The distance between the two atoms in an HCl | vedclass

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(C) Let the mass of the hydrogen atom be $m_1 = 1$ and the mass of the chlorine atom be $m_2 = 35.5$.Place the hydrogen atom at the origin $(0, 0)$,so

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$1.24$

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(C) Let the mass of the hydrogen atom be $m_1 = 1$ and the mass of the chlorine atom be $m_2 = 35.5$.Place the hydrogen atom at the origin $(0, 0)$,so its position is $\vec{r}_1 = 0$.The chlorine atom is at a distance of $1.27 \mathring{A}$ from the hydrogen atom along the $x$-axis,so its position is $\vec{r}_2 = 1.27 \hat{i} \mathring{A}$.The position of the center of mass $\vec{R}_{cm}$ is given by the formula:$\vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}$Substituting the values:$\vec{R}_{cm} = \frac{1 	imes 0 + 35.5 	imes 1.27 \hat{i}}{1 + 35.5}$$\vec{R}_{cm} = \frac{35.5 	imes 1.27}{36.5} \hat{i}$$\vec{R}_{cm} \approx 0.9726 	imes 1.27 \hat{i} \approx 1.235 \hat{i} \mathring{A}$Rounding to two decimal places,the position is approximately $1.24 \mathring{A}$ from the hydrogen atom.

The distance between the two atoms in an $HCl$ molecule is approximately $1.27 \mathring{A}$ $(1 \mathring{A} = 10^{-10} \text{ m})$. The chlorine atom is $35.5$ times heavier than the hydrogen atom. The position of the center of mass from the hydrogen atom is approximately ..... $\mathring{A}$.

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