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The ozone molecule O3 has a permanent dipole moment of 1.8×10−30 Cm. Although the molecule is very slightly bent-which is why it has a dipole moment-it can be modeled as a uniform rod of length 2.5×10−10 m with the dipole moment perpendicular to the axis of the rod. Suppose an ozone molecule is in a 8000 N/C uniform electric field. In equilibrium, the dipole moment is aligned with the electric field. But if the molecule is rotated by a small angle and released, it will oscillate back and forth in simple harmonic motion.

What is the frequency f of oscillation?

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Answer:

934701926.438 Hz

Explanation:

Mass of molecule

[tex]m=3\times 16\times 1.67\times 10^{-27}\ kg[/tex]

Moment of inertia is given by

[tex]I=\dfrac{1}{12}ml^2\\\Rightarrow I=\dfrac{1}{12}\times 3\times 16\times 1.67\times 10^{-27}\times (2.5\times 10^{-10})^2\\\Rightarrow I=4.175\times 10^{-46}\ kgm^2[/tex]

E = Electric field = 8000 N/C

p = Dipole moment = [tex]1.8\times 10^{-30}\ Cm[/tex]

l = Length of rod = [tex]2.5\times 10^{-10}\ m[/tex]

Frequency of oscillations is given by

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{pE}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{1.8\times 10^{-30}\times 8000}{4.175\times 10^{-46}}}\\\Rightarrow f=934701926.438\ Hz[/tex]

The frequency of oscillations is 934701926.438 Hz

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