conclusion   . 1. According that the polarization of irradiated deuteron polyethylene on the irradiation of 1×1016e-/cm2 is (16.7±1.7)%, we know that the spin density is (8.0±0.4)×1019spins/cc. 2. Five kind of irradiation (3.3-6.7)×1014 e-/cm2;4.3×1015 e-/cm2;5.0×1014 e-/cm2;1.7×1014 e-/cm2;4.8×1013 e-/cm2 were held on the target of polyethylene and deuteron polyethylene. By the first irradiation, the spin density of polyethylene and deuteron polyethylene is (1.02±0.05)×1019spins/cc and (8.12±0.41)×1018spins/cc and the width is 6.34mT and 3.76mT. By the second irradiation, the spin density is (5.06±0.25)×1019spins/cc and (3.18±0.16)×1019spins/cc and the width is 7.62mT and 4.02mT. By the third irradiation, the spin density is (3.02±0.16)×1018spins/cc and (1.98±0.10)×1018spins/cc and the width is 7.62mT and 4.02mT 3.

Abstract   In order to obtain high polarization of polyethylene and deuteron polyethylene in target, by the method of Dynamic Nuclear Polarization, unpaired electrons that can improve the polarization of hydrogen and deuteron by interaction of electrons and nucleon is very necessary to exist in target. As we hope to find the relationship between electrons density and polarization of target. Using electrons beam, 5 kinds of irradiation were held at each target of polyethylene and deuteron polyethylene. Electrons density of target can be found out by ESR test and polarization can be obtained by NMR test. From the result, we found that (16.7±1.7)% polarization of irradiated deuteron polyethylene on the irradiation of 1×1016e-/cm2 is related to the spin density of (8.0±0.4)×1019spins/cc.