If E_H represents the total energy of an electron in the second Bohr orbit of a hydrogen atom and E_ He represents the total energy of an electron in the third Bohr orbit of a singly ionized helium atom ( He^+ ), then the ratio E_H : E_ He is:

Step-by-step solution

The total energy of an electron in the n -th Bohr orbit of a hydrogen-like atom is given by E = -13.6 Z^2 n^2 eV . For the hydrogen atom ( Z = 1 ) in the second Bohr orbit ( n = 2 ): E_H = -13.6 1^2 2^2 = - 13.6 4 eV For the singly ionized helium atom ( He^+ , Z = 2 ) in the third Bohr orbit ( n = 3 ): E_ He = -13.6 2^2 3^2 = -13.6 4 9 eV Taking the ratio of the two energies: E_H E_ He = - 13.6 4 -13.6 4 9 = 1 4 9 4 = 9 16 The ratio E_H : E_ He is 9 : 16 . Answer: 9 : 16