In a water molecule the angle between the two H-O bonds is 105 $\circ$ and the distance between the oxygen and hydrogen atoms is 9 * 10^-11 m.The mass of the hydrogen atom m_H is 1.67 * 10^-27 kg and that of the oxygen atom m_O = 16 m_H.The moment of inertia of the water molecule about an axis along one of the H-O bonds is A)1.35 * 10^-47 kg m². B)1.26 * 10^-47 kg m². C)0.89 * 10^-47 kg m². D)0.15 * 10^-47 kg m².

Question

Quizplus · Accepted Answer

The moment of inertia of the water molecule about the H-O bond axis can be found using the formula:

I = μr^2
where μ = reduced mass of the molecule
      r = distance between the H and O atoms

The reduced mass is given by:

μ = mH*mO/(mH + mO)

Substituting the given values, we get:

μ = (1.67*10^-27)*(16*10^-27)/(1.67*10^-27 + 16*10^-27) = 1.49*10^-26 kg

The distance between the H and O atoms is given as 9*10^-11 m.

Therefore, the moment of inertia about the H-O bond axis is:

I = μr^2 = (1.49*10^-26)*(9*10^-11)^2 = 1.26*10^-47 kg m^2

Hence, the answer is B.

In a Water Molecule the Angle Between the Two H-O ∘\circ∘

In a Water Molecule the Angle Between the Two H-O $\circ$