Chapter 5, Problem 5.35P

(a)

To determine

The relative proportions of the two groups after passing through 25 cm of water.

Answer to Problem 5.35P

The relative proportions of the two groups after passing through 25 cm of water is, $1:1$

Explanation of Solution

Given:

Energy, $E_1=1\text{MeV}$

Energy, $E_2=10\text{MeV}$

Thickness, $x=25\text{cm}$

Cross-section, ${\sigma }_{H,1MeV}=3.4\times {10}^{-5}b=3.4\times {10}^{-29}{\text{cm}}^2$

Cross-section, ${\sigma }_{H,10MeV}=3.2\times {10}^{-5}b=3.2\times {10}^{-29}{\text{cm}}^2$

Cross-section, ${\sigma }_{O,1MeV}=1\times {10}^{-4}b=1\times {10}^{-28}{\text{cm}}^2$

Cross-section, ${\sigma }_{O,10MeV}=1\times {10}^{-4}b=1\times {10}^{-28}{\text{cm}}^2$

Formula used:

The intensity of gamma beam through the absorber curve is,

  $I=I_0{e}^{-N\sigma x}$

Where,

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

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

  $N$ =Number of atoms

  $x$ =Absorber thickness

x =Thickness

Calculation:

The number of Hydrogen atoms per cm³ is,

  $\begin{array}{l}{N}_H=\frac{1{\text{g H}}_2\text{O}}{{\text{cm}}^3}\times \frac{1{\text{mol H}}_2\text{O}}{18\text{g}{\text{H}}_2\text{O}}\times \frac{1\text{mol}H}{1\text{mol}{\text{H}}_2\text{O}}\times \frac{6.023\times {10}^{23}}{\text{mol}\text{H}}\\ N_H=6.692\times {10}^{22}\text{atoms}\text{of}\text{Hydrogen}\text{per}{\text{cm}}^3\end{array}$

The number of Oxygen atoms per cm³ is,

  $\begin{array}{l}{N}_O=\frac{1{\text{g H}}_2\text{O}}{{\text{cm}}^3}\times \frac{1{\text{mol H}}_2\text{O}}{18\text{g}{\text{H}}_2\text{O}}\times \frac{1\text{mol}\text{O}}{1\text{mol}{\text{H}}_2\text{O}}\times \frac{6.023\times {10}^{23}}{\text{mol}\text{H}}\\ N_O=3.346\times {10}^{22}\text{atoms}\text{of}\text{Oxygen}\text{per}{\text{cm}}^3\end{array}$

For Energy, $E_1=1\text{MeV}$

The intensity is given by

  $\begin{array}{l}\frac{I}{I_0}=e^{-{\sigma }_{H,1\text{MeV}}{N}_{H}x-{\sigma }_{O,1\text{MeV}}{N}_{H}x}\\ \frac{I}{I_0}=e^{\left(-3.4\times {10}^{-29}\times 6.692\times {10}^{22}-1.0\times {10}^{-28}\times 3.346\times {10}^{22}\right)\times 25}\\ \frac{I}{I_0}=0.9998\end{array}$

For Energy, $E_1=10\text{MeV}$

The intensity is given by

  $\begin{array}{l}\frac{I}{I_0}=e^{-{\sigma }_{H,1\text{MeV}}{N}_{H}x-{\sigma }_{O,1\text{MeV}}{N}_{H}x}\\ \frac{I}{I_0}=e^{\left(-3.2\times {10}^{-29}\times 6.692\times {10}^{22}-1.0\times {10}^{-28}\times 3.346\times {10}^{22}\right)\times 25}\\ \frac{I}{I_0}=0.9998\end{array}$

The relative proportions of the two groups after passing through 25 cm of water is,

  $\frac{0.9998}{0.9998}=1:1$

Conclusion:

The relative proportions of the two groups after passing through 25 cm of water is, $1:1$

(b)

To determine

The relative proportions of the two groups after passing through a slab of lead of the same density thickness.

Answer to Problem 5.35P

The relative proportions of the two groups after passing through 25 cm of water is $0.7344:1$ .

Explanation of Solution

Given:

Energy, $E_1=1\text{MeV}$

Energy, $E_2=10\text{MeV}$

Density thickness, $t=25{\text{g/cm}}^2$

Cross-section, ${\sigma }_{Pb-204,1MeV}=0.24b=0.24\times {10}^{-24}{\text{cm}}^2$

Cross-section, ${\sigma }_{Pb-204,10MeV}=1.79b=1.79\times {10}^{-24}{\text{cm}}^2$

Cross-section, ${\sigma }_{Pb-206,1MeV}=0.36b=0.36\times {10}^{-24}{\text{cm}}^2$

Cross-section, ${\sigma }_{Pb-206,10MeV}=1.47b=1.47\times {10}^{-24}{\text{cm}}^2$

Cross-section, ${\sigma }_{Pb-207,1MeV}=0.55b=0.55\times {10}^{-24}{\text{cm}}^2$

Cross-section, ${\sigma }_{Pb-207,10MeV}=0.98b=0.98\times {10}^{-24}{\text{cm}}^2$

Cross-section, ${\sigma }_{Pb-208,1MeV}=2.4\times {10}^{-4}b=2.4\times {10}^{-28}{\text{cm}}^2$

Cross-section, ${\sigma }_{Pb-208,10MeV}=1.15b=1.15\times {10}^{-24}{\text{cm}}^2$

Formula used:

The intensity of gamma beam through the absorber curve is,

  $I=I_0{e}^{-N\sigma x}$

Where,

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

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

  $N$ =Number of atoms

  $x$ =Absorber thickness

x =Thickness

The thickness is given by,

  $x=\frac{\text{Density thickness}}{\text{Density}}$

Calculation:

