Các ion ..................
Giải thích
\(d = 2r = \frac{{2m{v_0}}}{{|q|B}} \Rightarrow \frac{{{d_1}}}{{{d_2}}} = \frac{{{m_X}}}{{{m_{Au}}}} \Rightarrow \frac{{19,81}}{{40}} = \frac{{{m_X}}}{{3,27 \cdot {{10}^{ - 25}}}} \Rightarrow {m_X} \approx 1,62 \cdot {10^{ - 25}}\;{\rm{kg}}.{\rm{ }}\)Chọn B