In Bohr's atomic model, the energy of the electron is 'E' in the second orbit of hydrogen atom. So the energy of the electron in the third orbit of helium atom (Z = 2) will be

Step-by-step solution

The energy of an electron in the n -th orbit of a hydrogen-like atom is given by E_n Z^2 n^2 . For the second orbit of a hydrogen atom ( Z=1, n=2 ): E = k 1^2 2^2 = k 4 k = 4E For the third orbit of a helium atom ( Z=2, n=3 ): E' = k 2^2 3^2 = k 4 9 Substituting k = 4E : E' = 4E 4 9 = 16E 9 Answer: 16 ,E 9