In the $HCl$ molecule, the separation between the nuclet of the two atoms is about $1.27 \;\mathring A\left(1\; \mathring A=10^{-10} \;m \right) .$ Find the approximate location of the $CM$ of the molecule. given that a chlorine atom is about $35.5$ times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
In the $HCl$ molecule, the separation between the nuclet of the two atoms is about $1.27 \;\mathring A\left(1\; \mathring A=10^{-10} \;m \right) .$ Find the approximate location of the $CM$ of the molecule. given that a chlorine atom is about $35.5$ times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Distance between $H$ and $Cl$ atoms $=1.27 \mathring A$
Mass of $H$ atom $=m$
Mass of $Cl$ atom $=35.5 m$
Let the centre of mass of the system lie at a distance $x$ from the $Cl$ atom.
Distance of the centre of mass from the $H$ atom $=(1.27-x)$
Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
$\frac{m(1.27-x)+35.5 m x}{m+35.5 m}=0$
$m(1.27-x)+35.5 m x=0$
$1.27-x=-35.5 x$
$\therefore x=\frac{-1.27}{(35.5-1)}=-0.037 \mathring A$
Here, the negative sign indicates that the centre of mass lies at the left of the molecule. Hence, the centre of mass of the $HCl$ molecule lies $0.037 \mathring A$ from the $Cl$ atom.
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