The energy of the electron in the $n = 2$ orbit of hydrogen by Bohr's model is:
A$-13.6$ eV, the ground state energy of hydrogen always
B$-1.51$ eV, the $n = 3$ value of the hydrogen series
C$-3.4$ eV, since $E_n = -13.6/n^2$ gives $-13.6/4$
DZero, since the second level is above the binding energy
Answer & Solution
Correct answer: C. $-3.4$ eV, since $E_n = -13.6/n^2$ gives $-13.6/4$
$E_2 = -13.6/4 = -3.4$ eV.
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