The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is 1.1 $A^{0}$ . 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 center of mass of the carbon monoxide molecule

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0.63 $A^{0}$ from the carbon atom

1 $A^{0}$ from the oxygen atom

0.12 $A^{0}$ from the oxygen atom

6.3 $A^{0}$ from the carbon atom

answer is C.

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Detailed Solution

$m_{1} d_{1} = m_{2} d_{2}$ and hence $d_{1} : d_{2} = 16 : 12 = 4 : 3$

$∴ d_{1} = (\frac{4}{3 + 4}) 1.1 = 0.63$

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