$\frac{{{E_1}}}{4} - {E_1} = {Z^2}\left[ {\frac{{{E_1}}}{{16}} - \frac{{{E_1}}}{4}} \right]\,\,\, \Rightarrow \,\,\,\frac{{ - 3}}{4} = {Z^2}\left[ {\frac{{ - 12}}{{16 \times 4}}} \right]\,$
$\,\,\,\therefore \,\,\,\,\,{Z^2} = 4\,\, \Rightarrow \,\,Z = 2\,\,H{e^ + }$