In the $HCl$ molecule, the separation between the nuclei of the two atoms is about $1.5 \circ A (1 \circ A = 10^{- 10} m)$ . The approximate location of the centre of mass from the hydrogen atom assuming the chlorine atom to be about 35.5 times as massive as hydrogen is

1.45 A^{\circ}

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0.05 A^{\circ}

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0.72 A^{\circ}

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0.96 A^{\circ}

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Solution

The correct option is A $1.45 A^{\circ}$

Assuming mass of H atom, $m_{1} = 1 kg$
Then, mass of Cl atom, $m_{2} = 35.5 kg$
Taking position of $m_{1}$ as a reference, $x_{1} = 0$
Position of $m_{2}$ with respect to $m_{1}$ , $x_{2} = 1.5 A^{\circ}$
$∴$ Position of centre of mass,
$x_{com} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} = \frac{0 + 35. 5 \times 1.5}{1 + 35.5}$
$= 1.45 A^{\circ}$ from hydrogen atom.

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