Chapter 5, Problem 5.13P

(a)

To determine

The linear, mass and atomic attenuation coefficient.

Explanation of Solution

Given:

Atomic mass of Lead, $m=207\text{g/mol}$

Density of lead,

  $\rho =11.36{\text{g/cm}}^3$

Formula Used:

The attenuation of gamma ray can be described as,

  $I=I_0{e}^{-\mu x}$

Where,

  $I$=Intensity after attenuation

  $I_0$= Intensity of incident attenuation

  $x$= Thickness of observer

  $\mu$=Linear attenuation

Calculation:

The attenuation of gamma ray can be described as,

  $\begin{array}{l}I=I_0{e}^{-\mu x}\\ \therefore \mu =\frac{1}{x}\ln \left(\frac{I_0}{I}\right)\end{array}$

As intensity observed initially is 1000. So,

  $I_0=1000$

For $x=2\text{mm= 2}\times {\text{10}}^{-3}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{2\times {10}^{-3}}\ln \left(\frac{1000}{880}\right)\\ \mu =63.91{\text{m}}^{-1}\end{array}$

For $x=4\text{mm= 4}\times {\text{10}}^{-3}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{4\times {10}^{-3}}\ln \left(\frac{1000}{770}\right)\\ \mu =65.34{\text{m}}^{-1}\end{array}$

For $x=6\text{mm= 6}\times {\text{10}}^{-3}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{6\times {10}^{-3}}\ln \left(\frac{1000}{680}\right)\\ \mu =64.27{\text{m}}^{-1}\end{array}$

For $x=8\text{mm= 8}\times {\text{10}}^{-3}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{8\times {10}^{-3}}\ln \left(\frac{1000}{600}\right)\\ \mu =63.85{\text{m}}^{-1}\end{array}$

For $x=10\text{mm= 10}\times {\text{10}}^{-3}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{10\times {10}^{-3}}\ln \left(\frac{1000}{530}\right)\\ \mu =63.48{\text{m}}^{-1}\end{array}$

For $x=15\text{mm= 15}\times {\text{10}}^{-3}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{15\times {10}^{-3}}\ln \left(\frac{1000}{390}\right)\\ \mu =62.77{\text{m}}^{-1}\end{array}$

For $x=20\text{mm= 20}\times {\text{10}}^{-3}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{20\times {10}^{-3}}\ln \left(\frac{1000}{285}\right)\\ \mu =62.76{\text{m}}^{-1}\end{array}$

For $x=25\text{mm= 25}\times {\text{10}}^{-3}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{25\times {10}^{-3}}\ln \left(\frac{1000}{210}\right)\\ \mu =62.42{\text{m}}^{-1}\end{array}$

So, mean is,

  $$

  $\begin{array}{l}\mu =\frac{63.91+65.34+64.27+63.85+63.48+62.77+62.76+62.42}{8}\\ \mu =63.60{\text{m}}^{-1}\\ \mu =0.63{\text{cm}}^{-1}\end{array}$

The mass attenuation is given by,

  $\begin{array}{l}\sigma =\frac{\mu }{\rho }\\ \sigma =\frac{0.636{\text{cm}}^{-1}}{11.34{\text{g/cm}}^3}\\ \sigma =0.056{\text{cm}}^{\text{2}}\text{/g}\end{array}$

The mass of the lead per unit volume is, $m=11.34\text{g}$

So, mass of each atom is, $m_o=3.4\times {10}^{-22}\text{g}$

Thus, number of atoms per unit volume is, $\frac{11.34}{3.4\times {10}^{-22}}=3.33\times {10}^{22}\text{atoms}$

So, atomic attenuation coefficient is,

  $\begin{array}{l}\eta =\frac{\mu }{\text{number of atoms per unit volume}}\\ \eta =\frac{0.636}{3.33\times {10}^{22}}\\ \eta =1.91\times {10}^{-23}{\text{cm}}^{-1}\end{array}$

Conclusion:

The linear, mass and atomic attenuation coefficient are, $\mu =63.60{\text{m}}^{-1}$, $\sigma =0.056{\text{cm}}^{\text{2}}\text{/g}$and

  $\eta =1.91\times {10}^{-23}{\text{cm}}^{-1}$

(b)

To determine

The energy of the gamma rays.

Explanation of Solution

The mass attenuation coefficient is 0.055 cm²/g.

Total mass attenuation coefficient 0.0569 cm²/g is close to calculated in part (a) and this corresponds to a value of $1.25\text{MeV}$

Thus, the energy corresponding to it is, $E=1.25\text{MeV}$

Conclusion:

The energy of gamma rays is $E=1.25\text{MeV}$