Task 4. Compare all results. Task 3. The wavelength of primary photons is equal to 0.05 nm. 40000 photons passed through a region with an area 5cm^2 in a time 3 minutes. Find the Particle Fluence Rate after this radiation passes through thickness 8mm of beryllium. I Find the minimum thickness of matter Platinum in which electrons with energy 3 MeV, 30 MeV, and 300 MeV will lose all their energy. For them 11:

To solve the given problem, we need to calculate the particle fluence rate and the minimum thickness of matter for electrons with different energies.<br /><br />### Task 3: Particle Fluence Rate<br /><br />Given:<br />- Wavelength of primary photons (\(\lambda\)) = 0.05 nm<br />- Number of photons (\(N\)) = 40000<br />- Area (\(A\)) = 5 cm²<br />- Time (\(t\)) = 3 minutes<br />- Thickness of beryllium (\(d\)) = 8 mm<br /><br />First, let's find the energy of each photon using the relationship between wavelength and energy:<br />\[ E = \frac{hc}{\lambda} \]<br />where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)) and \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)).<br /><br />\[ E = \frac{6.626 \times 10^{-34} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{0.05 \times 10^{-9} \, \text{m}} \]<br />\[ E = 3.98 \times 10^{-18} \, \text{J} \]<br /><br />Next, calculate the total energy of the photons:<br />\[ \text{Total Energy} = N \times E \]<br />\[ \text{Total Energy} = 40000 \times 3.98 \times 10^{-18} \, \text{J} \]<br />\[ \text{Total Energy} = 1.592 \times 10^{-15} \, \text{J} \]<br /><br />Now, calculate the fluence rate (\(F\)), which is the total energy divided by the area and time:<br />\[ F = \frac{\text{Total Energy}}{A \times t} \]<br />\[ F = \frac{1.592 \times 10^{-15} \, \text{J}}{5 \times 10^{-4} \, \text{m}^2 \times 180 \, \text{s}} \]<br />\[ F = \frac{1.592 \times 10^{-15}}{9 \times 10^{-4}} \]<br />\[ F = 1.76 \times 10^{-11} \, \text{J/m}^2 \]<br /><br />### Task 4: Minimum Thickness of Matter for Electrons<br /><br />To find the minimum thickness of matter for electrons to lose all their energy, we use the concept of the range of electrons in a medium. The range \(R\) is given by:<br />\[ R = \frac{E}{\mu e} \]<br />where \(E\) is the energy of the electron, \(\mu\) is the mass attenuation coefficient of the material, and \(e\) is the electron's charge (\(1.602 \times 10^{-19} \, \text{C}\)).<br /><br />For simplicity, let's assume the mass attenuation coefficient (\(\mu\)) for beryllium is given or can be found from experimental data. For this example, let's assume \(\mu = 0.1 \, \text{m}^2/\text{kg}\).<br /><br />For each energy level:<br />1. **3 MeV:**<br />\[ R = \frac{3 \times 10^6 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV}}{0.1 \times 9.11 \times 10^{-31} \, \text{kg}} \]<br />\[ R = \frac{4.806 \times 10^{-13}}{0.1 \times 9.11 \times 10^{-31}} \]<br />\[ R = 5.27 \times 10^17 \, \text{m} \]<br /><br />2. **30 MeV:**<br />\[ R = \frac{30 \times 10^6 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV}}{0.1 \times 9.11 \times 10^{-31} \, \text{kg}} \]<br />\[ R = \frac{4.806 \times 10^{-12}}{0.1 \times 9.11 \times 10^{-31}} \]<br />\[ R = 5.27 \times 10^{19} \, \text{m} \]<br /><br />3. **300 MeV:**<br />\[ R = \frac{300 \times 10^6 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV}}{0