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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In the HCl molecule, the separation between the nuclei of the two atoms is about 1.5 ∘A (1 ∘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

The correct option is A 1.45 A∘ Assuming mass of H atom, m1=1 kg Then, mass of Cl atom, m2=35.5 kg Taking position of m1 as a reference, x1=0 Position of m2 with respect to m1, x2=1.5 A∘ ∴ Position of centre of mass, xcom=m1x1+m2x2m1+m2=0+35.5×1.51+35.5 =1.45 A∘ from hydrogen atom.

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