The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 ˚ A$ . Given mass of carbon atom is $12 a . m . u .$ and mass of oxygen atom is $16 a . m . u .,$ calculate the position of the centre of mass of the carbon monoxide molecule

6.3 ˚ A

from the carbon atom

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1 ˚ A

from the carbon atom

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0.63 ˚ A

from the carbon atom

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0.12 ˚ A

from the carbon atom

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Solution

The correct option is C $0.63 ˚ A$ from the carbon atom
The distance of centre of mass from carbon atom is,
$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
$= \frac{12 \times 0 + 16 \times 1.1}{12 + 16}$
$= 0.63 ˚ A$
Thus, the centre of mass is $0.63 ˚ A$ from carbon atom.

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The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is 1.1˚A . Given mass of carbon atom is 12a.m.u. and mass of oxygen atom is 16a.m.u., calculate the position of the centre of mass of the carbon monoxide molecule

The correct option is C 0.63˚A from the carbon atomThe distance of centre of mass from carbon atom is,xcm=m1x1+m2x2m1+m2 =12×0+16×1.112+16=0.63˚AThus, the centre of mass is 0.63˚A from carbon atom.

The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 ˚ A$ . Given mass of carbon atom is $12 a . m . u .$ and mass of oxygen atom is $16 a . m . u .,$ calculate the position of the centre of mass of the carbon monoxide molecule

The correct option is C $0.63 ˚ A$ from the carbon atom
The distance of centre of mass from carbon atom is,
$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
$= \frac{12 \times 0 + 16 \times 1.1}{12 + 16}$
$= 0.63 ˚ A$
Thus, the centre of mass is $0.63 ˚ A$ from carbon atom.