The energy of the $n$th Bohr orbit of a hydrogen-like atom is
A$E_n = -13.6\,n^2/Z^2\,eV$
B$E_n = -13.6\,Z^2/n^2\,eV$
C$E_n = -13.6\,Z/n\,eV$
D$E_n = +13.6\,Z/n\,eV$
Answer & Solution
Correct answer: B. $E_n = -13.6\,Z^2/n^2\,eV$
$E_n = -13.6 \times Z^2/n^2\,eV$. Negative because the electron is *bound*; $n=1, Z=1$ gives the ground-state H energy of $-13.6\,eV$.
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