A photon with an energy of 17,7eVknocked 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 of the photon 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 given by p = E/c, where c is the speed of light.<br /><br />p = E/c = (2.83 × 10^-18 J) / (3 × 10^8 m/s) = 9.43 × 10^-21 kg·m/s<br /><br />Since the photon and electron are initially at rest, the total momentum before the collision is zero. After the collision, the total momentum is the momentum of the electron plus the momentum of the hydrogen atom.<br /><br />Let v be the speed of the electron after the collision. The momentum of the electron is given by p_e = m·v.<br /><br />p_e = m·v = (9.11 × 10^-31 kg) × v<br /><br />The momentum of the hydrogen atom is given by p_H = M·v_H, where M is the mass of the hydrogen atom and v_H is the speed of the hydrogen atom.<br /><br />p_H = M·v_H = (1.67 × 10^-27 kg) × v_H<br /><br />Since the total momentum before the collision is zero, the total momentum after the collision is also zero:<br /><br />p_e + p_H = 0<br /><br />(9.11 × 10^-31 kg) × v + (1.67 × 10^-27 kg) × v_H = 0<br /><br />v_H = -(9.11 × 10^-31 kg) × v / (1.67 × 10^-27 kg)<br /><br />Now, we can use the conservation of energy to find the speed of the electron. The energy of the photon is converted into the kinetic energy of the electron and the hydrogen atom:<br /><br />E = (1/2)·m·v^2 + (1/2)·M·v_H^2<br /><br />Substituting the values, we get:<br /><br />2.83 × 10^-18 J = (1/2)·(9.11 × 10^-31 kg)·v^2 + (1/2)·(1.67 × 10^-27 kg)·v_H^2<br /><br />Substituting v_H from the previous equation:<br /><br />2.83 × 10^-18 J = (1/2)·(9.11 × 10^-31 kg)·v^2 + (1/2)·(1.67 × 10^-27 kg)·[(-(9.11 × 10^-31 kg) × v / (1.67 × 10^-27 kg))^2]<br /><br />Solving this equation for v, we get:<br /><br />v ≈ 2.18 × 10^6 m/s<br /><br />Therefore, the speed of the electron outside the atom is approximately 2.18 × 10^6 m/s.