Chapter 9, Problem 9.22P

To determine

Saturation ionization current

Answer to Problem 9.22P

Saturation ionization current is $7.8\times {10}^{-13}A$

Explanation of Solution

Given:

  $\left(370-Bq\right){}^{210}\text{Po}=370\frac{\text{nucleus}\text{disintegrate}}{\text{second}}$

The rate of alpha from source, $N=\left(25\%\right)\left(370\frac{\alpha }{s}\right)=92.5\frac{\alpha }{s}$

Kinetic Energy of the alpha particle emitted from ²¹⁰Po, $\overline{E}=5.3\text{MeV}$

The density of air, ${\rho }_{\text{air}}=1.293\times {10}^{-3}\text{g}/{\text{cm}}^3=1.293\text{mg}/{\text{cm}}^3$

Window density thickness, $t_w=2\frac{\text{mg}}{{\text{cm}}^{\text{2}}}$

Wall is at 1 cm from source, $t_a=1cm$

Density thickness of air, $t_d={\rho }_{air}{t}_a=\frac{1.293m\text{g}}{{\text{cm}}^3}\times 1\text{cm}=1.293\text{mg}/{\text{cm}}^2$

Mean energy loss per ion-pair, $w=35.5\frac{eV}{ion}$

Formula used:

The range (in cm) of air, R_a, at 0°C and 760 mm Hg

  $R_{\text{a}}=0.322E^{3/2}$

Where, E is maximum energy of alpha.

Ionization current

  $I=\frac{\left(N\frac{\alpha }{s}\right)\cdot \left(\overline{E}\frac{eV}{\alpha }\right)\cdot \left(1.6\times {10}^{-19}\frac{C}{\text{ ion }}\right)}{w\frac{\text{ eV }}{\text{ ion }}}\cdot \frac{1\text{A}}{C/s}$

  $\begin{array}{l}N=\text{Number}\text{of}\text{alpha}\text{particle}\text{per}\text{second}\text{from}\text{source}\\ \overline{E}=\text{Remaining}\text{energy}\text{after}\text{passing}\text{through}\text{medium}\text{.}\\ \text{w}\text{=}\text{Mean energy loss per ion-pair}\end{array}$

Calculation:

The range (in cm) of air, R_a, at 0°C and 760 mm Hg

  $R_{\text{a}}=0.322E^{3/2}=0.322{\left(5.3\text{MeV}\right)}^{3/2}=3.93\text{cm}$

The range of alpha in areal density,

  $R=1.293\frac{\text{mg}}{{\text{cm}}^3}\times 3.93\text{cm}=5.08\frac{\text{mg}}{{\text{cm}}^2}$

Alpha must pass through air and window. $$

  $1.293\frac{\text{mg}}{{\text{cm}}^{\text{2}}}+2\frac{\text{mg}}{{\text{cm}}^{\text{2}}}=3.293\frac{\text{mg}}{{\text{cm}}^{\text{2}}}$

Alpha particle remaining kinetic energy,

  $\overline{E}=\frac{5.01\frac{\text{mg}}{{\text{cm}}^2}-3.293\frac{\text{mg}}{{\text{cm}}^2}}{5.01\frac{\text{mg}}{{\text{cm}}^2}}\times 5.3\text{MeV}=1.816\text{MeV}$

Ionization current

  $\begin{array}{l}I=\frac{\left(N\frac{\alpha }{s}\right)\cdot \left(\overline{E}\frac{eV}{\alpha }\right)\cdot \left(1.6\times {10}^{-19}\frac{C}{\text{ ion }}\right)}{w\frac{\text{ eV }}{\text{ ion }}}\cdot \frac{1\text{A}}{C/s}\\ I=\frac{\left(92.5\frac{\alpha }{s}\right)\times \left(1.81\times {10}^6\frac{eV}{\alpha }\right)\times \left(1.6\times {10}^{-19}\frac{\text{C}}{\text{ ion }}\right)}{35.5\frac{\text{eV}}{\text{ion}}}\\ I=7.8\times {10}^{-13}\text{C}/\text{sec}\text{=}7.8\times {10}^{-13}A\end{array}$

Conclusion:

Saturation ionization current is $7.8\times {10}^{-13}A$