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

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.