A) \[-41\,\,kJ\,\,mo{{l}^{-1}}\]
B) \[-1312\,\,kJ\,\,mo{{l}^{-1}}\]
C) \[-164\,\,kJ\,\,mo{{l}^{-1}}\]
D) \[-82\,\,kJ\,\,mo{{l}^{-1}}\]
Correct Answer: D
Solution :
The energy of second Bohr orbit of hydrogen atom\[({{E}_{2}})\]is\[-328\,\,kJ\,\,mo{{l}^{-1}}\] \[-328=\frac{E}{{{2}^{2}}}\Rightarrow E=-328\times 4=-1312\,\,kJ\,\,mo{{l}^{-1}}\] \[\therefore \] \[{{E}_{n}}=\frac{-1312}{{{n}^{2}}}kJ\,\,mo{{l}^{-1}}\] If \[n=4,\] \[\therefore \] \[{{E}_{4}}=\frac{1312}{{{4}^{2}}}kJ\,\,mo{{l}^{-1}}=-82\,\,kJ\,\,mo{{l}^{-1}}\]
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