Question

In the 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 atom (which is $1.27\times {10}^{-10}$ m), the atoms can be treated as point masses. The centre of mass of HCl molecule will be on the line joining the two atoms. Let us say the 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 Cl atoms, then from the definition of centre of mass,
we have
$x_{CM}=\frac{m_1\times 0+m_{2}x}{0m_1+m_2}=\frac{m_{2}x}{m_1+m_2}$
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 atom is
$1.27\stackrel{\circ }{A}-\frac{35.5\times 1.27}{36.5}=\frac{1.27\stackrel{\circ }{A}}{36.5}$
Hence the correct choices are (a) and (d).