MCQ
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 $Be^{3+}$ ion in $J$ is
- A$-\frac{x}{9}$
- B$-4 x$
- C$-\frac{4}{9} x$
- ✓$-x$
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}$
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$\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - Br}\\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{{C_2}{H_5}\,\,\,\,\,\,\,\,\,}
\end{array}$