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Chemistry Question on Structure of atom

The energy of second Bohr orbit of the hydrogen atom is −328 kJ mol−1,-328\, kJ\, mol^{-1}, hence the energy of fourth Bohr orbit would be

A

−41 kJ mol−1- 41\, kJ \,mol^{-1}

B

−1312 kJ mol−1-1312\, kJ\, mol^{-1}

C

−164 kJ mol−1-164\, kJ\, mol^{-1}

D

−82 kJ mol−1-82 \,kJ\, mol^{-1}

Answer

−82 kJ mol−1-82 \,kJ\, mol^{-1}

Explanation

Solution

The energy of second Bohr orbit of hydrogen
atom (E2)(E_2) is −328 kJ mol−1- 328\, kJ \,mol^{-1} because
\hspace15mm E_n=- \frac{1312}{n^2}kJ mol^{-1}
∴          E2=−131222kJmol−1\therefore\, \, \, \, \, \, \, \, \, \, E_2=-\frac{1312}{2^2}kJ mol^{-1}
If n=4n = 4
∴          E4=−131242kJmol−1\therefore\, \, \, \, \, \, \, \, \, \, E_4=-\frac{1312}{4^2}kJ mol^{-1}
          =−82 kJ mol−1\, \, \, \, \, \, \, \, \, \, =-82\, kJ\, mol^{-1}