For \(\mathrm{He}^{+}(n=1)\),
\(E_n \)\( =-x=-R_H\left(\frac{2^2}{1^2}\right)=-4 R_H \)
\(\therefore \quad R_H \)\(=\frac{x}{4}\)
For \(B^{3+}(n=2), \)
\(E_n \)\( =-R_H\left(\frac{z^2}{n^2}\right) J \)
\( =-\frac{x}{4} \times\left(\frac{4 \times 4}{2 \times 2}\right)=-\times \mathrm{J}\)