Question

In HCl molecule, the separation between the nuclei of hydrogen and chlorine atoms is $1.27\stackrel{\circ }{A}$. If the mass of a chlorine atom is 35.5 times that of a hydrogen atom, the centre of mass of the HCl molecule is at a distance of

• A. $\frac{35.5\times 1.27}{36.5}\stackrel{\circ }{A}$ from the hydrogen atom.
• B. $\frac{35.5\times 1.27}{36.5}\stackrel{\circ }{A}$ from the chlorine atom.
• C. $\frac{1.27}{36.5}\stackrel{\circ }{A}$ from the hydrogen atom.
• D. $\frac{1.27}{36.5}\stackrel{\circ }{A}$ from the chlorine atom.

Answer 1

Answer: (A), (D)

$\frac{1.27}{36.5}\stackrel{\circ }{A}$ from the chlorine atom.

Since most of the mass of an atom is concentrated in its nucleus and the size of a nucleus (which is of the order of ${10}^{-15}m$) is very small compared to the separation between the hydrogen and chlorine atoms (which is $1.27\times {10}^{-10}m$), the atoms can be treated as point masses.
The centre of mass of HCI molecule will be on the line joining the two atoms. Let us say that the H atom is located at, say, x = 0 and the Cl atom at x = x. Let $x_{CM}$ be the position of the centre of mass between x = 0 and x = x. If $m_1$ and $m_2$, are the masses of H and CI atoms, then from the definition of centre of mass, we have
$x_{CM}=\frac{m_1\times 0+m_{2}x}{m_1+m_2}=\frac{m_{2}x}{m_1+m_2}$
Now x = separation between H and Cl atoms $=1.27\stackrel{\circ }{A}$ and $m_2=35.5m_1$.
Hence, $x_{CM}=\frac{35.5m_1\times 1.27\stackrel{\circ }{A}}{m_1+35.5m_1}=\frac{35.5\times 1.27\stackrel{\circ }{A}}{36.5}$
This gives the distance of the centre of mass from the hydrogen atom. The distance of the centre of mass from the chlorine is
$1.27\stackrel{\circ }{A}-\frac{35.5\times 1.27}{36.5}\stackrel{\circ }{A}=\frac{1.27\stackrel{\circ }{A}}{36.5}$
Hence, the correct choices are (a) and (d).