## 3 Results and Discussion
\(\rho_{[112]}\) is the largest at 10 NN and 7 NN, and is significantly larger than \(\rho_{[110]}\) at 10 NN. Figure 8: A selected reference site Z, its 5 NN, 6 NN, 10 NN, and 7 NN positions, and the planes decided by them in the \(\alpha\)-Fe single crystal. (**a**) The 5 NN and 6 NN positions of Z (\(\rm{s}^{\rm{5 NN-6 NN}}\)) and the maximum density planes of them. (**b**) Two types of [110] and one type of [112] plane decided by \(\rm{s}^{\rm{5 NN-6 NN}}\). (**c**) The 5 NN and 10 NN positions of Z (\(\rm{s}^{\rm{5 NN-10 NN}}\)) and the maximum density planes of them. (**d**) Three types of [110] and one type of [112] plane decided by \(\rm{s}^{\rm{5 NN-10 NN}}\). (**e**) The 7 NN positions of Z (\(\rm{s}^{\rm{7 NN}}\)) and the maximum density planes of them. (**f**) Two types of [110] and three types of [112] planes decided by \(\rm{s}^{\rm{7 NN}}\), in which the type I and II planes of [110] are parallel but with different shapes. As shown in Figure 8b,d,f and listed in Table 4, most of the points in the DP are also situated on the [110] and [112] planes; thus, the slips of the [110] and [112] planes will affect the structures of the DPs. The DPs of s7 NN completely surround \(Z\) to form a closed convex decahedron structure, which makes the effect of hydrogen distributed on the [110] and [112] planes tend to be more isotropic. It can be seen from Figure 7b that, at 7 NN, \(\rho_{[112]}\) is the second largest compared with other NNs. \(\rho_{[110]}>\rho_{[112]}\), which shows that the shears and slips were promoted by the increase of hydrogen atoms situated on these positions. Combined with the phenomenon of the DP surrounding structure, these shears and slips may eventually act on the volume, resulting in the reduction of B and making the material easier to compress. Figure 8b,d,f shows the positions of s5 NN-6 NN, s5 NN-10 NN, and s7 NN, as well as the plane types in the [110] and [112] families where they are distributed. These positions form different polygons on different types of crystal planes, and the areal densities of these positions are also different. The classification of the crystal planes is listed in Table 4. It can be seen from Figures 7b and 8b that compared with s6 NN, s5 NN-6 NN newly determines a group of [112] faces and a group of [110] faces, which makes \(\rho_{[112]}\) increase from 0 sites/A\({}^{2}\) to 0.2814 sites/A\({}^{2}\), and \(\rho_{[110]}\) increases from 0.3625 sites/A\({}^{2}\) to 0.4028 sites/A\({}^{2}\), that is, the increase of hydrogen atoms on 5 NN will significantly increase the amount of hydrogen atoms on the [112] planes, which will bring greater resistance to shears and slips and weaken the softening effect. From Figure 8d, the addition of s5 NN will also increase the \(\rho_{[110]}\) of the type III [110] planes determined by s10 NN, which will enhance the shears and slips. However, although the total \(\rho_{[110]}=0.2637\) sites/A\({}^{2}\) determined by s5 NN-10 NN is higher than the original \(|\rho_{[110]}=0.2372\) sites/A\({}^{2}\) determined by s10 NN, it is still less than the areal density of \(\rho_{[112]}=0.2772\) sites / A\({}^{2}\) of [112] planes, that is, the occupation of s5 NN by hydrogen atoms will not eliminate the possibility of hardening caused by occupation of s10 NN. Therefore, hydrogen can weaken the softening effect of 6 NN by occupying the 5 NN positions, and can achieve hardening by occupying the 10 NN positions.