Question

If $E_n$ is the energy of electron in nth orbit for hydrogen like species, then which is the correct relationship ?

• A. $E_{1}ofH=\frac{1}{2}{E}_{2}ofH{e}^{+}=\frac{1}{3}{E}_{3}ofL{i}^{2+}=\frac{1}{4}{E}_{4}ofB{e}^{3+}$
• B. $E_1\left(H\right)=E_2\left(H{e}^{+}\right)=E_3\left(L{i}^{2+}\right)=E_4\left(B{e}^{3+}\right)$
• C. No relation
• D. $E_1\left(H\right)=2E_2\left(H{e}^{+}\right)=3E_3\left(L{e}^{2+}\right)=4E_4\left(B{e}^{3+}\right)$

Answer 1

The correct option is B $E_1\left(H\right)=E_2\left(H{e}^{+}\right)=E_3\left(L{i}^{2+}\right)=E_4\left(B{e}^{3+}\right)$

$E=-13.6\times \frac{Z^2}{n^2}eV{E}_1\left(H\right)=-13.6\times \frac{1^2}{1^2}=-13.6eV;E_2\left(H{e}^{+}\right)=-13.6\times \frac{2^2}{2^2}=-13.6eV{E}_3\left(L{i}^{2+}\right)=-13.6\times \frac{3^2}{3^2}=-13.6eV{E}_4\left(B{e}^{3+}\right)=-13.6\times \frac{4^2}{4^2}=-13.6eV\therefore E_1\left(H\right)=E_2\left(H{e}^{+}\right)=E_3\left(L{i}^{2+}\right)=E_4\left(B{e}^{3+}\right)$