In the Bohr model of the hydrogen atom, an electron orbits a proton (the nucleus
ID: 1895056 • Letter: I
Question
In the Bohr model of the hydrogen atom, an electron orbits a proton (the nucleus) in a circular orbit of radius 0 53 times 10-10m What is the electric potential at the electron's orbit due to the proton? What is the kinetic energy of the electron? Express your answer in joules? What is the kinetic energy of the electron? Express your answer in eV? What is the total energy of the electron in its orbit? Express your answer in joules What is the ionization energy-that is, the energy required to remove the electron from the atom and take it to r = omega, at rest? Express your answer in joulesExplanation / Answer
a) v = k (q/r) V = 27 volts b) Change in potential energy is the opposite of change in kinetic energy. Potential energy is QV, or 2.2 x 10^-18. b & e) The force on the electron is the Coulomb force between the proton and the electron. It pulls the electron towards the proton. For the electron to move in a circular orbit, the Coulomb force must equal the centripetal force. We need kee2/r2 = mv2/r. This yields mv2 = kee2/r, so the kinetic energy of the electron is KE(r) = (1/2)mv2 = (1/2) kee2/r. The potential energy of the electron in the field of the positive proton point charge is U(r) = -eV(r) = - kee2/r. The total energy is the sum of the electron's kinetic energy and its potential energy. KE(r) + PE(r) = -(1/2)kee2/r = (-1/2 (9*109)(1.60*10-19) /(5.29*10-11) J Total Energy = -2.18*10-18 J. This is usually stated in energy units of electron volts (eV). An eV is 1.60*10-19J, so dividing by this gives an energy of -13.6 eV. To remove the electron from the atom, i.e. to move it very far away and give it zero kinetic energy, 13.6 eV of work must be done by an external force. 13.6 eV is the ionization energy hydrogen. ionization energy hydrogen = 13.6 eV