Answer :
Answer:
[tex]E_s = 7.35\times 10^{-22}\ J[/tex]
Explanation:
given,
angular speed, ω = 3.90 x 10¹² rad/s
bond length, L = 1.10 Å
molar mass = 28.0 g/mole
Rotational Kinetic energy
[tex]E_s = \dfrac{1}{2} I \omega^2[/tex]
now,
moment of inertia ,[tex]I = \dfrac{1}{2}MR^2[/tex]......(1)
mass of O₂ = m
mass of O = M = m/2
now, mass
[tex]M = \dfrac{28\times 1.67\times 10^{-27}}{2}[/tex]
form equation(1)
[tex]I = \dfrac{1}{2}\times \dfrac{28\times 1.67\times 10^{-27}}{2}\ R^2[/tex]
[tex]I = \dfrac{1}{2}\times \dfrac{28\times 1.67\times 10^{-27}}{2}\ (1.10\times 10^{-10})^2[/tex]
I = 9.66 x 10⁻⁴⁷ kg.m²
putting value in Rotational kinetic energy equation
[tex]E_s = \dfrac{1}{2}\times 9.66\times 10^{-47}(3.9\times 10^{12})^2[/tex]
[tex]E_s = 7.35\times 10^{-22}\ J[/tex]
Hence, rotational kinetic energy of the molecule is equal to [tex]E_s = 7.35\times 10^{-22}\ J[/tex]