Question

Please fully answer ALL parts! :-) Thank you so much in advance!!

1) This week, we introduced a new model, the rigid rotor, that can be used to describe rotating molecules. Before we get into problems that describe the quantum properties of a rigid rotor, let's quickly review the classical behavior of a rotating molecule. Consider a single atom with mass m rotating about the origin in a circular orbit with radius r. If the particle's linear momentum is p - mv, show the particle's kinetic energy is: a. 21 where L is the atom's angular momentum and I is the atom's moment of inertia b. Consider now two atoms, one with mass m, and a second with mass m2, that are chemically bonded to one another to form a diatomic molecule with a bond length R. This molecule rotates about its center of mass, which is located at the origin. The rotational kinetic energy for our molecule is just the sum of the rotational kinetic energy for each of its atoms: KE - n.r 2'2 Show this two-body kinetic energy can be written as a single term that describes the kinetic energy of a single atom with a mass equal to the reduced mass of the two atoms, u mm2 (m, +m2), rotating about the origin with a radius Rr

Answer 1

Part (a) p = m v (translation momentum) K. E = 1/ 2 mv^2 = p^2/ 2m K. E = p^2/ 2m K. E = l^2/ 2 mR^2 I = m R^2 (about 0 rotation axes) K. E = L^2/ 2I part (b) K. E = (K.E)_1 + (K. E)_2 K.E = 1/2 m_1 v^2_1 K.E = P^2_1/2m_1 + P^2/ 2m_2 (total K.E) V-1 P_1 = L_1 V_2 P_2 = L_2 K. E = L^2_1/ 2 V^2_1 m_1 + L^2_2/ 2 V^2_2 m_2 thin in what actually by individual kinetic energy sum net kinetic energy about which rotation happen (L) = angular momentum of system about 0 L = L_1 + L_2 = m_1 v^2_1 w + m_2 v^2_2 w w = L/ (m_1 v^2_1 + m_2 v^2_2) rightarrow (1) as about rotation happen R = v_1 + v_2 m_1 v^rightarrow_1 + m_2 v^rightarrow_2 = 0 m_1/m_2 = v_2/ v_1