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Malignant tumors are commonly treated with targeted ray radiation therapy. To generate these medical rays, a linear accelerator directs a high energy beam of electrons toward a metal target-typically tungsten. As they near the tungsten nuclei, the electrons are deflected and accelerated, emitting high-energy photons via bremsstrahlung. The resulting rays are collimated into a beam that is directed at the tumor. The photons can deposit energy in the tumor through Compton and photoelectric interactions. A typical tumor has ${\mathbf{10}}^{\mathbf{8}}\mathbf{cells}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}\mathbf{,}$and in a full treatment, 4-MeV photons may produce a dose of 70 Gy in 35 fractional exposures on different days. The gray (Gy) is a measure of the absorbed energy dose of radiation per unit mass of tissue:1Gy =1 J/kg.

While interacting with molecules (mainly water) in the tumor tissue, each Compton electron or photoelectron causes a series of ionizations, each of which takes about . Estimate the maximum number of ionizations that one photon generated by this linear accelerator can produce in tissue.$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{100}\mathbf{;}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{1000}\mathbf{;}\mathbf{\left(}\mathbf{c}\mathbf{\right)}{\mathbf{10}}^{\mathbf{4}}\mathbf{;}\mathbf{\left(}\mathbf{d}\mathbf{\right)}{\mathbf{10}}^{\mathbf{5}}$

Step by step solution

01

Define the total number of ionizations.

When the energy of one photon is E and the ionization energy is ${\mathit{E}}_{\mathbf{i}}$, then the total number of ionizations is given by:

$\mathit{N}\mathbf{=}\frac{\mathbf{E}}{{\mathbf{E}}_{\mathbf{i}}}$

02

Determine the total number of ionizations.

Given that, the energyE of one photon 4.00 MeV and ionization of each event is 40.0 eV.

So,$$\begin{gathered} \mathrm{N}=\frac{\mathrm{E}}{{\mathrm{E}}_{\mathrm{i}}} \\ =\frac{4.00×{10}^6}{40.0\mathrm{eV}/\mathrm{event}} \\ ={10}^5\mathrm{event} \end{gathered}$$

Hence, the option (d) is correct answer.

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