Question

Classical 2p. Somtmes it is llustrative to consider some aspects of a quantum system the radius of maximum probability Tmp - 4ao. Assuming the electron's radius is (tem- porarily, for the sake of this part of the calculation) fixed at exactly rmp-4a0 (i.e., a classical orbit), what is the kinetic energy due to rotation of the electron about the nu- cleus? (Hint: the "centrifugal potential" is actually the kinetic energy due to rotation.) What is the potential energy at this distance? Using these two pieces of information, along with the energy of the 2p state, what is the radial portion of the kinetic energy?

Answer 1

potential energy for hydrogen atom = -e^2/4 pi elementof_0 r for r_m p = 4 a_0 v (r), P.E = -e^2/4 pi elementof_0 times 4 times a_0 centrifugal potential = L vector^2/2 u r^2 = l (l + 1) h^2/2 u r^2 L Psi = l (l + 1) h^2 Psi eigen value for 2 p state l = 1 rightarrow l (l + 1) h^2/2 u r^2 = 2 h^2/2 u r^2 = h^2/u r^2 } rotational kinetic energy u rightarrow reduced mass mu = m_c m_p/m_c + m_p mu almostequalto m_e V_effective (r) = L^2/2 mu r^2 + v (r) = h^2/mu r^2 + -e^2/16 pi elementof_0 a_0 = h^2/mu 16 a_0^2 + -e^2/16 pi elementof_0 a_0 since = r = 4 a_0 total energy, E_Tot = -13.6 eV/n^2 = -13.6 times 1.6 times 10^-19/n^2 for 2 p state n = 2 = E_t-t = -13.6 times 1.6 times 10^-19/4 = 5.44 times 10^-19 Joules