Question

An oxygen molecule consists of two oxygen nuclei, each of massm= 2.7×10−26kg,separated by a distance 1.2×10−10m, and surrounded by a cloud of electrons, each ofmass 9.1×10−31kg.(a) Ignoring the small mass of the electron, what is the moment of inertia for rotationof the nuclei around their center of mass? If this rotation has angular momentumequal to~, that is, Planck’s constant divided by 2π, what is the correspondingenergy of rotation? Express your answer in units of electron volts (eV), where 1eV = 1.6×10−19J.(b) Suppose one of the electrons orbits in a circular orbit perpendicular to the axisjoining the nuclei, with radius 1×10−10m. This motion contributes a momentof inertia of the molecule along the internuclear axis. How big is this momentof inertia? If the electron has angular momentum~, what is the correspondingenergy of rotation, in eV? (This illustrates, very roughly, the discrepancy betweenelectronic and rotational energy scales in molecules.)

Answer 1

Answer:

a)  I = 1,944 10⁻⁴⁶ Kg m²

, b)   I = 7,915 10⁻⁵¹ Kg m²

Explanation:

a) The moment of inertia of point masses is

                 I = m r²

   

The nuclei have a very small size (10-15 m) so we can consider them punctual.

The distance to the center of mass passing through the middle of the two nuclei of equal mass

       r = 0.6 10⁻¹⁰ m

The moment of total inertia

       I = I₁ + I₂ = 2 I₀

       I = 2 2.7 10⁻²⁶ (0.6 10⁻¹⁰)²

       I = 1,944 10⁻⁴⁶ Kg m²

The kinetic energy of the rotation is

      w = h / 2π

      K = ½ I w²

      K = ½ 1,944 10⁻⁴⁶ (h / 2π)² = ½ 1.944 10⁻⁴⁶ (6.63 10⁻³⁴ / 2π)²

      K = 2.16 10⁻¹¹⁴ J (1eV / 1.6 10⁻¹⁹ J)

      K = 2.16 10⁻⁹⁵ eV

B) the moment of inertia of the electron in the orbit, we can calculate it with the parallel axis theorem

       I =$I_{cm}$ + m R²

       I = $m_{e}$ r² +  $m_{e}$ R²

Where R we calculate it by Pythagoras

      R² = (0.6 10⁻¹⁰)² + (0.5 10⁻¹⁰) 2

      R = √ (0.61 10⁻²⁰)

      R = 0.78 10⁻¹⁰ m

      I = 9.1 10⁻³¹ (0.5 10⁻¹⁰)² + 9.1 10⁻³¹ (0.78 10⁻¹⁰)²

      I = (2,275 +5.54) 10⁻⁵¹

      I = 7,915 10⁻⁵¹ Kg m²

The rotation energy of the electron

      K = ½ I w²

If the angular velocity is the electrons outside the core, its kinetic energy is much lower by an order 10⁵, but the angular velocity of the electrons is much higher.