A photon with an energy of 17 ,7 eV knocked out an electron from a hydrogen atom in the ground state.Find the speed of an electron outside the atom (Mm/s)

To find the speed of the electron outside the atom, we can use the conservation of momentum and energy principles.Given:- Energy of the photon, E = 17.7 eV- Mass of the electron, m = 9.11 × 10^-31 kgFirst, we need to convert the energy from eV to Joules:E = 17.7 eV × 1.602 × 10^-19 J/eV = 2.83 × 10^-18 JNext, we can use the conservation of momentum to find the speed of the electron. The momentum of the photon is equal to the momentum of the electron:p_photon = p_electronE = p_electron × cp_electron = E / cv_electron = p_electron / mv_electron = (E / c) / mv_electron = (2.83 × 10^-18 J) / (3 × 10^8 m/s) / (9.11 × 10^-31 kg)v_electron ≈ 1.01 × 10^7 m/sTherefore, the speed of the electron outside the atom is approximately 1.01 × 10^7 m/s.