Question

inter atomic distance of NO molecule is 115.1pm.calculate its moment of inertia and the energy for transition J=2toJ=3

Answer 1

m1 = 14 u (Nitrogen)        m2 = 16 u  (Oxygen)

i)  Reduced mass =  μ = m1 * m2 / (m1 + m2) = 112/15 u
                                = 1.239 * 10⁻²⁶ kg

      Given d = 115.1 pm

ii) I = moment of Inertia = μ d2   (derived from I = m1 d12 + m2 d22 , m1 d1 = m2 d2)

    I = 1.239 * 10-26 * (115.1*10-12)2  kg-m2 =  1.641 *10-46  kg-m2

    h = 6.626 * 10^-34/2π  units

E = Energy in Jth state: J(J+1) h2 / (2I) = J(J+1) h2 / (8 π2 μ d2)

EJ->J+1 = EJ+1  - EJ = (J+1) h2 / I  = 2 (J+1) B  ,

          B = Rotational constant of bond = h2 / (2 I)

          B = 6.626^2*10^-68 /(4*π2* 2 *1.641*10^-46)  J = 3.388 *10^-23 J

 

iii) Transition from J = 0 to J = 1 (lowest Absorption)

    E_J=0->1 = 2 (0+1) 3.388 * 10^-23 J = 6.776 * 10^-23 J

     n_J=0->1 = frequency = E /h = 1.0078 * 10^11  Hz

     Wave number  in m^-1 = n / c =

                  = 1.0078 * 10^11 Hz / (3 * 10^8 m/s) = 335.9 m^-1

 

iv) Transition from J = 2 to J = 3

      E_J=2->3  = 2 (J+1) B = 2 * (2+1) * 3.388 * 10^-23 J

                   = 2.033 * 10^-22 J

     frequency corresponding = E/h

      wave number = Energy in units of m^-1   = E/(hc)

               = 2.033 * 10⁻²² /1.988 *10⁻²⁶   m⁻¹

               = 1.022 * 10⁴  m⁻¹