The thickness is given by,

  $\begin{array}{l}x=\frac{\text{Density thickness}}{\text{Density}}\\ x=\frac{25}{11.34}\\ x=2.20\text{cm}\end{array}$

For energy 1 MeV

Number of Pb-204 atoms are,

  $\begin{array}{l}{N}_{Pb-204}=\frac{11.34}{204}\times 6.023\times {10}^{23}\\ N_{Pb-204}=3.34\times {10}^{22}\end{array}$

So,

  $\begin{array}{l}{\sigma }_{204}{N}_{204}=0.24\times {10}^{-24}\times 3.34\times {10}^{22}\\ {\sigma }_{204}{N}_{204}=0.008016\end{array}$

Number of Pb-206 atoms are,

  $\begin{array}{l}{N}_{Pb-206}=\frac{11.34}{206}\times 6.023\times {10}^{23}\\ N_{Pb-206}=3.32\times {10}^{22}\end{array}$

So,

  $\begin{array}{l}{\sigma }_{206}{N}_{206}=0.36\times {10}^{-24}\times 3.32\times {10}^{22}\\ {\sigma }_{206}{N}_{206}=0.01195\end{array}$

Number of Pb-207 atoms are,

  $\begin{array}{l}{N}_{Pb-206}=\frac{11.34}{207}\times 6.023\times {10}^{23}\\ N_{Pb-206}=3.29\times {10}^{22}\end{array}$

So,

  $\begin{array}{l}{\sigma }_{207}{N}_{207}=0.55\times {10}^{-24}\times 3.29\times {10}^{22}\\ {\sigma }_{207}{N}_{207}=0.0180\end{array}$

Number of Pb-208 atoms are,

  $\begin{array}{l}{N}_{Pb-208}=\frac{11.34}{208}\times 6.023\times {10}^{23}\\ N_{Pb-208}=3.28\times {10}^{22}\end{array}$

So,

  $\begin{array}{l}{\sigma }_{208}{N}_{208}=2.4\times {10}^{-28}\times 3.28\times {10}^{22}\\ {\sigma }_{208}{N}_{208}=7.872\times {10}^{-6}\end{array}$

Thus,

  $\begin{array}{l}\sigma N={\left(\sigma N\right)}_{204}+{\left(\sigma N\right)}_{206}+{\left(\sigma N\right)}_{207}+{\left(\sigma N\right)}_{208}\\ \sigma N=0.008016+0.01195+0.0180+7.872\times {10}^{-6}\\ \sigma N=0.03797\end{array}$

The intensity is given by:

  $\begin{array}{l}\frac{I}{I_0}=e^{-\sigma Nx}\\ \frac{I}{I_0}=e^{\left(0.03797\right)\times 2.20}\\ \frac{I}{I_0}=1.087\end{array}$

For energy 10 MeV

Number of Pb-204 atoms are,

  $\begin{array}{l}{N}_{Pb-204}=\frac{11.34}{204}\times 6.023\times {10}^{23}\\ N_{Pb-204}=3.34\times {10}^{22}\end{array}$

So,

  $\begin{array}{l}{\sigma }_{204}{N}_{204}=1.79\times {10}^{-24}\times 3.34\times {10}^{22}\\ {\sigma }_{204}{N}_{204}=0.05978\end{array}$

Number of Pb-206 atoms are,

  $\begin{array}{l}{N}_{Pb-206}=\frac{11.34}{206}\times 6.023\times {10}^{23}\\ N_{Pb-206}=3.32\times {10}^{22}\end{array}$

So,

  $\begin{array}{l}{\sigma }_{206}{N}_{206}=1.47\times {10}^{-24}\times 3.32\times {10}^{22}\\ {\sigma }_{206}{N}_{206}=0.0488\end{array}$

Number of Pb-207 atoms are,

  $\begin{array}{l}{N}_{Pb-206}=\frac{11.34}{207}\times 6.023\times {10}^{23}\\ N_{Pb-206}=3.29\times {10}^{22}\end{array}$

So,

  $\begin{array}{l}{\sigma }_{207}{N}_{207}=0.98\times {10}^{-24}\times 3.29\times {10}^{22}\\ {\sigma }_{207}{N}_{207}=0.0322\end{array}$

Number of Pb-208 atoms are:

  $\begin{array}{l}{N}_{Pb-208}=\frac{11.34}{208}\times 6.023\times {10}^{23}\\ N_{Pb-208}=3.28\times {10}^{22}\end{array}$

So,

  $\begin{array}{l}{\sigma }_{208}{N}_{208}=1.15\times {10}^{-24}\times 3.28\times {10}^{22}\\ {\sigma }_{208}{N}_{208}=0.03772\end{array}$

Thus,

  $\begin{array}{l}\sigma N={\left(\sigma N\right)}_{204}+{\left(\sigma N\right)}_{206}+{\left(\sigma N\right)}_{207}+{\left(\sigma N\right)}_{208}\\ \sigma N=0.05978+0.0488+0.0322+0.03772\\ \sigma N=0.1785\end{array}$

The intensity is given by

  $\begin{array}{l}\frac{I}{I_0}=e^{-\sigma Nx}\\ \frac{I}{I_0}=e^{\left(0.1785\right)\times 2.20}\\ \frac{I}{I_0}=1.480\end{array}$

The relative proportions of the two groups after passing through 25 cm of water is,

  $\frac{1.087}{1.480}=0.7344:1$

Conclusion:

The relative proportions of the two groups after passing through 25 cm of water is:

  $0.7344:1$