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.<br /><br />Given:<br />- Energy of the photon, E = 17.7 eV<br />- Mass of the electron, m = 9.11 × 10^-31 kg<br /><br />First, we need to convert the energy from eV to Joules:<br />E = 17.7 eV × 1.602 × 10^-19 J/eV = 2.83 × 10^-18 J<br /><br />Next, 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:<br /><br />p_photon = p_electron<br /><br />E = p_electron × c<br /><br />p_electron = E / c<br /><br />v_electron = p_electron / m<br /><br />v_electron = (E / c) / m<br /><br />v_electron = (2.83 × 10^-18 J) / (3 × 10^8 m/s) / (9.11 × 10^-31 kg)<br /><br />v_electron ≈ 1.01 × 10^7 m/s<br /><br />Therefore, the speed of the electron outside the atom is approximately 1.01 × 10^7 m/s.