Question

The HCl molecule consists of a hydrogen atom (mass 1u) and a chlorine atom (mass 35u). The centres of the two atoms are separated by 127pm (=1.27x10-10m). What is the moment of inertia, I, about an axis perpendicular to the line joining the two atoms which passes through the centre of mass of the HCl molecule ?

Answer 1

Answer on Question #50965 – Physics – Other

The HCl molecule consists of a hydrogen atom (mass $m_H = 1\,\mathrm{u}$) and a chlorine atom (mass $m_{\mathrm{Cl}} = 35\,\mathrm{u}$). The centres of the two atoms are separated by $l = 127\,\mathrm{pm}$ (=1.27×10⁻¹⁰ m). What is the moment of inertia, $I$, about an axis perpendicular to the line joining the two atoms which passes through the centre of mass of the HCl molecule?

Solution:

The centre of mass of HCl molecule locates at a distance of $\frac{l}{36}$ from the chlorine atom and at a distance of $\frac{35}{36} l$ from the hydrogen atom due to its masses. The moment of inertia of a point particle with mass $m$ located at a distance $r$ from the axis is given by

$$I = m r^2$$

So the moment of inertia of the HCl molecule is given by

$$\begin{aligned} I_{\mathrm{HCl}} &= m_{\mathrm{H}} \left(\frac{35}{36} l\right)^2 + m_{\mathrm{Cl}} \left(\frac{l}{36}\right)^2 = \frac{35^2 \cdot 1\,\mathrm{u} + 35\,\mathrm{u}}{36^2} l^2 = \frac{35}{36} \mathrm{u} l^2 = \\ &= \frac{35}{36} \cdot 1.66 \cdot 10^{-27} \,\mathrm{kg} \cdot (1.27 \cdot 10^{-10})^2 \mathrm{m}^2 = 2.6 \cdot 10^{-47} \,\mathrm{kg} \cdot \mathrm{m}^2 \end{aligned}$$

**Answer:** $I_{\mathrm{HCl}} = m_{\mathrm{H}} \left(\frac{35}{36} l\right)^2 + m_{\mathrm{Cl}} \left(\frac{l}{36}\right)^2 \cdot 2.6 \cdot 10^{-47} \,\mathrm{kg} \cdot \mathrm{m}^2$.

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