The ground state energy of the electron in the hydrogen atom is -13.6 eV. What will be its velocity around the nucleus?

2.2 × 10⁶ m/s 

The electrostatic force of attraction (F_e = mv²/r) between the revolving electrons and the nucleus provides the required centripetal force (F_c = e²/4πε₀) to keep the electrons in their orbits, which gives the relation between v and r as

r = e²/4πε₀mv²                                                                                                                                             ..............(1)

The total energy (E) of the electron is the sum of kinetic energy (mv²/2) and potential energy (-e²/4πε₀r)

E = mv²/2 + (-e²/4πε₀r)

Substituting the value of r from the equation....(1) becomes

E = -mv²/2 

E = -13.6 eV = -13.6 × 1.6 × 10^-19 J

m = 9.1 × 10^-31 kg

v = \(\sqrt {-2E/m}\)

v = 2.2 × 10⁶ m/s