{-1}{43}\cdot \left(-1.2039\right)\left[\ln \left(0.3\right)\approx -1.2039\right]\end{array}$

Thus,

  $x\approx 0.028$ centimeters.

Conclusion:

Hence, the thickness of lead shielding is $x\approx 0.028$ centimeters.

d.

To find: the result from (a) − (c) and also explain your best material.

  $\approx 2.7717$ centimeters.

Given:

The equation is $I(x)=I_0(x)\cdot e^{-\mu \cdot x}$ . Here $I_0(x)$ is the initial intensity which $\mu$ is a number that depends on the type of material and the wavelength of the $X-$ rays.

Concept used: A exponential model or function is defined as:

  $y=a\cdot {(b)}^{k\cdot x}$ ; where $a>0$ .

Calculation:

According to this question; part(a) − part(c)

Here, in part(a) thickness of aluminum material is $x\approx 2.7997$ centimeters. And in part(c) thickness of the lead material is $x\approx 0.028$ centimeters.

So,

  $2.7997-0.028$

  $\approx 2.7717$ centimeters.

Less lead is required to protect from the same amount of radiation for would need an aluminum apron $100$ times heavier than a lead apron to provide the same protection. A copper one would only be $13.4$ times as heavy.

Conclusion:

Hence, the required result is $\approx 2.7717$ centimeters. And much less lead is needed than the other two for the same amount of protection.