• # question_answer The ionisation energy of Hydrogen atom is 13.6 eV. Following Bohr's theory the energy corresponding to a transition between the 3rd and the 4th orbit is A)  3.40eV                                 B)  1.51eV   C)  0.85 eV                                D)  0.66 eV

${{E}_{3}}=-\frac{13.6}{{{3}^{2}}}=-1.51eV$ ${{E}_{4}}=-\frac{13.6}{{{4}^{2}}}=-0.85eV$ $\Delta E={{E}_{4}}-{{E}_{3}}=-0.85-(-1.51)$$=0.66eV$