According to Bohr's theory, the energy of an electron in hydrogen atom starts corresponding to n=1,2,3,4 is -13.6 eV ,-3.4 eV ,-1.51 eV and -0.85 eV respectively. In order to obtain emission of Hβ line of Balmer series, the energy that has to be given to a normal hydrogen atom is

Question

According to Bohr's theory, the energy of an electron in hydrogen atom starts corresponding to n=1,2,3,4 is -13.6 eV ,-3.4 eV ,-1.51 eV and -0.85 eV respectively. In order to obtain emission of Hβ line of Balmer series, the energy that has to be given to a normal hydrogen atom is (A) 12.75 eV (B) 14.45 eV (C) 2.55 eV (D) 4.25 eV

Tardigrade Admin · Accepted Answer

The hydrogen atom in the ground state (-13.6 eV ) must be excited to the energy level with n=4( i.e., -0.85 eV ) so as to have Hβ line.

Q. According to Bohr's theory, the energy of an electron in hydrogen atom starts corresponding to n=1,2,3,4 is −13.6eV,−3.4eV,−1.51eV and −0.85eV respectively. In order to obtain emission of Hβ​ line of Balmer series, the energy that has to be given to a normal hydrogen atom is

12.75 eV

14.45 eV

2.55 eV

4.25 eV

The hydrogen atom in the ground state (−13.6eV) must be excited to the energy level with n=4( i.e., −0.85eV) so as to have Hβ​ line.

The hydrogen atom in the ground state $(- 13.6 e V)$ must be excited to the energy level with $n = 4 ($ i.e., $- 0.85 e V)$ so as to have $H_{β}$ line.