Task 2. Find the thickness of matter Calcium required attenuating photon radiation with an energy of 0.3 MeV, and 3 Me V by a factor of 10.I Compare all results.

To find the thickness of matter required to attenuate photon radiation with a certain energy by a factor of 10, we can use the formula for attenuation:<br /><br />\[ I = I_0 e^{-\mu x} \]<br /><br />where:<br />- \( I \) is the intensity of the photon radiation after passing through a thickness \( x \) of matter,<br />- \( I_0 \) is the initial intensity of the photon radiation,<br />- \( \mu \) is the attenuation coefficient of the matter for the given photon energy,<br />- \( x \) is the thickness of the matter.<br /><br />We want to find the thickness \( x \) such that \( I = \frac{I_0}{10} \). Rearranging the formula, we get:<br /><br />\[ \frac{I_0}{10} = I_0 e^{-\mu x} \]<br /><br />Dividing both sides by \( I_0 \):<br /><br />\[ \frac{1}{10} = e^{-\mu x} \]<br /><br />Taking the natural logarithm of both sides:<br /><br />\[ \ln\left(\frac{1}{10}\right) = -\mu x \]<br /><br />\[ -\ln(10) = -\mu x \]<br /><br />\[ x = \frac{\ln(10)}{\mu} \]<br /><br />Now, we need to find the attenuation coefficient \( \mu \) for calcium and the given photon energies.<br /><br />The attenuation coefficient \( \mu \) is given by:<br /><br />\[ \mu = \sigma N \]<br /><br />where:<br />- \( \sigma \) is the cross-section for attenuation,<br />- \( N \) is the number density of the matter (for simplicity, we assume 1 cm\(^{-3}\) for this calculation).<br /><br />For calcium, the cross-section \( \sigma \) can be found using the atomic number \( Z = 20 \) and the energy \( E \) of the photon. The cross-section can be approximated using the following formula:<br /><br />\[ \sigma \approx \frac{Z^2}{E} \]<br /><br />Now, let's calculate \( \mu \) for the given photon energies:<br /><br />For \( E = 0.3 \) MeV:<br /><br />\[ \sigma \approx \frac{20^2}{0.3} \approx 4444.44 \text{ cm}^{-2}/\text{MeV} \]<br /><br />\[ \mu = 4444.44 \times 1 \approx 4444.44 \text{ cm}^{-1} \]<br /><br />For \( E = 3 \) MeV:<br /><br />\[ \sigma \approx \frac{20^2}{3} \approx 444.44 \text{ cm}^{-2}/\text{MeV} \]<br /><br />\[ \mu = 444.44 \times 1 \approx 444.44 \text{ cm}^{-1} \]<br /><br />Now, we can calculate the thickness \( x \) for each energy:<br /><br />For \( E = 0.3 \) MeV:<br /><br />\[ x = \frac{\ln(10)}{4444.44} \approx 0.0015 \text{ cm} \]<br /><br />For \( E = 3 \) MeV:<br /><br />\[ x = \frac{\ln(10)}{444.44} \approx 0.015 \text{ cm} \]<br /><br />Comparing the results, we can see that the thickness required for attenuation by a factor of 10 is inversely proportional to the attenuation coefficient \( \mu \). As the photon energy increases, the attenuation coefficient decreases, resulting in a larger thickness required for the same level of attenuation.