Home › NEET UG › Chemistry › Atomic combinations › The energy of the electron in the $n = 2$ orbit …
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.
Related questions
The strength of a bond depends chiefly on the extent of:Sidewise overlapping produces charged clouds shaped like a:In a pi bond, the axes of the overlapping orbitals remain:Head-on overlap along the internuclear axis is also called:A sigma bond is formed by overlap that is:In a molecule such as HF, the covalent bond is:A bond where the electron pair sits exactly between identical nuclei is:The nitrogen bond enthalpy of 946 kJ per mol is one of the: