Question

A radiation shield is used to attenuate a neutron beam of initial intensity L=1010 [neutrons / (cm².sec)]. If the shield's macroscopic total cross section is , = 0.10 cm-1, what would be the thickness of the shield for attenuating the neutron beam’s intensity by a factor of 1 million (106) times, considering an exponential attenuation of the beam? Using the exponential attenuation law, calculate the e-folding distance (the thickness of a shield that would attenuate a beam of neutrons by a factor equal to the base of the natural logarithm e = 2.718). Use: 2, = 0.1cm-1

Answer 1

According to the exponential attenuation law -

$I = I_{o} * exp[-\Sigma_{t} t ]$

where, I = beam intensity after shielding.

Io = beam intensity before shielding = 10^10 neutrons/cm2.sec

$\Sigma t$ = neutron removal cross section are = 0.1 cm-1.

t = thickness of the shield, cm.

now, attenuating factor = Io / I = 10^6.

or, I/ Io = 10^(-6) = exp [-t*$\Sigma t$] = exp [ - 0.1*t]

or, ln( 10^(-6)) = -0.1*t

or, t = ln( 10^(-6)) / (-0.1) = 138.1551 cm = thickness of the shield.