Question

Assuming Rydberg constants are equal, the ground state energy of the electron in hydrogen atom is equal to

• A. the ground state energy of the electron in $H{e}^{+}$
• B. the first excited state energy of the electron in $H{e}^{+}$
• C. the first excited state energy of the electron in $L{i}^{2+}$
• D. the ground state energy of the electron in $B{e}^{3+}$

Answer 1

Answer: (B)

$E_n=-R_H\cdot \frac{Z^2}{n^2}$

For groùnd state hydrogen atom : $Z=1,n=1$

$\therefore E_1=-13.6\times \frac{1}{1}=-13.6eV$

For the ground state in $H{e}^{+};Z=2,n=1$

$E_1=-13.6\times \frac{4}{1}=-54.4eV$

For the first excited state in $H{e}^{+};Z=2,n=2$;

$\therefore E_2=-13.6\times \frac{4}{4}=-13.6eV$

For the first excited state in $L{i}^{2+};Z=3,n=2$

$E_1=-13.6\times \frac{9}{4}=-30.6eV$

For the ground state energy of the electron in $B{e}^{3+}$

$Z=4,n=1$

$\therefore E_1=-13.6\times 16=-217.6eV$