Question is attached Determine the thickness of iron needed to achieve the following shielding goal. Irondensity is 7.8 g/cm³. Fe atomic mass is 55.845u Iron mass attenuation coefficient for 1.5MeV photon is 0.0535 cm²/g.(a) Stop 1.5 MeV alpha particle.(b) Stop 1.5 MeV beta particle.(c) Reduce the 1.5 MeV photon intensity to 1%.

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question is attached

Determine the thickness of iron needed to achieve the following shielding goal. Iron
density is 7.8 g/cm³. Fe atomic mass is 55.845u Iron mass attenuation coefficient for 1.5
MeV photon is 0.0535 cm²/g.
(a) Stop 1.5 MeV alpha particle.
(b) Stop 1.5 MeV beta particle.
(c) Reduce the 1.5 MeV photon intensity to 1%.

Expert Solution

Step 1

we have to find out the range of alpha particle in the medium so the range in the particle is defined as $\begin{array}{rcl} R & = & \frac{(3.2 \times 10^{- 4}) \sqrt{A} \times R_{a}}{ρ} \\ w h e r e A i s 55.845 u, R_{a} i s r a n g e o f α p a r t i c l e, ρ d e n s i t y i s 7.8 g m / c m^{3} \\ R_{a} & = & 0.318 \times E^{\frac{3}{2}} w h e r e E i s e n e r g y o f α p a r t i c l e E i s 1.5 \\ R_{a} & = & 0.318 \times 1 . 5^{\frac{3}{2}} \Rightarrow 0.584 \\ R & = & \frac{(3.2 \times 10^{- 4}) \sqrt{55.845} \times 0.584}{7.8} \\ b) F o r b e t a p a r t i c l e u \sin g f e a t h e r r u l e b e t a s h a v i n g r a n g e 0.6 M e v \\ r a n g e (g / c m^{2}) & = & R \Rightarrow (0.542 \times 1.5) - 0133 \\ = & 0.680 g / c m^{2} \\ R_{c m} & = & \frac{R (g / c m^{2})}{ρ} \\ R_{c m} & = & \frac{0.680}{7.8} \\ = & 0.08718 c m \\ t h i c k n e s s r e q u i r e d \end{array}$

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