The inter atomic distance of 14^N^16O molecule is 115.1 pm. Calculate: ( i) its reduced mass ii) its moment of inertia iii) the wave number of the line corresponding to lowest absorption in m^-1 unit, and iv) the energy in m-1 unit for the transition J = 2 to J = 3.
Answers
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' = h/2π = 6.626 * 10⁻³⁴/2π J-s
E = Energy in Jth state: J(J+1) h' ² / (2I) = J(J+1) h' ² / (8 π² μ d²)
E_J->J+1 = E_J+1 - E_J = (J+1) h' ² / I = 2 (J+1) B ,
B = Rotational constant of bond = h' ² / (2 I)
B = 6.626² *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
f_J=0->1 = frequency = E / h = 1.0078 * 10¹¹ Hz
Wave number in m^-1 = n / c =
= 1.0078 * 10¹¹ Hz / (3 * 10⁸ 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