Question

The nuclear reactions resulting from thermal neutron absorption in boron and cadmium are 10B5 + 1 n0 ï  7Li3 + 4He2 113Cd48 + 1 n0 ï  114Cd48 + γ[5 MeV] The microscopic thermal absorption cross sections for B-10 and Cd-113 are 3841 b and 20,600 b respectively. Which of these two materials would be the more effective radiation shield? Explain

Answer 1

Solution :

The nuclear reaction for boron is given as :

$^{10}\textrm{B}_5 + ^{1}\textrm{n}_0 \rightarrow ^{7}\textrm{Li}_3 + ^{4}\textrm{He}_2$

And the reaction for Cadmium is :

$^{113}\textrm{Cd}_48 + ^{1}\textrm{n}_0 \rightarrow ^{114}\textrm{Cd}_48 + \gamma [5 \ \textrm{MeV}]$

We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.