The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 Å$. Given,…

The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 Å$. Given, mass of carbon atom is $12 \mathrm{amu}$ and mass of oxygen atom is 16 amu. Calculate the position of the centre of mass of the carbon monoxide molecule.
  1. $6.3 Å$ from the carbon atom
  2. $1.0 Å$ from the oxygen atom
  3. $0.63 Å$ from the carbon atom
  4. $0.12 Å$ from the oxygen atom

Solution

Given that, mass of oxygen, $m_{\mathrm{O}}=16 \mathrm{amu}$ Mass of carbon, $m_{\mathrm{C}}=12 \mathrm{amu}$ Distance between them, $r=1.1 Å$ Let $x$ be the distance of centre of mass from carbon atom
By using expression of centre of mass for two particle system, $X_{\mathrm{COM}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}$ Assuming centre of mass at orgin, we get $X_{\mathrm{COM}}=0=\frac{m_{\mathrm{C}}(-x)+m_{\mathrm{O}}(r-x)}{m_{\mathrm{C}}+m_{\mathrm{O}}}$ $m_{\mathrm{C}} x=m_{\mathrm{O}}(r-x)$ $m_{\mathrm{C}} x+m_{\mathrm{O}} x=m_{\mathrm{O}} r$ $x=\frac{m_{\mathrm{O}}}{m_{\mathrm{C}}+m_{\mathrm{O}}} r$ By substituting the values, we get Hence, position of centre of mass is $0.63 Å$ from carbon atom.

Asked in: AP EAMCET 2021 (24 Aug Shift 2)

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