The energy of an electron in the ground state $(n=1)$ for $\mathrm{He}^{+}$ion is $-x J$, then that for an electron in $n=2$ state for $\mathrm{Be}^{3+}$ ion in J is

Question

Aicharya · Accepted Answer

$E_n=-R_H\left(\frac{z^2}{n^2}ight) J$ For $\mathrm{He}^{+}(n=1)$, $\begin{aligned} & E_n=-x=-R_H\left(\frac{2^2}{1^2}ight)=-4 R_H \ & 	herefore \quad R_H=\frac{x}{4} \ & \end{aligned}$ For $\mathrm{Be}^{3+}(n=2)$, $\begin{aligned} E_n & =-R_H\left(\frac{z^2}{n^2}ight) J \ & =-\frac{x}{4} 	imes\left(\frac{4 	imes 4}{2 	imes 2}ight)=-x J \end{aligned}$

The energy of an electron in the ground state $(n=1)$ for $\mathrm{He}^{+}$ion is $-x J$, then that for an…

Solution