Chapter 5, Problem 5.21P

Calculate the probability that a 2-MeV photon in a narrow, collimated beam will be removed from the beam by each of the following shields:
(a) Lead, 1-cm thick
(b) Iron, 1-cm thick
(c) Lead, ${\text{1-g/cm}}^{\text{2}}$ thick
(d) Iron, ${\text{1-g/cm}}^{\text{2}}$ thick

To determine

The probability of 2 MeV photon in a narrow collimated beam will be removal from the beam by 1 cm thick lead shield.

Answer to Problem 5.21P

The probability of 2 MeV photon in a narrow, collimated beam will be removal from the beam by 1 cm thick lead shield is, $P_r=0.4066$

Explanation of Solution

Given:

Energy, $E=2\text{MeV}$

The thickness of shield, $x=1\text{cm}$

Formula Used:

The intensity of gamma beam through the absorber curve is,

  $I=I_0{e}^{-\mu x}$

Where,

  $I$ = The intensity of gamma beam through the absorber curve

  $I_0$ =The intensity of gamma beam through zero thickness

  $\mu$ =Linear attenuation

  $x$ =Absorber thickness

Calculation:

The linear attenuation of Lead having energy 2 MeV is, $\mu =0.522{\text{cm}}^{-1}$

The probability of transmission is given by,

  $P=\frac{I}{I_0}$

So, the probability of removal is given by,

  $\begin{array}{l}{P}_r=1-\frac{I}{I_0}=1-e^{-\mu x}\\ P_r=1-e^{-0.522\times 1}\\ P_r=1-0.5933\\ P_r=0.4066\end{array}$

Conclusion:

The probability of 2 MeV photon in a narrow, collimated beam will be removal from the beam by 1 cm thick lead shield is, $P_r=0.4066$

To determine

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 cm thick Iron shield.

Answer to Problem 5.21P

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 cm thick iron shield is, $P_r=0.3064$

Explanation of Solution

Given:

Energy, $E=2\text{MeV}$

The thickness of shield, $x=1\text{cm}$

Formula Used:

The intensity of gamma beam through the absorber curve is,

  $I=I_0{e}^{-\mu x}$

Where,

  $I$ = The intensity of gamma beam through the absorber curve

  $I_0$ =The intensity of gamma beam through zero thickness

  $\mu$ =Linear attenuation

  $x$ =Absorber thickness

Calculation:

The linear attenuation of Iron having energy 2 MeV is, $\mu =0.366{\text{cm}}^{-1}$

The probability of transmission is given by,

  $P=\frac{I}{I_0}$

So, the probability of removal is given by,

  $\begin{array}{l}{P}_r=1-\frac{I}{I_0}=1-e^{-\mu x}\\ P_r=1-e^{-0.366\times 1}\\ P_r=1-0.6935\\ P_r=0.3064\end{array}$

Conclusion:

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 cm thick iron shield is, $P_r=0.3064$

To determine

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 g/cm² thick lead shield.

Answer to Problem 5.21P

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 g/cm²thick lead shield is, $P_r=0.0409$

Explanation of Solution

Given:

Energy, $E=2\text{MeV}$

Thickness density of shield, $t_d=1{\text{g/cm}}^2$

Formula Used:

The intensity of gamma beam through the absorber curve is,

  $I=I_0{e}^{-\mu x}$

Where,

  $I$ = The intensity of gamma beam through the absorber curve

  $I_0$ =The intensity of gamma beam through zero thickness

  $\mu$ =Linear attenuation

  $x$ =Absorber thickness

The thickness of the shield is,

  $x=\frac{t_d}{\rho }$

Where,

  $x$ =Thickness

  $t_d$ =Thickness density

  $\rho$ =Density

Calculation:

The thickness of the Lead shield is,

  $\begin{array}{l}x=\frac{t_d}{\rho }\\ x=\frac{1}{11.34}\\ x=0.088\text{cm}\end{array}$

The linear attenuation of Lead having energy 2 MeV is, $\mu =0.522{\text{cm}}^{-1}$

The probability of transmission is given by,

  $P=\frac{I}{I_0}$

So, the probability of removal is given by:

  $\begin{array}{l}{P}_r=1-\frac{I}{I_0}=1-e^{-\mu x}\\ P_r=1-e^{-0.522\times 0.088}\\ P_r=1-0.9590\\ P_r=0.0409\end{array}$

Conclusion:

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 g/cm² thick lead shield is, $P_r=0.0409$

To determine

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 g/cm²thick iron shield.

Answer to Problem 5.21P

The probability of 2 MeV photon in a narrow, collimated beam will be removal from the beam by 1 g/cm² thick Iron shield is, $P_r=0.0454$

Explanation of Solution

Given:

Energy, $E=2\text{MeV}$

Thickness density of shield, $t_d=1{\text{g/cm}}^2$

Formula used:

The intensity of gamma beam through the absorber curve is,

  $I=I_0{e}^{-\mu x}$

Where,

  $I$ = The intensity of gamma beam through the absorber curve

  $I_0$ =The intensity of gamma beam through zero thickness

  $\mu$ =Linear attenuation

  $x$ =Absorber thickness

The thickness of the shield is,

  $x=\frac{t_d}{\rho }$

where,

  $x$ =Thickness

  $t_d$ =Thickness density

  $\rho$ =Density

Calculation:

The thickness of the Iron shield is,

  $\begin{array}{l}x=\frac{t_d}{\rho }\\ x=\frac{1}{7.87}\\ x=0.1270\text{cm}\end{array}$

The linear attenuation of Lead having energy 2 MeV is, $\mu =0.366{\text{cm}}^{-1}$

The probability of transmission is given by:

  $P=\frac{I}{I_0}$

So, the probability of removal is given by,

  $\begin{array}{l}{P}_r=1-\frac{I}{I_0}=1-e^{-\mu x}\\ P_r=1-e^{-0.366\times 0.1270}\\ P_r=1-0.9545\\ P_r=0.0454\end{array}$

Conclusion:

The probability of 2 MeV photon in a narrow, collimated beam will be removal from the beam by 1 g/cm² thick lead shield is, $P_r=0.0454$