Attenuation of Gamma rays by matter Intensity of Parallel gamma rays as a Function of absorber thickness Gamma-ray intensity Intensity, I decreases exponentially as the thickness of the absorber increases I=e Intensity at distance x absorption constant X: thickness Thickness x u(E)enoch(e)+oc(e)+pp(e
Attenuation of Gamma Rays by Matter Intensity of Parallel Gamma Rays as a Function of Absorber Thickness. Thickness x Intensity, I Gamma-ray intensity decreases exponentially as the thickness of the absorber increases. I = Io e –μx I: Intensity at distance x μ: absorption constant x: thickness
10 10 concrete 10 lead 10 10 10 photon energy(Mev) Comparison of the total mass interaction coefficients
Average Trave/ Distance Before An Interaction the interaction probability P(x that a particle interacts somewhere along a path of length x is P(x)=1 °(x)=1°(0)e-H The probability P(a)I that a particle does not interact while traveling a distanceⅩ P(a=l-P(r=ekt p(xdx be the probability that a particle interacts for the first time between x and x + dx p(xdx=prob. particle travels a distance x without interaction] prob. it interacts in the next dxj P(x)HP(dx))=(e-utxHurdx)=Hte-Ht*dx
the interaction probability P(x) that a particle interacts somewhere along a path of length x is The probability th that a particle does not interact while traveling a distance x Average Travel Distance Before An Interaction p(x)dx be the probability that a particle interacts for the first time between x and x + dx
the average distance: the average distance such a particle travels before it interacts xp(x)dx=ut xp(x)dx=ut xe-utxdx mean-free-path length Half-Thickness: the thickness of a medium required for half of the incident radiation to undergo an interaction 110(x1/2) e-ktx1/ /2 20(0) ut
the average distance: the average distance such a particle travels before it interacts. mean-free-path length Half-Thickness: the thickness of a medium required for half of the incident radiation to undergo an interaction
What is the thickness of a water shield and of a lead shield needed to reduce a normally incident beam of 1 Mev photons to one-tenth of the incident intensity? For water ux(1 Mev)=0.07066 cm" and for lead ux(1 Mev=0.7721 cm-1 c1/10=ln10/ for water x1/10= 32.59 cm, for lead X,10=2. 98 cm X1=0.898cm 2 1/100 5.96cm
What is the thickness of a water shield and of a lead shield needed to reduce a normally incident beam of 1 MeV photons to one-tenth of the incident intensity? For water μx (1 MeV) = 0.07066 cm-1 and for lead μx (1 MeV) = 0.7721 cm-1 for water x1/10 = 32.59 cm, for lead x1/10 = 2.98 cm x1/2 = 0.898 cm x1/100 = 5.96 cm