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 $He ^{+}$
• B. the first excited state energy of the electron in $He ^{+}$
• C. the first excited state energy of the electron in $Li ^{2+}$
• D. the ground state energy of the electron in $Be ^{3+}$

Answer 1

Answer: (B)

$E_{n}=-R_{ H } \cdot \frac{Z^{2}}{n^{2}}$

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

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

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

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

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

$\therefore E_{2}=-13.6 \times \frac{4}{4}=-13.6\, eV$

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

$E_{1}=-13.6 \times \frac{9}{4}=-30.6\, eV$

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

$Z=4, n=1$

$\therefore E_{1}=-13.6 \times 16=-217.6\, eV$